Ragged Hierarchies and extra joins

We have an ORG_DIM with a three-level logical hierarchy (LVL-All, LVL-1, LVL-2, LVL-3 -- the leaf/detail level). The primary key is LVL3_KEY; LVL2_KEY and LVL1_KEY are level keys. We have a SYS_DIM with the trival (all/detail) logical hierarchy and SYS_KEY primary key.
We have a base fact and two aggregate facts -- LVL1_FACT, LVL2_FACT and LVL3_FACT (level 3 is the base); the logical Fact - Lvls has an LTS for each of these physical fact tables. The base fact (LVL3_FACT) has LVL13_KEY and SYS_KEY as its primary key and also has columns LVL2_KEY and LVL1_KEY; the aggregate facts lose LVL3_KEY and LVL2_KEY, respectively.
ORG_DIM represents a flattened skip-level hierarchy. (There are actually more levels and more tables, but I am simplifying the problem for this description.)
We also have an ORG_STACK table (ORGNAME, ORGKEY, PARENTORGKEY, LFT, RT, STACKLVL), following a method for implementing ragged hierarchies proposed by Kurt Wolff and Joe Celko. The ORG_STACK table has three aliases -- ORG_PARENTS, ORG_CHILDREN and ORG_LEAVES. The first two are joined on (ORG_CHILDREN.PARENTORGKEY=ORG_PARENTS.ORGKEY). The latter two are joined on (ORG_CHILDREN.LFT BETWEEN ORG_LEAVES.LFT AND ORG_LEAVES.RT) -- a complex physical join. ORG_LEAVES joins to LVL3_FACT on ORG_LEAVES.ORG_KEY=LVL3_FACT.LVL3_KEY. The logical Fact - Lvls has a 4th LTS in which ORG_LEAVES and LVL3_FACT are joined.
We run a query in answers that invokes columns from logical tables Dim - OrgChildren, Dim - OrgParents and Fact - Lvls; one of the columns from Fact - Lvls is mapped only in the LTS that joines ORG_LEAVES with LVL3_FACT. This query spins for a long time without returning any result, so we end up cancelling it. We intend to have a DBA look at it.
The generated SQL includes those tables (ORG_STACK aliases and LVL3_FACT) plus ORG_DIM and SYS_DIM. We can see nothing in the query that requires ORG_DIM or SYS_DIM; how can we determine why these are included in the SQL?
Queries that used columns from Dim - Org also showed poor performance after the other Dim - OrgXXX (the stack suff) were added. Presumably this was because the LTS for Fact - Lvls that maps to LVL3_FACT also joined the (much larger) ORG_STACK table via ORG_LEAVES. Adding the 4th LTS for that (leaving the 3rd LTS mapped only to LVL3_FACT) should have resolved this. However, the interesting this is that removing the logical join between Dim - OrgChildren and Fact - Lvls resolved this problem. If the problem was ORG_LEAVES and LVL3_FACT joined in the LTS, then removing a logical join should not have affected it.

Chandra Kanth,
Hope this helps you
http://blogs.simba.com/.a/6a00e3932f172e88340133f4d758b5970b-800wi
Thanks,
Saichand.v

Similar Messages

Hierarchy with Intervals and Temporal Join

Hi,
I am loading a hierarchy structure (Time dependant and with Temporal Join and Intervals)
in the flat file structure for hier loading we have leaf to and leaf from fields which means Page of the hierarchy to and page of the hierarchy from.
can anyone let me know what is the use of these fields, and the values we have to pass for these fields with an example.
i haven't come across Hier loading with intervals and termporal join.
can anyone please help me in this?
Regards,
p423168.

Hi,
BI Process (Loading Hierarchies from Flat file to BI)
Go to RSA1
Create info object for customer as zcust_b5 and activate
Create info object as zcust_b5 and activate it
Select the info object and double click on it
Select the hierarchies tab and select the option with hierarchies
And select Hierarchies not time dependent
Click external characteristics in hierarchies
Select your Infoobjects in the list and pass it and activate it
After activating the objects we will see tables of hierarchies at maintain Hierarchies
Create an application component
Create the transfer rules for the info object and activate (for hierarchies, texts, attributes)
This screen will show the communication structure.
After creating the data source at info source level the field attributes are shown in above
Create an excel sheet with required values and save it in CSV file.
Uploading the Hierarchy Data into BW
a) Create Info Package for Hierarchy Upload. Select the Hierarchy to be uploaded
and also the Full Update button as we are uploading the entire Hierarchy.
b) There are three update methods for uploading the hierarchy.
Full Update: For Uploading the entire hierarchy structure
Insert Sub tree: For appending the hierarchy with new nodes.
Update Sub tree: For modifying the nodes of the hierarchy.
Create info package and schedule the data loads
Select the Hierarchy and name it before scheduling
Schedule and Start the data loads
After creating the info package of the info object the above screen shows the initial hierarchy maintenance
Hierarchy is done successfully by extracting data from flat file.
time dependent :
a) For making the hierarchy time dependent select the u2018Time-Dependent Hierarchy structureu2019 radio button and activate the info object ZSREP
b) In the transfer structure there will be 2 more additional fields date from and date to for the time dependent nodes.
c) Consider that Sales Representative u2018SREP2u2019 was in England up to 31/12/1995 and moved to USA
d) Flat file structure for the above hierarchy is shown below . It contains two extra columns Date-to and Date-from.
e) In the info-package Update tab , select Fixed Time interval radio button and give start date as 01/01/1000 and End date as 31/12/9999 so that all nodes are uploaded.
f) Upload the hierarchy data from flat-file to BW.
g) Display the loaded hierarchy.
Regards
Sudheer
Edited by: SUDHEER on Dec 30, 2008 12:52 PM

Ragged Hierarchy and Skip Level Hierarchy

Hi all,
I am confused with ragged and skip level hierarchy, as i read some documents related to obiee 11g.
Will you please explain with an example the difference between ragged hierarchy and skip level hierarchy and also when to use these hierarchies?
regards
chandra kanth.

Chandra Kanth,
Hope this helps you
http://blogs.simba.com/.a/6a00e3932f172e88340133f4d758b5970b-800wi
Thanks,
Saichand.v

Why Left and Right Joins if only one of them can serve the same purpose?

I understand that left join is used to display all the records from left table whereas the right join is used to display all the records from right table in a join.
However, if we switch the positions of the join tables and perform left join we can get all the records of the left table.
A left join B -> will give all records from A
A right join B ->will give all records from B
B left join A -> will again give all records from B.
The order in which the data is displayed(ie the field order) can also be maintained in the SELECT clause.
So, Why do we have left and right joins separately in existance, when only one of them can serve the same purpose?

Hi,
I think it's because it is sometime more natural to outer join on left or right.
In your data model (or the query you are writing), there is some kind of hierarchy between tables : I feel more natural to have the table with lowest hierarchical level (the root table) on the left. So I sometimes prefer to right outer join...
This is a very personal feeling, I must admit.
To me what is strange is to have chosen left and right as they refer to a "latin" reading direction => forward join and backward join would be more meaningful for people that read from right to left.
+And I prefer the new joining syntax that doesn't us the {noformat}(+){noformat} sign...+

Support for ragged hierarchies?

Are ragged (i.e. variable depth) hierarchies supported in Universe Designer? Do any of the Business Objects tools provide support for ragged hierarchies or is this considered an ETL/data modeling issue?
We are embarking on a project using the Data Services, Universe Designer and Xcelsius tools and we are trying to figure out how to handle ragged hierarchies.
Thanks!

Hi Didier ,
I have gone thought the document of Universe design tool :- SAP Business Objects XI 4.0 created on 2010-12-24 on SAP Portal .
This document dont have any information regarding the Ragged Hierarchy .
Can you please let me know whether downloads of SAP Business Objects BI 4.0 which are available on SAP Portal Supports ragged Hierarchy or We need to wait for next patch or release?
Regards ,
Jeevan

Hierarchical and summary 2

Hi
In the last week I post a thread about SUM hierarchical ( sum in levels in the tree )
Re: hierarchical and summary and division : Query Challenge 2
With help , I got to do , but the cost is very high, I explain again what I must to do
I have a table (GRUPO_RELACAO) that save information about parent and child, but in this relation
each group can to be different sectors, when I must to sum a Sector I can not to sum groups of other
sector. look example below with respective values
               1[3]
            /     \ nm_ciclo_fim is not null then 4 do not have parent
             2[1]      4[1]
         5[3] 6[4] 8[2] 7[2]
         |
         |
         11[6]
      9[5]   12[0]
         13[8]   16[4] 1,2,4,5,6,7,8,9 are in SECTOR 16
11,12,13,16 are in SECTOR 18
In structure of table (GRUPO_RELACAO) there is a column NM_CICLO_FIM that must to be NULL
if is not null, then Child is not more relation with parent , in Example 4 was child of 1, but The
column is filled
In example below , to show only SECTOR 16
     GROUP     SUM     LEVEL    SECTOR
     1     1     1     16   TEAM OF GROUP 1 ONLY 1 (GROUP 2) because group 4 is not child
     1     12     2     16   NET      (GROUPS 5 AND 6 AND 9 , but no 11 because 11 is sector 18)
     1     16     3     16   NETWORK   (GROUPS 1,2,5,6,9) = 3+1+3+4+5 = 16
     2     7     1     16   TEAM OF GROUP 2
     2     5     2     16   NET OF GROUP 2
     2     8     3     16   NETWORK OF GROUP 2
     3     0     1     16   IS NOT IN TREE BUT IS IN T_ESTRUTURA_COMERCIAL
     3     0     2     16   IDEM ABOVE
     3     0     3     16   IDEM ABOVE
     4     3     1     16   TEAM OF GROUP 4
     4     0     2     16   NET OF GROUP 4
     4     4     3     16   NETWORK OF GROUP 4
     5     0     1     16   TEAM OF GROUP 5 (11 IS SECTOR 18) NO SUM
     5     5     2     16   NET OF GROUP 5 (9 IS SECTOR 16 ) SUM
     5     3     3     16   NETWORK OF GROUP 5 (5 + 3 ) SELF VALUE
     6     0     1     16
     6     0     2     16
     6     4     3     16
     7     0     1     16
     7     0     2     16
     7     1     3     16
     8     0     1     16
     8     0     2     16
     8     2     3     16
     9     0     1     16
     9     0     2     16
     9     5     3     16
     165     0     1     16 Others groups that are not in tree must to show too
     165     0     2     16 in example only group 16
     165     0     3     16
     193     0     1     16
     193     0     2     16
     193     0     3     16
     194     0     1     16
     194     0     2     16
     194     0     3     16
     705     0     1     16
     705     0     2     16
     705     0     3     16
     706     0     1     16
     706     0     2     16
     706     0     3     16
     714     0     1     16
     714     0     2     16
     714     0     3     16
     715     0     1     16
     715     0     2     16
     715     0     3     16
     803     0     1     16
     803     0     2     16
     803     0     3     16
     804     0     1     16
     804     0     2     16
     804     0     3     16
     805     0     1     16
     805     0     2     16
     805     0     3     16
Connected to Oracle Database 10g Enterprise Edition Release 10.2.0.2.0
Connected as PROCSCNOB
SQL>
SQL> SELECT T1.CD_EC_MADRINHA,
2         T1.CD_EC_AFILHADA,
3         ID_TIPO_RELACAO,
4         NM_CICLO_FIM
5   FROM SISSCNOB.GRUPO_RELACAO T1
6 ORDER BY 1;
CD_EC_MADRINHA CD_EC_AFILHADA ID_TIPO_RELACAO NM_CICLO_FIM
             1              2               1
             1              4               1       200802
             2              5               1
             2              6               1
             4              7               1
             4              8               1
             5             11               1
            11              9               1
            11             12               1
            12             13               1
            12             16               1
11 rows selected
SQL> SELECT T2.CD_ESTRUTURA_COMERCIAL, T2.CD_SETOR, T2.ID_ESTRUTURA_ATIVA
2     FROM T_ESTRUTURA_COMERCIAL T2
3    WHERE T2.CD_TIPO_ESTRUTURA_COMERCIAL = 5
4      AND T2.CD_ESTRUTURA_COMERCIAL IN (1,2,3,4,5,6,7,8,9,11,12,13,16);
CD_ESTRUTURA_COMERCIAL CD_SETOR ID_ESTRUTURA_ATIVA
                     1       16                  1
                     2       16                  1
                     3       16                  0
                     4       16                  1
                     5       16                  1
                     6       16                  1
                     7       16                  1
                     8       16                  1
                     9       16                  1
                    11       18                  1
                    12       18                  1
                    13       18                  1
                    16       18                  1
13 rows selected
SQL> SELECT T3.CD_ESTRUTURA_COMERCIAL,
2         T3.NM_CICLO_OPERACIONAL,
3         T3.CD_INDICADOR,
4         T3.VL_INDICADOR
5     FROM T_EC_SETOR_PERFIL_CICLO T3
6    WHERE T3.CD_TIPO_ESTRUTURA_COMERCIAL = 5
7     AND T3.CD_ESTRUTURA_COMERCIAL IN (1,2,3,4,5,6,7,8,9,11,12,13,16)
8     AND T3.NM_CICLO_OPERACIONAL = 200712
9     AND T3.CD_PERFIL = 1
10     AND T3.CD_INDICADOR = 63;
CD_ESTRUTURA_COMERCIAL NM_CICLO_OPERACIONAL CD_INDICADOR      VL_INDICADOR
                     1               200712           63              3.00
                     2               200712           63              1.00
                     4               200712           63              1.00
                     5               200712           63              3.00
                     6               200712           63              4.00
                     7               200712           63              1.00
                     8               200712           63              2.00
                     9               200712           63              5.00
                    11               200712           63              6.00
                    13               200712           63              8.00
                    16               200712           63              4.00
11 rows selected
THE TABLES
-- In this table ID_TIPO_RELACAO must to be 1 and NM_CICLO_FIM IS NULL
--- When NM_CICLO_FIM not is null the relation between cd_ec_madrinha and cd_ec_afilhada is break
WITH GRUPO_RELACAO AS(
select   1 CD_EC_MADRINHA,
         2 CD_EC_AFILHADA,
         1 ID_TIPO_RELACAO ,
         TO_NUMBER(NULL,0) NM_CICLO_FIM FROM DUAL UNION
SELECT   1,4,1,200802 FROM DUAL UNION
SELECT   2,5,1,TO_NUMBER(NULL,0) FROM DUAL UNION
SELECT   2,6,1,TO_NUMBER(NULL,0) FROM DUAL UNION
SELECT   4,7,1,TO_NUMBER(NULL,0) FROM DUAL UNION
SELECT   4,8,1,TO_NUMBER(NULL,0) FROM DUAL UNION
SELECT   5,11,1,TO_NUMBER(NULL,0) FROM DUAL UNION
SELECT   11,9,1,TO_NUMBER(NULL,0) FROM DUAL UNION
SELECT   11,12,1,TO_NUMBER(NULL,0) FROM DUAL UNION
SELECT   12,13,1,TO_NUMBER(NULL,0) FROM DUAL UNION
SELECT   12,16,1,TO_NUMBER(NULL,0) FROM DUAL )
--- Here table master all CD_EC_MADRINHA and CD_EC_AFILHADA from GRUPO_RELACAO
--- is in this table , There is data here that are not in table GRUPO_RELACAO
---- in JOIN with GRUPO_RELACAO, the column   ID_ESTRUTURA_ATIVA must to be 1
----- When to sum a SECTOR it must to sum only values of the SECTOR
--- I can to have Group of a sector that is parent of the other group
--- In example I showed Group 5(group 16) is parent of the group 11 (sector 18)
-- that is parent of the group 9 (group 16)
WITH T_ESTRUTURA_COMERCIAL AS(
select    1 CD_ESTRUTURA_COMERCIAL,      16 CD_SECTOR,1 ID_ESTRUTURA_ATIVA FROM DUAL union
select    2, 16,1 FROM DUAL UNION
select    3,16,0 FROM DUAL UNION
select    4,16,1 FROM DUAL UNION
select    5,16,1 FROM DUAL UNION
select    6,16,1 FROM DUAL UNION
select    7,16,1 FROM DUAL UNION
select    8,16,1 FROM DUAL UNION
select    9,16,1 FROM DUAL UNION
select    11,18,1 FROM DUAL UNION
select    12,18,1 FROM DUAL UNION
select    13,18,1 FROM DUAL UNION
select    16,18,1 FROM DUAL )
SELECT * FROM T_ESTRUTURA_COMERCIAL
-- The 3rd and last table
-----this Table have the values that I must to sum in LEVELS
WITH T_EC_SETOR_PERFIL_CICLO AS (
SELECT      1 CD_ESTRUTURA_COMERCIAL, 200712 NM_CICLO_OPERACIONAL,63 CD_INDICADOR ,3.00 VL_INDICADOR FROM DUAL UNION
SELECT   2,200712,63,1.00 FROM DUAL UNION
SELECT   4,200712,63,1.00 FROM DUAL UNION
SELECT   5,200712,63,3.00 FROM DUAL UNION
SELECT   6,200712,63,4.00 FROM DUAL UNION
SELECT   7,200712,63,1.00 FROM DUAL UNION
SELECT   8,200712,63,2.00 FROM DUAL UNION
SELECT   9,200712,63,5.00 FROM DUAL UNION
SELECT   11,200712,63,6.00 FROM DUAL UNION
SELECT   13,200712,63,8.00 FROM DUAL UNION
SELECT   16, 200712,63,4.00 FROM DUAL )
SELECT * FROM T_EC_SETOR_PERFIL_CICLOSomebody can help me, because my solution is with COST high
TIA

That plan is better, don't know why you did not post
this follow up in your old thread, where I asked
whether
[url=http://forums.oracle.com/forums/message.jspa?m
essageID=2330090#2330090]42 billion rows
was roughly the size of the results you expected.
If you widen your sqlplus window and set linesize
140 it would look like the output below.
Unfortunately then we can see things like this
| 50 |     VIEW                               |
|   120T| 5474T|
|   148G (96)|999:59:59 |
Where I believe the T in 120T stands for
tera[
/b] or 120 trillion rows, that is 120,000,000,000,000
or a right
shedfull as it is sometimes known. The next step is
to sort these rows, which could take some time,
999:59:59
is the estimate which looks suspiciously like just
the biggest number the optimizer can produce.
In your query you have several connect by statements
but no start withs. This will generate every possible
permutation
of the hierarchy, I think in one place the results
are used as the input of another query that then
regenerates every
possible permutation of that hierarchy.
In these cases I am not really sure what you think
you are trying to do apart from generate the biggest
result set
ever known to mankind. This is what happens to the 14
row emp table when subjected to some of these
queries. What
are the original row counts of the tables you are
doing this to in the query?
SQL> select count(*)
2 from emp
3 start with mgr is null
4 connect by mgr = prior empno;
COUNT(*)
14
elect count(*)
2 from emp
3 connect by mgr = prior empno;
COUNT(*)
39
di
Wrote file afiedt.buf
1 select count(*) from
2    (
3    select mgr, empno
4    from emp
5    connect by mgr = prior empno
6    )
7* connect by mgr = prior empno
L> /
COUNT(*)
261
The formatted plan
[pre]
Execution Plan
Plan hash value: 3341212558
| Id | Operation                              | Name
| Rows | Bytes |TempSpc| Cost
(%CPU)| Time     |
|   0 | SELECT STATEMENT                       |
|    48G| 2910G|       |
150G (95)|999:59:59 |
1 | TEMP TABLE TRANSFORMATION             |
|       |       |       |
|          |
OAD AS SELECT                       |
|       |       |       |
|          |
HASH GROUP BY                       |
|    13 | 2366 |       |
14 (29)| 00:00:01 |
4 |     VIEW                               |
|    13 | 2366 |
|    13 (24)| 00:00:01 |
|      NESTED LOOPS OUTER                |
|    13 |   949 |       |
13 (24)| 00:00:01 |
6 |       VIEW                             |
|    13 |   845 |
|     8 (38)| 00:00:01 |
|        SORT UNIQUE                     |
|    13 | 1150 |       |
8 (63)| 00:00:01 |
8 |         UNION-ALL                      |
|       |       |       |
|          |
VIEW                          |
|    12 | 1092 |       |
3   (0)| 00:00:01 |
0 |           NESTED LOOPS                 |
|    12 |   516 |       |
3   (0)| 00:00:01 |
1 |            VIEW                        |
|    11 |   319 |       |
3   (0)| 00:00:01 |
2 |             TABLE ACCESS FULL          |
GRUPO_RELACAO               |    11 |    88 |       |
3   (0)| 00:00:01 |
3 |            TABLE ACCESS BY INDEX ROWID |
M_ESTRUTURA_COMERCIAL_CAD   |     1 |    14 |       |
0   (0)| 00:00:01 |
4 |             INDEX UNIQUE SCAN          |
I0_ESTRUTURA_COMERCIAL      |     1 |       |       |
0   (0)| 00:00:01 |
5 |          HASH JOIN ANTI                |
|     1 |    58 |       |
3 (34)| 00:00:01 |
6 |           VIEW                         |
|    12 |   660 |       |
1   (0)| 00:00:01 |
7 |            NESTED LOOPS                |
|    12 |   204 |       |
1   (0)| 00:00:01 |
8 |             VIEW                       |
|    11 |    33 |       |
1   (0)| 00:00:01 |
9 |              INDEX FULL SCAN           |
GRP_RELAC_PK                |    11 |    33 |       |
1   (0)| 00:00:01 |
0 |             TABLE ACCESS BY INDEX ROWID|
M_ESTRUTURA_COMERCIAL_CAD   |     1 |    14 |       |
0   (0)| 00:00:01 |
1 |              INDEX UNIQUE SCAN         |
I0_ESTRUTURA_COMERCIAL      |     1 |       |       |
0   (0)| 00:00:01 |
2 |           INDEX FULL SCAN              |
GRP_RELAC_PK                |    11 |    33 |       |
1   (0)| 00:00:01 |
3 |       VIEW                             |
|     1 |     8 |       |
0   (0)| 00:00:01 |
4 |        INLIST ITERATOR                 |
|       |       |       |
|          |
TABLE ACCESS BY INDEX ROWID    |
T_EC_SETOR_PERFIL_CICLO     |     1 |    23 |       |
0   (0)| 00:00:01 |
6 |          INDEX RANGE SCAN              |
I0_SETOR_PERFIL_CICLO       |     1 |       |       |
0   (0)| 00:00:01 |
7 |   LOAD AS SELECT                       |
|       |       |       |
|          |
SORT UNIQUE                         |
|     2 |   129 |       |
4818K (57)| 16:03:40 |
29 |     UNION-ALL                          |
|       |       |       |
|          |
VIEW                              |
|     1 |    59 |       |
2409K (13)| 08:01:50 |
31 |       SORT UNIQUE                      |
|     1 |    81 |       |
2409K (13)| 08:01:50 |
32 |        WINDOW SORT                     |
|     1 |    81 |       |
2409K (13)| 08:01:50 |
* 33 |         VIEW                           |
|     1 |    81 |       |
2409K (13)| 08:01:50 |
34 |          SORT UNIQUE                   |
|     1 |    91 |       |
2409K (13)| 08:01:50 |
* 35 |           CONNECT BY WITHOUT FILTERING |
|       |       |       |
|          |
COUNT                       |
|       |       |       |
|          |
VIEW                       |
|   316M|    26G|       |
2193K (4)| 07:18:39 |
38 |              TABLE ACCESS FULL         |
SYS_TEMP_0FD9D6A2F_44B054AC |   316M|    26G|       |
2193K (4)| 07:18:39 |
39 |      VIEW                              |
|     1 |    70 |       |
2409K (13)| 08:01:50 |
40 |       SORT UNIQUE                      |
|     1 |    65 |       |
2409K (13)| 08:01:50 |
41 |        WINDOW SORT                     |
|     1 |    65 |       |
2409K (13)| 08:01:50 |
* 42 |         VIEW                           |
|     1 |    65 |       |
2409K (13)| 08:01:50 |
43 |          SORT UNIQUE                   |
|     1 |    91 |       |
2409K (13)| 08:01:50 |
* 44 |           CONNECT BY WITHOUT FILTERING |
|       |       |       |
|          |
COUNT                       |
|       |       |       |
|          |
VIEW                       |
|   316M|    26G|       |
2193K (4)| 07:18:39 |
47 |              TABLE ACCESS FULL         |
SYS_TEMP_0FD9D6A2F_44B054AC |   316M|    26G|       |
2193K (4)| 07:18:39 |
48 |   WINDOW BUFFER                        |
|    48G| 2910G|       |
150G (95)|999:59:59 |
49 |    MERGE JOIN SEMI                     |
|    48G| 2910G|
|   148G (96)|999:59:59 |
|     VIEW                               |
|   120T| 5474T|       |
148G (96)|999:59:59 |
51 |      SORT ORDER BY                     |
|   120T|     9P|
23P|   148G (96)|999:59:59 |
52 |       MERGE JOIN OUTER                 |
|   120T|     9P|       |
6506M(100)|999:59:59 |
53 |        SORT JOIN                       |
| 1900M|    42G|   128G|
38M (9)|126:57:43 |
54 |         VIEW                           |
| 1900M|    42G|       |
13M (13)| 46:09:55 |
55 |          SORT UNIQUE                   |
| 1900M|    23G|    71G|
13M (71)| 46:09:55 |
56 |           UNION-ALL                    |
|       |       |       |
|          |
VIEW                        |
|   633M| 7853M|       |
91649 (100)| 00:18:20 |
| 58 |             TABLE ACCESS FULL          |
SYS_TEMP_0FD9D6A30_44B054AC |   633M|    50G|       |
91649 (100)| 00:18:20 |
| 59 |            VIEW                        |
|   633M| 7853M|       |
91649 (100)| 00:18:20 |
| 60 |             TABLE ACCESS FULL          |
SYS_TEMP_0FD9D6A30_44B054AC |   633M|    50G|       |
91649 (100)| 00:18:20 |
| 61 |            VIEW                        |
|   633M| 7853M|       |
91649 (100)| 00:18:20 |
| 62 |             TABLE ACCESS FULL          |
SYS_TEMP_0FD9D6A30_44B054AC |   633M|    50G|       |
91649 (100)| 00:18:20 |
|* 63 |        SORT JOIN                       |
|   633M|    40G|    99G|
18M (4)| 60:03:28 |
64 |         VIEW                           |
|   633M|    40G|       |
91649 (100)| 00:18:20 |
| 65 |          TABLE ACCESS FULL             |
SYS_TEMP_0FD9D6A30_44B054AC |   633M|    50G|       |
91649 (100)| 00:18:20 |
|* 66 |     SORT UNIQUE                        |
|     1 |    14 |       |
1 (100)| 00:00:01 |
7 |      TABLE ACCESS BY INDEX ROWID       |
M_ESTRUTURA_COMERCIAL_CAD   |     1 |    14 |       |
0   (0)| 00:00:01 |
8 |       INDEX RANGE SCAN                 |
I0_ESTRUTURA_COMERCIAL      |   811 |       |       |
0   (0)| 00:00:01 |
Predicate Information (identified by operation id):
14 - access("W1"."CD_TIPO_ESTRUTURA_COMERCIAL"=5
AND
"W2"."CD_EC_AFILHADA"="W1"."CD_ESTRUTURA_COMERCIAL")
15 - access("CD_EC_AFILHADA"="W2"."CD_EC_MADRINHA")
21 - access("W1"."CD_TIPO_ESTRUTURA_COMERCIAL"=5 AND
"W2"."CD_EC_MADRINHA"="W1"."CD_ESTRUTURA_COMERCIAL")
26 - access("W"."CD_TIPO_ESTRUTURA_COMERCIAL"=5 AND
"T"."CD_EC_AFILHADA"="W"."CD_ESTRUTURA_COMERCIAL"
AND
"W"."NM_CICLO_OPERACIONAL"=200712 AND
"W"."CD_PERFIL"=1 AND ("W"."CD_INDICADOR"=0 OR
"W"."CD_INDICADOR"=63))
33 - filter("CD_EC_AFILHADA"<>"PARENT" AND
"CD_SETOR"=16 AND "ID_ESTRUTURA_ATIVA"=1)
35 - access("CD_EC_MADRINHA"=PRIOR
"CD_EC_AFILHADA")
filter("ID_ESTRUTURA_ATIVA"=1)
filter("CD_SETOR"=16 AND "ID_ESTRUTURA_ATIVA"=1)
44 - access("CD_EC_MADRINHA"=PRIOR
"CD_EC_AFILHADA")
filter("ID_ESTRUTURA_ATIVA"=1)
access("T1"."LVL"="T2"."LVL"(+))
filter("T1"."PARENT"="T2"."PARENT"(+) AND
"T1"."LVL"="T2"."LVL"(+))
66 - access("T3"."CD_ESTRUTURA_COMERCIAL"="PARENT")
filter("T3"."CD_ESTRUTURA_COMERCIAL"="PARENT")
filter("T3"."CD_SETOR"=16 AND
"T3"."ID_ESTRUTURA_ATIVA"=1)
68 - access("T3"."CD_TIPO_ESTRUTURA_COMERCIAL"=5)
/pre]Hi thank
In your query you have several connect by statements
but no start withs. This will generate every possible
permutationPlease, [b] in my example how can put a start with, that walk all tree ?

I HAVE JUST BOUGHT AN I PHONE 4S AND SUCCESSFULLY JOINED I CLOUD, ALSO ON MY PC. BUT I HAVE USED A DIFFERENT SIGN IN E MAIL ADDRESS THAN ON MY ORIGINAL I PAD. HOW CAN I CHANGE THAT TO THE NEW E MAIL SIGN IN ?

I have just bought an i phone 4s and successfully joined i cloud, i have also successfully linked up my pc with i cloud. However my i pad is a little older, and when i signed on to i cloud some time ago i used a different e mail 'sign in' address. How do I change this in order to link all three devices. Thanks in anticipation, David.

Go to Settings>iCloud, tap Delete Account and provide the password (if using iOS 7), then sign back in with the ID ("email sign in") you want to use.

What is the different between Logical complex join and Physical join?

hi,
Do somebody know what is the function different between logical complex join in BMM layer and physical join in physical layer?
Thanks.

Thank you very much1
I understand their differentiation:
In the BMM Complex join (intelligent), means OBI picks the best way from possibly many different ways to join using physical join. FK join forces the same join always, which limits flexibility.

Logical And Physical Joins Of SH_PartTwo.rpd provided by oracle OBIEE

Hi,
I am new the OBIEE tool.After downloading SH_PartTwo.rpd .Planned to recreate the physical and logical joins.
If any one has already tried this, kindly explain why we are creating Alias tables of (SALES,TIMES)and joining COSTS to them instead of directly joining to Sales fact table. Below is the table structure.My question again,previously in SH.rpd we were joining Sales fact with other dimensions.If I need cost data then why should'nt I join cost with sales.
Thanks for your cooperation in advance.
Regards,
Sreekanth
Costs Sales Times Customers Countries Products Promotions Channels
Timeid Timeid Timeid
Productid Productid Productid
Promotionid Promotionid Promotionid
Channelid Channelid Channelid
Custid Custid
Countryid Countryid
Unitcost Amount_sold

Well don't know the example anymore ;)
But you normally will make aliasses to reuse the object.
So if you have a time dimension, and you have all kind of fields, like date_sold, date_bought, date_paid...
Now if you connect those dates directly to the time dimension, then you can get very odd reports... if you use aliasses, then the times have nothing to do with each other...
So it is always better to use aliasses... you never know if you need a table only once..

HT5098 How can I read on my Apple Arabic files of Times New Roman or Arial in PC? when I get the files the Arabic letters are separate and not joined. Is there a common font in Arabic that can convert the PC Windows files to my Apple?

How can I read on my Apple Arabic files of Times New Roman or Arial fonts made in PC? when I get the files the Arabic letters are separate and not joined. Is there a common font in Arabic that can convert the Arabic PC Windows files to my Apple?
The same with email messages ,
And when I get Power Point files made in PC Windows, they are like PDF, not Power Point!
Anybody with a good suggestion?

Yewtree wrote:
when I get the files the Arabic letters are separate and not joined
Do not use MS Word for Mac, it does not support Arabic. Instead use Mellel, TextEdit, Nisus Writer or OpenOffice. Try the font Geeza Pro if others do not work.
What are you using to read email?
What are you using to read powerpoint?

[solved] pacman apart from core and extra repos

I just can't get pacman to update any other repos apart from core and extra since I am on arch64 - tried to change servers directly in pacman.conf and in the include files, unpacked the databases manually to /var/lib/pacman - can't seem to persue it though - what am I missing here?
Last edited by mykey (2008-01-07 14:19:05)

sure! - pacman.conf:
# /etc/pacman.conf
# See the pacman manpage for option directives
# GENERAL OPTIONS
[options]
LogFile = /var/log/pacman.log
NoUpgrade = etc/passwd etc/group etc/shadow etc/sudoers
NoUpgrade = etc/fstab etc/raidtab etc/ld.so.conf
NoUpgrade = etc/rc.conf etc/rc.local
NoUpgrade = etc/modprobe.conf etc/modules.conf
NoUpgrade = etc/lilo.conf boot/grub/menu.lst
HoldPkg = pacman glibc
IgnorePkg = kernel26
#XferCommand = /usr/bin/wget --passive-ftp -c -O %o %u
# REPOSITORIES
# - can be defined here or included from another file
# - pacman will search repositories in the order defined here
# - local/custom mirrors can be added here or in separate files
# - repositories listed first will take precedence when packages
# have identical names, regardless of version number
#[testing]
#Include = /etc/pacman.d/testing
[core]
# Add your preferred servers here, they will be used first
#Server = ftp://archlinux.puzzle.ch/core/os/x86_64
Include = /etc/pacman.d/core
[extra]
# Add your preferred servers here, they will be used first
#Server = ftp://archlinux.puzzle.ch/extra/os/x86_64
Include = /etc/pacman.d/extra
#[community]
# Add your preferred servers here, they will be used first
#Server = ftp://archlinux.puzzle.ch/community/os/x86_64
Include = /etc/pacman.d/community
#[unstable]
# Add your preferred servers here, they will be used first
#Server = ftp://archlinux.puzzle.ch/unstable/os/x86_64
Include = /etc/pacman.d/unstable
# An example of a custom package repository. See the pacman manpage for
# tips on creating your own repositories.
#[custom]
#Server = file:///home/custompkgs
and - pacman -Sy:
:: Synchronizing package databases...
core is up to date
extra 303.3K 373.6K/s 00:00:01 [#####################] 100%
local database is up to date

New line break and extra blank space characters disappear after submit form?

Hello,
I have a PDF form with a Submit button that is dynamically created in my code to send the form data to the server in HTML format.
After the form data is received on the server side, all strings with new line break and extra blank spaces are gone.
For example, if I enter string in a text field as shown below on the form:
Hello   ,
this is
    just a
test
After the form data is sent to the server, this string would become:
Hello , this is just a test
New line breaks are gone. Also, if there is more than 1 blank space character between 2 characters, the extra blank space characters would be removed as well.
It does not only happen to multi-line text field, even with single-line text field. If I have a string like this in a single-line text field:
Hello         this is just              a         test
After the form data is sent to the server, it would become:
Hello this is just a test
The form is created in OpenOffice then converted to PDF. The Submit button is created in my program using iText. I have no idea it is iText that trims my string or PDF itself does it.
Can anyone give me any possible explanation? Thanks.

That is not what I get. Since it's URL-encoded, spaces are represented by the "+" character and carriage returns are represented by the "%0d%0a" (cr/lf) sequence.
Are you looking at the actual data that getting sent to the server or the output from the server after it processes it?

Facts and dimensions joins

Hi,
Can we use three fact tables in rpd and not join them? We have three facts and four dimensions. All the four are common dimensions to the three fact tables. So in the physical joins can we just join the facts to the dimensions? will it be accepatble if we do not have a join between fact tables. Is it necessary that all the tables (fact and dimensions) need to be joined to some tabler? Can we leave the joins between the facts which are not necessary? Please advice.

We do not do joins between fact tables, the joins between facts will be happened through dimensions (conformed dimensions).
In your case the 4 dimension will act as an conformed one.
Is it necessary that all the tables (fact and dimensions) need to be joined to some tabler? You cannot have a table in BMM which is not joined to any other table.
Can we leave the joins between the facts which are not necessary?Yes, Dim1 - fact1 -Dim 2 in this fact 1 is connected with Dim1, Dim2 and there is no relationship for fact1 with Dim3 so we don't join it.
Dim1,Dim2,Dim3 is joined with fact2 in this case Dim 1 and Dim2 will act as conformed dimesion between fact1 and fact2, where as dim3 will be a non conformed dimesion.
Thanks,
Vino

Difference between inner join and outer join

1.Difference between inner join and outer join
2.wht is the difference in using hide and get crusor value in interactive.
3. Using join is better or views in writting program . Which is better.

Table 1                      Table 2
A
B
C
D
D
E
F
G
H
a1
b1
c1
1
1
e1
f1
g1
h1
a2
b2
c2
1
3
e2
f2
g2
h2
a3
b3
c3
2
4
e3
f3
g3
h3
a4
b4
c4
3
|--|||--|
    Inner Join
    |--||||||||--|
    | A | B | C | D | D | E | F | G | H |
    |--||||||||--|
    | a1 | b1 | c1 | 1 | 1 | e1 | f1 | g1 | h1 |
    | a2 | b2 | c2 | 1 | 1 | e1 | f1 | g1 | h1 |
    | a4 | b4 | c4 | 3 | 3 | e2 | f2 | g2 | h2 |
    |--||||||||--|
Example
Output a list of all flights from Frankfurt to New York between September 10th and 20th, 2001 that are not sold out:
DATA: DATE   LIKE SFLIGHT-FLDATE,
      CARRID LIKE SFLIGHT-CARRID,
      CONNID LIKE SFLIGHT-CONNID.
SELECT FCARRID FCONNID F~FLDATE
    INTO (CARRID, CONNID, DATE)
    FROM SFLIGHT AS F INNER JOIN SPFLI AS P
           ON FCARRID = PCARRID AND
              FCONNID = PCONNID
    WHERE P~CITYFROM = 'FRANKFURT'
      AND P~CITYTO   = 'NEW YORK'
      AND F~FLDATE BETWEEN '20010910' AND '20010920'
      AND FSEATSOCC < FSEATSMAX.
WRITE: / DATE, CARRID, CONNID.
ENDSELECT.
If there are columns with the same name in both tables, you must distinguish between them by prefixing the field descriptor with the table name or a table alias.
Note
In order to determine the result of a SELECT command where the FROM clause contains a join, the database system first creates a temporary table containing the lines that meet the ON condition. The WHERE condition is then applied to the temporary table. It does not matter in an inner join whether the condition is in the ON or WHEREclause. The following example returns the same solution as the previous one.
Example
Output of a list of all flights from Frankfurt to New York between September 10th and 20th, 2001 that are not sold out:
DATA: DATE   LIKE SFLIGHT-FLDATE,
      CARRID LIKE SFLIGHT-CARRID,
      CONNID LIKE SFLIGHT-CONNID.
SELECT FCARRID FCONNID F~FLDATE
    INTO (CARRID, CONNID, DATE)
    FROM SFLIGHT AS F INNER JOIN SPFLI AS P
           ON FCARRID = PCARRID
    WHERE FCONNID = PCONNID
      AND P~CITYFROM = 'FRANKFURT'
      AND P~CITYTO   = 'NEW YORK'
      AND F~FLDATE BETWEEN '20010910' AND '20010920'
      AND FSEATSOCC < FSEATSMAX.
WRITE: / DATE, CARRID, CONNID.
ENDSELECT.
Note
Since not all of the database systems supported by SAP use the standard syntax for ON conditions, the syntax has been restricted. It only allows those joins that produce the same results on all of the supported database systems:
Only a table or view may appear to the right of the JOIN operator, not another join expression.
Only AND is possible in the ON condition as a logical operator.
Each comparison in the ON condition must contain a field from the right-hand table.
If an outer join occurs in the FROM clause, all the ON conditions must contain at least one "real" JOIN condition (a condition that contains a field from tabref1 amd a field from tabref2.
Note
In some cases, '*' may be specified in the SELECT clause, and an internal table or work area is entered into the INTO clause (instead of a list of fields). If so, the fields are written to the target area from left to right in the order in which the tables appear in the FROM clause, according to the structure of each table work area. There can then be gaps between table work areas if you use an Alignment Request. For this reason, you should define the target work area with reference to the types of the database tables, not simply by counting the total number of fields. For an example, see below:
Variant 3
... FROM tabref1 LEFT [OUTER] JOIN tabref2 ON cond
Effect
Selects the data from the transparent database tables and/or views specified in tabref1 and tabref2. tabref1 und tabref2 both have either the same form as in variant 1 or are themselves join expressions. The keyword OUTER can be omitted. The database tables or views specified in tabref1 and tabref2 must be recognized by the ABAP-Dictionary.
In order to determine the result of a SELECT command where the FROM clause contains a left outer join, the database system creates a temporary table containing the lines that meet the ON condition. The remaining fields from the left-hand table (tabref1) are then added to this table, and their corresponding fields from the right-hand table are filled with ZERO values. The system then applies the WHERE condition to the table.
Left outer join between table 1 and table 2 where column D in both tables set the join condition:
Table 1                      Table 2
A
B
C
D
D
E
F
G
H
a1
b1
c1
1
1
e1
f1
g1
h1
a2
b2
c2
1
3
e2
f2
g2
h2
a3
b3
c3
2
4
e3
f3
g3
h3
a4
b4
c4
3
|--|||--|
    Left Outer Join
    |--||||||||--|
    | A | B | C | D | D | E | F | G | H |
    |--||||||||--|
    | a1 | b1 | c1 | 1 | 1 | e1 | f1 | g1 | h1 |
    | a2 | b2 | c2 | 1 | 1 | e1 | f1 | g1 | h1 |
    | a3 | b3 | c3 | 2 |NULL|NULL|NULL|NULL|NULL|
    | a4 | b4 | c4 | 3 | 3 | e2 | f2 | g2 | h2 |
    |--||||||||--|
Regards
Prabhu

Include structure and extra fields in the same Internal Table

Hi developers,
im trying to declare an internal table with include structure and extra fields,
something like this:
data: BEGIN OF it_loans occurs 0,
      include structure zthrca006,
      status(10),
      pernr   like pa0001-pernr,
      sname   like pa0001-pernr,
      tipomov(20),
      monto   like zthrca006-monto,
data: END of it_loans.
zthrca006 is huge so i dont want to type everithing.

What is the issue?
data: BEGIN OF it_loans occurs 0.
             include structure zthrca006.
data :     status(10),
data :     pernr like pa0001-pernr.
data :     sname like pa0001-pernr.
data :     tipomov(20).
data :     monto like zthrca006-monto,.
data: END of it_loans.
Regards,
Ravi
Note - Please mark all the helpful answers

Ragged Hierarchies and extra joins

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