How to create index attributes in CAF Entity Service

When one creates an entity service with external persistency (local persistency on the persistency tab is unchecked) you have to create at least one "key" attribute. However, how do you create "index" attributes to improve performance. Currently, if you add custom key you must select "key" checkbox in attribute wizard when creating an attribute and this key field must be unique. This is not what I am looking for. I'd like to create a few custom keys that the contents allow duplicate.

Hi Tom,
This is a very valuable question. Yes, you can add an index for specific fields to the generated database table. You can do in the NWDS, navigate to the dictionnary view and open the corresponding database table. Now you can add an index and assign database field(s) to this index.
However: you must be aware that, if the entity is changed (i.e. new parameters are added), the database table is re-generated and the indices are lost.
Regards,
Jan

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Hi All
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of the CAF project and added it to the public part of this DC.But I can't find it in the list of data structures while trying to mention datatype for a complex attribute in an entity service. Can any body suggest how to do it.
Cheers
Sudip

Hello Sudip,
You can create a complex structure by doing the following:
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Regards,
Austin.

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hello,
here is my question.
how to design CAF Entity service?
i got employees data and departments data.
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in tradition RDBMS(relational database management system), i'll create three tables and named "Employee", "Department" and "DepEmp". the table "DepEmp" is a kind of table for N to N relaction.
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=======================================
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name
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=======================================
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Reward points if useful.
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How to create indexes on ODS ?

Hello friends ,
Need some help .
Could any one please let me know how to create indexes on ODS ?
How Indexes are useful on ODS ?
Thanks in advance
Regards

Dear Akshay,
Below is the information about indexes and there creation for ODS.
You can search a table for data records that satisfy certain search criteria faster using an index.
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Leaving content frame.
With Regards,
Prafulla Singh

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Hi,
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Can anyone please tell me how to create indexes using CREATE TABLE staement?e.g. creating a primary key or a unique constraint will generate indexes along with the create table syntax:
SQL> create table t (a integer primary key, b integer unique)
Table created.
SQL> select   index_name, index_type, uniqueness
from   user_indexes
where   table_name = 'T'
INDEX_NAME                     INDEX_TYPE                  UNIQUENES
SYS_C0016575                   NORMAL                      UNIQUE
SYS_C0016574                   NORMAL                      UNIQUE
2 rows selected.

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how to create/add attributes in element list in Screen (example 100)
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Removing/updating data through CAF entity service

Does anyone know a way to remove/update data through a CAF entity service from a web dynpro which uses the webdynpro model of the CAF project ?

Hi Nicolaij,
This example describes how UPDATE and DELETE works under SP8.
Hopefully it helps.
Regards
Kamil
UPDATE of entity called "Bank"
=======================
ABank recordBank;
recordBank = BankServiceProxy.read(000000024);
recordBank.setCountryId(Germany);
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aspectList.sendChanges();
DELETE from entity called "TransferID"
============================
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Problem when creating CAF Entity Service finder Methods

Guys,
Can any one please suggest me appropriatly to the below problem.
I am using NWDS 7.06. I have no problem when creating CAF project, Application Service and Entity Service even.
But in the Entity Service:
1. Add a string or longtext attribute.
2. try to create a custom finder method in operations tab, during that operation i am not finding the attribute which i have created in step1.
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I was able to do that in some version long ago but i have forgotten that NWDS version.
summary of Problem Is: " Cannot create custom methods with custom parameters in Entity Service"
Can you please suggest me the right version or a solution to this problem on urgent basis please.
I promise you that i will give you full points to you who ever gives me the right solution.
Please mail your suggestions to [email protected]

Guys,
I have solved the problem successfully. Thanks for your attention.

How to create index for a column of a view

Hi,
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As mentioned this is a java forum not a mysql forum, but as I know the answer - you can't create an index directly on a view in mysql.

How to create index not unique at the time of the CREATE TABLE

Hi,
I am trying to find out how in Oracle create secondary indexes that can accept duplicated into the CREATE TABLE statement, without have to execute a CREATE INDEX separately.
As far I can see the only syntax accepted by Oracle 9i to create more than one index at the time of the table creation is:
CREATE TABLE test_tab (x INTEGER, y INTEGER, z INTEGER PRIMARY KEY(x,y), UNIQUE(z))
But, in my case I need to have the unicity only for the primary key, but not for the second index, that I would like to have not unique.
How to do that inside of the CREATE TABLE statement?
Any help?
Thanks a lot in advance.

To create an index automatically (not constraint related) , you will need to have an 'event' trigger on the schema that will (IE: detecting a table create/drop matching ARCHIVE_%) , generate the appropriate SQL required (create index sql) and pass THAT to a DBMS_JOB.
Make sure you have job queue's enabled.
This was the ONLY way that I could re-create a view automatically as new tables were created that matched a criteria IE ARCHIVE_JAN06, ARCHIVE_FEB06 etc.
The application's "archiving" method created and was thus aware of these tables and permitted searching within 'archived' data, but 3rd party reporting applications needed to see a view encompassing all data, regardless of the tables involved. The view automatically created was a 'union_all' of all tables concerned.

How to create index on a single parition

Hi,
I have a table which has got 7 partitons,I want to index only one partition how do it?
i am using oracle 11g.Also i want to know if it is possible to create an index within a o partiotn only fr some values..
Edited by: user8731258 on Jul 13, 2012 12:35 AM

Hi,
I've been thinking a bit about your requirement.
One way could be to use the fact that index do not "store" nulls.
Let's make a table with data :SQL> create table t
2 (
3 grp integer,
4 id integer,
5 val varchar2(10)
6 )
7 partition by range(grp)
8 (
9 partition p01 values less than (2)
10 ,partition p02 values less than (3)
11 ,partition p03 values less than (4)
12 ,partition pXX values less than (maxvalue)
13 );
Table created.
SQL> insert into t select 1+mod(level,4), level, dbms_random.string('a',10) from dual connect by level <= 200000;
200000 rows created.
SQL> select grp, count(*) cnt from t group by grp order by grp;
       GRP       CNT
      1     50000
      2     50000
      3     50000
      4     50000
SQL> select grp, count(*) cnt from t group by grp order by grp;
       GRP       CNT
      1     50000
      2     50000
      3     50000
      4     50000
SQL> begin
2 dbms_stats.gather_table_stats(
3 ownname => user,
4 tabname => 'T',
5 method_opt => 'for all columns size 1',
6 granularity => 'global and partition',
7 estimate_percent => null);
8 end;
9 /
PL/SQL procedure successfully completed.
SQL> select table_name, partition_name, num_rows from user_tab_partitions where table_name='T' order by partition_name;
TABLE_NAME                 PARTITION_NAME               NUM_ROWS
T                      P01                       50000
T                      P02                       50000
T                      P03                       50000
T                      PXX                       50000Now let's say you only want to index P01.
You could do it as follows :SQL> create index t_idx on t(decode(grp,1,id,null)) local;
Index created.The index would only have IDs from partition P01.
SQL> select INDEX_NAME, PARTITION_NAME, DISTINCT_KEYS
2 from user_ind_partitions
3 where index_name='T_IDX'
4 order by partition_name;
INDEX_NAME                 PARTITION_NAME                DISTINCT_KEYS
T_IDX                      P01                          50000
T_IDX                      P02                           0
T_IDX                      P03                           0
T_IDX                      PXX                           0
SQL> select SEGMENT_NAME, PARTITION_NAME, BYTES
2 from user_segments
3 where SEGMENT_NAME='T_IDX'
4 order by partition_name;
SEGMENT_NAME                                               PARTITION_NAME                BYTES
T_IDX                                                    P01                        1048576
T_IDX                                                    P02                          65536
T_IDX                                                    P03                          65536
T_IDX                                                    PXX                          65536See, only the keys from P01 are here, and only the segment for P01 partition has grown <i>(the 65k segment for the others partitions come from default values of storage clause which I didn't specify)</i>.
Now you can query your table in the following way.
Instead of writing (for P01) : select * from t where grp=1 and id between 30 and 40;If you want to use the index on P01, you would do :SQL> select * from t where grp=1 and decode(grp,1,id,null) between 30 and 40;
       GRP        ID VAL
      1        32 hSrmmTYXXW
      1        36 xofKlsxORE
      1        40 vazRGnWzWc
Execution Plan
Plan hash value: 1186642050
| Id | Operation                  | Name | Rows | Bytes | Cost (%CPU)| Time        | Pstart| Pstop |
|   0 | SELECT STATEMENT             |        |   125 | 2375 |      3   (0)| 00:00:01 |        |        |
|   1 | PARTITION RANGE SINGLE         |        |   125 | 2375 |      3   (0)| 00:00:01 |      1 |      1 |
|* 2 |   TABLE ACCESS BY LOCAL INDEX ROWID| T        |   125 | 2375 |      3   (0)| 00:00:01 |      1 |      1 |
|* 3 |    INDEX RANGE SCAN             | T_IDX |   225 |        |      2   (0)| 00:00:01 |      1 |      1 |
Predicate Information (identified by operation id):
   2 - filter("GRP"=1)
   3 - access(DECODE("GRP",1,"ID",NULL)>=30 AND DECODE("GRP",1,"ID",NULL)<=40)The index is being used.
+(also note that the grp value is "given" twice in the query : once explicitly, the other one within the decode.)+
Now for a query like : SQL> select * from t where grp=2 and id between 30 and 40;
       GRP        ID VAL
      2        33 bhFrCYCiDb
      2        37 jnPauHJiyo
Execution Plan
Plan hash value: 2931986080
| Id | Operation            | Name | Rows | Bytes | Cost (%CPU)| Time     | Pstart| Pstop |
|   0 | SELECT STATEMENT       |      |     5 |    95 |    68     (0)| 00:00:01 |       |       |
|   1 | PARTITION RANGE SINGLE|      |     5 |    95 |    68     (0)| 00:00:01 |     2 |     2 |
|* 2 |   TABLE ACCESS FULL    | T    |     5 |    95 |    68     (0)| 00:00:01 |     2 |     2 |
Predicate Information (identified by operation id):
   2 - filter("ID"<=40 AND "ID">=30 AND "GRP"=2)The optimizer goes for a FULL TABLE (partition) SCAN.
<b>NOTE :</b> That would also work with a global index.
Of course, it remains a rather dirty tweak...
;-)

How to sort web dynpro table wich data bind to CAF Entity Service

Hi
I created UI Table based on Model Node (CAF Entity).
When I try to sort this Table with using TableSorter, I get following in the trace log:
The error is: com.sap.caf.rt.exception.CAFBaseRuntimeException: Aspect does not support changing rows via its list interface
Have I met CAF limitation, or do I do something wrong?

Hi Nikolai,
if you use the implementation
<code>
IServiceFacade serviceFacade = CAFServiceFactory.getServiceFacade(ts2Definition.class);
</code>
it means you use the typed access.
Possible in this sneak the
<code>
     IServiceFacade serviceFacade = CAFServiceFactory.getServiceFacade();
</code>
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Try the following way - it must work.
<code>
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node.sort(your_comparator) ;
the comparator like this:
<code>
     public class YourComparator implements Comparator
       public int compare(Object o1, Object o2)
          IWDNodeElement nodeElement1 = (IWDNodeElement) o1 ;
          IWDNodeElement nodeElement2 = (IWDNodeElement) o2 ;
          String name1 = nodeElement1.getAttributeAsText(_attrName) ;
          String name2 = nodeElement2.getAttributeAsText(_attrName) ;
          if (name1 == null)
               return -1 ;
          if (name2 == null)
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          if (_direction == "up")
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          else if (_direction == "down")
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          return 0 ;
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</code>
Best regards,
Aliaksei.

How to create index attributes in CAF Entity Service

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