Intend hierarchy in the colomn

Hi everybody,
does anybody know how i can intend hierarchy nodes in the colomn in the integrated planning. I have a big hierarchy with so many nodes and childnodes. I am not able to realise which node where is. I could not find any possibilities to intend this hierarchy in order to realise the position of each nodes in the hierarchy easier. Has anyone an idea how it is possible?
Thanks in advance.
Ali

Thanks for your Reply. Infact I am aware that free characterstics will come into effect when your drill down. When i use a variable on hierarchy and place the info object at rows,Query is working fine and when we kwwp the same info object in free characterstic the report is displaying no applicable data found. Ideally the info object restricted by a variable and kept in free characterstic section will come into effect when we drill down only.
I do not have clue why the report is running in unique way.

Similar Messages

Can we have a Hierarchy in the Prompt of webi? promot as key-text?

Hello all,
Can we have a Hierarchy in the Prompt of webi? We are using Fund Center hierarchy and the user wants to select the top node to get all the nodes under it in hierarchies?
Also along same lines can we display the Long text and Key ofthe FUnd center in prompt (hierarchy or not)
Thanks in advance for all the replies.

Hi david, Gowtham
Thanks for the replies. I was able to get a hierarchy in the prompt. I created mandatory hierarchy variable in the bex query which is feeding the universe and it automatically came over as hierarchy. I am still not able to get how would you create cascade on same object, I understand the market and region (2 different object) example. So if you can please explain.
With that said I still am not sure how you would display your prompt with description of objects rather than technical names. Any insight on that please?
Thanks a lot

Display key date of hierarchy in the query

Hello,
we have a query that uses a hierarchy with the time depending structure. The key date of this hierarchy has been restricted with a variable and can be entered by user in the query selection screen.
Is there any way to display this key date variable in the query result ? Otherwise the users do not see the key date they had selected before.
Thanks in advance !
Regards
Igor

Hi Chandamita,
Thank you so much for your quick response. I'm not sure if it will fix the issue because it has the same properties for other object which we do not encounter the issue.
Thank you!

How to return list of lowest level elements in a hierarchy given the node o

We would like to return a list of the lowest level Elements in a Hierarchy from a given node of the hierarchy. The level of the node that is given to us is unknown, but regardless, we need the ability to return the lowest level elements. We are trying to find a non-ABAP solution to our needs and trying to use a BEx query as a source for our APD.
In query designer we are able to obtain the correct list when using the Hierarchy and Condition with a Leaf value of '1' to return the lowest level. However, We have come to learn APDs do no support the BEx hierarchies. When we de-selected the 'Activate Hierarchy' display in the query, the entire structure under the given node is returned.
Does anybody know of a way to return just the lowest level elements in a hierarchy so we can feed our APD?
Thanks,
Andrea

expac works only with official and unofficial repos, not with the AUR, but it's very nice.
Adding '-q' should help, grep to weed out false positives.
yaourt -Sa $(yaourt -Ssaq pulse)
works, but
$ yaourt -Sa $(yaourt -Ssaq chrome)
resolving dependencies...
looking for inter-conflicts...
warning: removing 'chromium' from target list because it conflicts with 'chromium-no-sse2'
error: unresolvable package conflicts detected
error: failed to prepare transaction (conflicting dependencies)
:: chromium-no-sse2 and chromium-scroll-pixelGs are in conflict
so at least
$ yaourt -Sa $(yaourt -Ssaq chrome|grep google)
is needed.
Edit: Also:
$ LC_ALL=C TZ=GMT0 diff -Naur /usr/bin/pacsearch /usr/local/bin/pacsearch
--- /usr/bin/pacsearch 2014-11-21 11:20:37.000000000 +0000
+++ /usr/local/bin/pacsearch 2014-12-21 08:21:14.758856006 +0000
@@ -84,7 +84,7 @@
my %allpkgs = ();
-my $syncout = `pacman -Ss '@ARGV'`;
+my $syncout = `pacman -Ssq '@ARGV'`;
# split each sync search entry into its own array entry
my @syncpkgs = split(/\n^(?=\w)/m, $syncout);
# remove the extra \n from the last desc entry
@@ -110,7 +110,7 @@
$allpkgs{$pkgfields[1]} = [ @pkgfields ];
-my $queryout = `pacman -Qs '@ARGV'`;
+my $queryout = `pacman -Qqs '@ARGV'`;
# split each querysearch entry into its own array entry
my @querypkgs = split(/\n^(?=\w)/m, $queryout);
# remove the extra \n from the last desc entry
$ /usr/local/bin/pacsearch pulse
libpulse
pulseaudio
libao
libcanberra-pulse
libpulse
paprefs
pavucontrol
pulseaudio
pulseaudio-alsa
floyd
libcec
mate-media-pulseaudio
mate-settings-daemon-pulseaudio
ponymix
projectm-pulseaudio
Although why not simply use pacman or expac?
Last edited by karol (2014-12-21 08:26:48)

How to put the product hierarchy in the bw?

I want to have the product hierarchy in the reporting, in a way, that the users can open and shut the elements of the tree.
So i added the foreign hierarchy criteria 0PRODH1, 2, 3 to the InfoObject 0MATERIAL.
Then I used InfoSource 0MATERIAL_LPRH_HIER to put data in 0MATERIAL.
In the result 0MATERIAL HAS the hierarchy, but without any materials added to the leafs.
What did i do wrong?
Tobias

Hi ,
Have a look on note: 407033
It says 2 main points with regarding to your problem.
1) Material master records with shared master data in the sales area cannot be assigned to a product hierarchy in BW.
This solution generates the thinnest possible hierarchy, but leads to some restrictions in the navigation options in the BW queries because you can no longer expand the hierarchy to the material numbers.
2) The 0PROD_HIER attribute of 0-MATERIAL must be active as a navigation attribute in the cube.When you expand the hierarchies by product hierarchy in the query, you can display this hierarchy.
With rgds,
Anil Kuma Sharma .P
Kindly assign points , if it really helps you.

Display a particular node of Hierarchy in the Report.

Hi , all
I have an urgent requirement that i have to display only a particular node(5) of Hierarchy in the Bw Report , I have tried different options available in the Hierarchy tab (That is Expand to level 5 , Position of lower node - Above , Value of posted node - Hide ) . But stil it is displaying 5 and then 4 node , but i don't want that 4 node .
Can anybody gine me some useful suggestion .
Thanks in advance..

Hi Akash,
- To show only particular nodes of a hierarchy in the Excel sheet:
Step 1: include the base characteristic
Step 2: select the hierarchy (properties of the characteristic) - (display key only)
Step 3: user restrict (from context menu of the characteristic) on the nodes that you want to display.
Step 4: expand to level: 1 (properites of the characteristics)
This should serve the purpose. I know it is a little late, but it might be helpful in the future

Product Hierarchy in the t-code F-02 - Profitability Segment section.

Dear All,
I am trying to control the input of the Product Hierarchy in the t-code F-02 - Profitability Segment section. May i know how i can do that? Currently i am using field exits but it is not working in this program.
Please help.
thank you

Hi Jasmina,
When you post with PK 31 you create you post other line with PK 40.
In the 40 line you fill in the Profit segment. A Derivation rule is based on 1 posting line and not 2 or more line's
In the derivation rule from the 40 you don't have the vendor number (this is in the 31 line).
Perhaps it is possible with a user exit in CO-PA.
Regards,
Paul

Stored Procedure- Multil level employee hierarchy in the same row

I have an employee table where I have employee details which includes the following properties:
EmployeeID, EmployeeName, ManagerID, DesignationCode.
ManagerID maps to EmployeeID in the Emp table. DesignationCode is to identify the employee designation.
The topmost person of the organization could be identified by : whose DesignationCode is 'XX' and ManagerID=EmplyeeID.
Here, 'XX' is fixed and will not change.
Also, we know there could be a maximum of 10 level for each hierarchy.
Example: Employee1 reports to Manager1 who reports to Manager2 who reports to Manager 3 ..... who reports to ManagerN.
I need to pull a hierarchy in the below format:
EmpName MgrName0 MagrName1 MgrName2 ..........MgrName7 MgrName8
MgrName9
SAMRAT
      XXX    XXX
SUDHAKAR
XXX         XXX    XXX
SATESWAR
        XXX    XXX
SRINI               XXX      XXX
XXX         XXX    XXX
IMPORTANT POINT: We need to identify the reporting hierarchy level for each employee's manager and then place the manager names accordingly based on the columns.
Example:
If an employee's manager has only 1 reporting manager(who is at top level), then he would be placed at the column "ManagerName8" and ManageName9 would be the topmost employee of the organization
In short, we need to identify the reporting level at which the employees manager belongs and then start filling the columns values for that row ( employee details)
I am stuck and unable to do the same. Please help.
Thank you.

Please post DDL, so that people do not have to guess what the keys, constraints, Declarative Referential Integrity, data types, etc. in your schema are. Learn how to follow ISO-11179 data element naming conventions and formatting rules. Temporal data should
use ISO-8601 formats. Code should be in Standard SQL as much as possible and not local dialect.
This is minimal polite behavior on SQL forums.
>> I have an employee table where I have employee details <<
An SQL programmer would have a Personnel table instead. What you have said is that you have only one employee!
>> which includes the following properties:
emp_id, emp_name, manager_emp_id, designation_code.<<
Why do you think that a manager is an attribute of an employee? This is absurd! They have a relationship; where is the table that models that relationship? It is missing from this non-normalized, improperly designed table.
There are many ways to represent a tree or hierarchy in SQL. YOU are trying to use an adjacency list model. This approach will wind up with really ugly code -- CTEs hiding recursive procedures, horrible cycle prevention code, etc.
Another way of representing trees is to show them as nested sets.
Since SQL is a set oriented language, this is a better model than the usual adjacency list approach you see in most text books. Let us define a simple OrgChart table like this.
CREATE TABLE OrgChart
(emp_name CHAR(10) NOT NULL PRIMARY KEY,
lft INTEGER NOT NULL UNIQUE CHECK (lft > 0),
rgt INTEGER NOT NULL UNIQUE CHECK (rgt > 1),
CONSTRAINT order_okay CHECK (lft < rgt));
OrgChart
emp_name lft rgt
======================
'Albert' 1 12
'Bert' 2 3
'Chuck' 4 11
'Donna' 5 6
'Eddie' 7 8
'Fred' 9 10
The (lft, rgt) pairs are like tags in a mark-up language, or parens in algebra, BEGIN-END blocks in Algol-family programming languages, etc. -- they bracket a sub-set. This is a set-oriented approach to trees in a set-oriented language.
The organizational chart would look like this as a directed graph:
Albert (1, 12)
Bert (2, 3) Chuck (4, 11)
/ | \
/ | \
/ | \
/ | \
Donna (5, 6) Eddie (7, 8) Fred (9, 10)
The adjacency list table is denormalized in several ways. We are modeling both the Personnel and the Organizational chart in one table. But for the sake of saving space, pretend that the names are job titles and that we have another table which describes the
Personnel that hold those positions.
Another problem with the adjacency list model is that the boss_emp_name and employee columns are the same kind of thing (i.e. identifiers of personnel), and therefore should be shown in only one column in a normalized table. To prove that this is not
normalized, assume that "Chuck" changes his name to "Charles"; you have to change his name in both columns and several places. The defining characteristic of a normalized table is that you have one fact, one place, one time.
The final problem is that the adjacency list model does not model subordination. Authority flows downhill in a hierarchy, but If I fire Chuck, I disconnect all of his subordinates from Albert. There are situations (i.e. water pipes) where this is true, but
that is not the expected situation in this case.
To show a tree as nested sets, replace the nodes with ovals, and then nest subordinate ovals inside each other. The root will be the largest oval and will contain every other node. The leaf nodes will be the innermost ovals with nothing else inside them
and the nesting will show the hierarchical relationship. The (lft, rgt) columns (I cannot use the reserved words LEFT and RIGHT in SQL) are what show the nesting. This is like XML, HTML or parentheses.
At this point, the boss_emp_name column is both redundant and denormalized, so it can be dropped. Also, note that the tree structure can be kept in one table and all the information about a node can be put in a second table and they can be joined on employee
number for queries.
To convert the graph into a nested sets model think of a little worm crawling along the tree. The worm starts at the top, the root, makes a complete trip around the tree. When he comes to a node, he puts a number in the cell on the side that he is visiting
and increments his counter. Each node will get two numbers, one of the right side and one for the left. Computer Science majors will recognize this as a modified preorder tree traversal algorithm. Finally, drop the unneeded OrgChart.boss_emp_name column
which used to represent the edges of a graph.
This has some predictable results that we can use for building queries. The root is always (left = 1, right = 2 * (SELECT COUNT(*) FROM TreeTable)); leaf nodes always have (left + 1 = right); subtrees are defined by the BETWEEN predicate; etc. Here are
two common queries which can be used to build others:
1. An employee and all their Supervisors, no matter how deep the tree.
SELECT O2.*
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O1.emp_name = :in_emp_name;
2. The employee and all their subordinates. There is a nice symmetry here.
SELECT O1.*
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O2.emp_name = :in_emp_name;
3. Add a GROUP BY and aggregate functions to these basic queries and you have hierarchical reports. For example, the total salaries which each employee controls:
SELECT O2.emp_name, SUM(S1.salary_amt)
FROM OrgChart AS O1, OrgChart AS O2,
Salaries AS S1
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND S1.emp_name = O2.emp_name
GROUP BY O2.emp_name;
4. To find the level and the size of the subtree rooted at each emp_name, so you can print the tree as an indented listing.
SELECT O1.emp_name,
SUM(CASE WHEN O2.lft BETWEEN O1.lft AND O1.rgt
THEN O2.sale_amt ELSE 0.00 END) AS sale_amt_tot,
SUM(CASE WHEN O2.lft BETWEEN O1.lft AND O1.rgt
THEN 1 ELSE 0 END) AS subtree_size,
SUM(CASE WHEN O1.lft BETWEEN O2.lft AND O2.rgt
THEN 1 ELSE 0 END) AS lvl
FROM OrgChart AS O1, OrgChart AS O2
GROUP BY O1.emp_name;
5. The nested set model has an implied ordering of siblings which the adjacency list model does not. To insert a new node, G1, under part G. We can insert one node at a time like this:
BEGIN ATOMIC
DECLARE rightmost_spread INTEGER;
SET rightmost_spread
= (SELECT rgt
FROM Frammis
WHERE part = 'G');
UPDATE Frammis
SET lft = CASE WHEN lft > rightmost_spread
THEN lft + 2
ELSE lft END,
rgt = CASE WHEN rgt >= rightmost_spread
THEN rgt + 2
ELSE rgt END
WHERE rgt >= rightmost_spread;
INSERT INTO Frammis (part, lft, rgt)
VALUES ('G1', rightmost_spread, (rightmost_spread + 1));
COMMIT WORK;
END;
The idea is to spread the (lft, rgt) numbers after the youngest child of the parent, G in this case, over by two to make room for the new addition, G1. This procedure will add the new node to the rightmost child position, which helps to preserve the idea
of an age order among the siblings.
6. To convert a nested sets model into an adjacency list model:
SELECT B.emp_name AS boss_emp_name, E.emp_name
FROM OrgChart AS E
LEFT OUTER JOIN
OrgChart AS B
ON B.lft
= (SELECT MAX(lft)
FROM OrgChart AS S
WHERE E.lft > S.lft
AND E.lft < S.rgt);
7. To find the immediate parent of a node:
SELECT MAX(P2.lft), MIN(P2.rgt)
FROM Personnel AS P1, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
AND P1.emp_name = @my_emp_name;
I have a book on TREES & HIERARCHIES IN SQL which you can get at Amazon.com right now. It has a lot of other programming idioms for nested sets, like levels, structural comparisons, re-arrangement procedures, etc.
--CELKO-- Books in Celko Series for Morgan-Kaufmann Publishing: Analytics and OLAP in SQL / Data and Databases: Concepts in Practice Data / Measurements and Standards in SQL SQL for Smarties / SQL Programming Style / SQL Puzzles and Answers / Thinking
in Sets / Trees and Hierarchies in SQL

View dimension hierarchy using the filter property of data form

Hi,
I created a data form where I want to show the dimension hierarchy. But when I am filtering the form using attribute dimension then the hierarchy is not being shown in the form.Is their any way so that I can use the filter property also and make the dimension hierarchy visible in the data form?
Thanks.

847833 wrote:
Hi Endy,
Thanks for your reply. I have used the "equal" function as to set that specific attribute as I am choosing the descendants function against the attribute dimension,it's showing error.Hi,
What's the error ? You have to use a function, otherwise it will only display the member you picked.
Actually in our dimension structure we have a 3 generation hierarchy and the attribute is only associated with the 3rd generation member. So we want to display the 3rd generation having the specific attribute "xx" wit the entire hierarchical structure in our data form.
Can you please tell me if I have missed anything?
Thanks.If you want to display the attribute value in the form, edit the form, go to the "Layout" tab, and click on your row. On the right side of the screen, you have a section called Display Properties. Check the box called Enable custom attributes, then check your attribute name. Save the form and you will have the attribute value displayed at the right of the member name.

Product hierarchy in the material master record

Hi,
I am handig Data migration project.
This is related to Product Hirerarchy.
I one business Product Hirerarchy have been configured. But in another buisiness, Product Hirerarchy has not been configured.
There are some common materials using by both business processes.
For coomon mateials if I am maintaining Product Hirerarchy value in the material master record, is there any impact on business which is not being used Product Hirerarchy. Whether it will stop / impact on business processes.
Can you explain plz
Thanks in advance
Regards

Hi karthink,
Thanks for your inputs.
we are meging two servers. In that in one server, product highrechy have been configuredand in another server not configured.
There are some materials with the numbers calling as a common parts exists in both servers.
ex: plat 0001 in one server 1 & plant 0002 in server2
Common part is : m1
plant 0001 not using product hierarchy and plant 0002 not using product hierarchy,.
After golive, all the common parts are being used by both plants and same basic data 1 view.
If we are maintaing product hierarchy in the basic data 1 view, it will flow into sales view as well.
Then what about impact on the plant 0001 business since it is not using product hierarchy.
In the configuration, is there any settings to activate / deactive product hierarchy at sales area level.
Is there any impact on MM / FICO side.
Please explain.
THANKS IN ADVANCE.
Regards
KRK

Exports a hierarchy from the web to Excel

Hi Experts
When the user exports a hierarchy from the web to Excel, it prompts the user for their password as the excel session opens. If they donu2019t enter it u2013 it constantly prompts for a password when the excel file is used.
Anyone got the same issue? or anyway I can get over this?
thanks

Not possible - at the present time anyway.
If you want such a photo available for viewing on your iPhone, you need to download/save the photo on your computer - used for syncing/transferring data from your computer to the iPhone, and transfer this photo from your computer to your iPhone - the same for any photos transferred from your computer to your iPhone.

Hi. Good morning, I intend to purchase the Photography Photoshop program. The Rs 499/month one. I currently use an Mac Air. But intend to buy a iMac in about 2 months time. Will I be able to transfer this program or continue using this on my new machine?

Hi. Good morning, I intend to purchase the Photography Photoshop program. The Rs 499/month one. I currently use an Mac Air. But intend to buy a iMac in about 2 months time. Will I be able to transfer this program or continue using this on my new machine?

My eyes just glazed over...Please in the future break down each of your issues with paragraphs separated by two carriage returns. It would be much easier when trying to address your issues.
Go to Apple menu -> System Preferences -> Keyboard and Mouse -> Mouse
And edit your mouse settings to do what you want it to do.
Secondly, this is not the place to vent. If you have a complaint, there is:
http://www.apple.com/feedback/
or http://www.apple.com/contact/
We are just end users here helping other end users.
Third, from my understanding, it would appear you are concerned about the noise the hard drive makes when it falls asleep? Why not put your machine in screen saver mode instead? Apple menu -> System Preferences -> Energy Saver turn off all Energy Saver settings, or set them to run Never.
Fourth, if your machine was purchased just a few days ago, you may still be able to get an exchange from the store, quicker than you can get a repair done. You may want to look into that possibility.
Fifth, it does appear you found the Logic forum. I would persist in asking there how to solve your technical issue with Logic regarding the audio. It may be you don't have to do anything special to the hard drive. Remember audio can be transmitted by wire, avoiding ambient sounds.
Good luck!

Downloaded my hierarchy using the program z_sap_hierarchy_download

hi
I want download downloaded my hierarchy using the program z_sap_hierarchy_download but I have any problem when I using idoc method, I have this message:
Node [00001350, 00001580]: Leaf 'E010000000' already exists as child of node 00001501
Best regard
Francoise

hi
The problem is that 2 id differents has the same description.
my program is:
begin
SAP Consulting BW Tools:
Download hierarchy into a flat file. The file has the correct format
Text elements:
P_DATES Include from/to dates
P_DATETO Valid-to date
P_FNAME File name
P_HIENM Hierarchy name
P_INTER Include from/to leaves
P_IOBJNM InfoObject
P_LANGU Language
P_VERS Hierarchy version
REPORT z_sap_hierarchy_download.
TYPE-POOLS: rs, rsdm, rrh1.
SELECTION-SCREEN BEGIN OF BLOCK b1 WITH FRAME.
PARAMETERS:
p_iobjnm TYPE rsdiobjnm MEMORY ID rsc.
SELECTION-SCREEN END OF BLOCK b1.
SELECTION-SCREEN BEGIN OF BLOCK b2 WITH FRAME.
PARAMETERS:
p_hienm TYPE rshiedir-hienm,
p_vers TYPE rshiedir-version,
p_dateto TYPE rshiedir-dateto,
p_langu TYPE rshiedirt-langu.
SELECTION-SCREEN END OF BLOCK b2.
SELECTION-SCREEN BEGIN OF BLOCK b3 WITH FRAME.
PARAMETERS:
p_fname LIKE rlgrap-filename,
p_dates AS CHECKBOX DEFAULT 'X',
p_inter AS CHECKBOX DEFAULT 'X'.
SELECTION-SCREEN END OF BLOCK b3.
File structure
TYPES:
No dates/intervals
BEGIN OF y_s_hierfile_1,
nodeid TYPE rshienodid,
iobjnm TYPE rsiobjnm,
nodename TYPE rsnodename,
tlevel TYPE rstlevel,
link TYPE rslink,
parentid TYPE rsparent,
childid TYPE rschild,
nextid TYPE rsnext,
langu TYPE langu,
txtsh TYPE rstxtsh,
txtmd TYPE rstxtmd,
txtlg TYPE rstxtlg,
END OF y_s_hierfile_1,
y_t_hierfile_1 TYPE STANDARD TABLE OF y_s_hierfile_1,
With dates
BEGIN OF y_s_hierfile_2,
nodeid TYPE rshienodid,
iobjnm TYPE rsiobjnm,
nodename TYPE rsnodename,
tlevel TYPE rstlevel,
link TYPE rslink,
parentid TYPE rsparent,
childid TYPE rschild,
nextid TYPE rsnext,
dateto TYPE rsdateto,
datefrom TYPE rsdatefrom,
langu TYPE langu,
txtsh TYPE rstxtsh,
txtmd TYPE rstxtmd,
txtlg TYPE rstxtlg,
END OF y_s_hierfile_2,
y_t_hierfile_2 TYPE STANDARD TABLE OF y_s_hierfile_2,
With intervals
BEGIN OF y_s_hierfile_3,
nodeid TYPE rshienodid,
iobjnm TYPE rsiobjnm,
nodename TYPE rsnodename,
tlevel TYPE rstlevel,
link TYPE rslink,
parentid TYPE rsparent,
childid TYPE rschild,
nextid TYPE rsnext,
leafto TYPE rsleafto,
leaffrom TYPE rsleaffrom,
langu TYPE langu,
txtsh TYPE rstxtsh,
txtmd TYPE rstxtmd,
txtlg TYPE rstxtlg,
END OF y_s_hierfile_3,
y_t_hierfile_3 TYPE STANDARD TABLE OF y_s_hierfile_3,
With dates/intervals
BEGIN OF y_s_hierfile_4,
nodeid TYPE rshienodid,
iobjnm TYPE rsiobjnm,
nodename TYPE rsnodename,
tlevel TYPE rstlevel,
link TYPE rslink,
parentid TYPE rsparent,
childid TYPE rschild,
nextid TYPE rsnext,
dateto TYPE rsdateto,
datefrom TYPE rsdatefrom,
leafto TYPE rsleafto,
leaffrom TYPE rsleaffrom,
langu TYPE langu,
txtsh TYPE rstxtsh,
txtmd TYPE rstxtmd,
txtlg TYPE rstxtlg,END OF y_s_hierfile_4,
y_t_hierfile_4 TYPE STANDARD TABLE OF y_s_hierfile_4.
Hierarchy definition
DATA:
g_s_hiesel TYPE rsndi_s_hiesel,
g_s_hiedir TYPE rsndi_s_hiedir,
g_subrc TYPE sy-subrc,
g_t_hiedirt TYPE TABLE OF rshiedirt,
g_s_hierstruc TYPE rssh_s_htab,
g_t_hierstruc TYPE TABLE OF rssh_s_htab,
g_s_thiernode TYPE rsthiernode,
g_t_thiernode TYPE TABLE OF rsthiernode WITH KEY langu hieid objvers
nodename,
g_s_hierintvl TYPE rssh_s_jtab,
g_t_hierintvl TYPE TABLE OF rssh_s_jtab WITH KEY hieid objvers nodeid,
g_s_message TYPE rsndi_s_message,
g_t_message TYPE TABLE OF rsndi_s_message,
g_s_chavlinfo TYPE rsdm_s_chavlinfo,
g_t_chavlinfo TYPE rsdm_t_chavlinfo.
File
DATA:
g_fname TYPE string,
g_struct_s TYPE string,
g_struct_t TYPE string,
gr_s_file TYPE REF TO data,
gr_t_file TYPE REF TO data.
FIELD-SYMBOLS:
<g_langu> TYPE ANY,
<g_s_file> TYPE ANY,
<g_t_file> TYPE STANDARD TABLE.
AT SELECTION-SCREEN ON VALUE-REQUEST FOR p_iobjnm.
CALL FUNCTION 'RSD_IOBJ_F4'
    EXPORTING
      i_show_cha    = rs_c_true
      i_objvers     = rs_c_objvers-active
      i_hietabfl    = rs_c_true
    CHANGING
      c_iobjnm      = p_iobjnm
    EXCEPTIONS
      illegal_input = 1.
CHECK sy-subrc = 0.
AT SELECTION-SCREEN ON VALUE-REQUEST FOR p_hienm.
DATA:
l_s_hiertxt TYPE rrh1_s_hiertxt,
l_t_hiertxt TYPE rrh1_t_hiertxt.
CALL FUNCTION 'RRH1_HIERARCHY_HELP_VALUES_GET'
    EXPORTING
      i_iobjnm          = p_iobjnm
      i_dateto          = p_dateto
      i_hienm           = p_hienm
      i_version         = p_vers
    IMPORTING
      e_t_hiertxt       = l_t_hiertxt
    EXCEPTIONS
      no_f4_available   = 1
      dialogue_canceled = 2
      OTHERS            = 3.
CHECK sy-subrc = 0.
READ TABLE l_t_hiertxt INTO l_s_hiertxt INDEX 1.
CHECK NOT l_s_hiertxt IS INITIAL.
p_hienm = l_s_hiertxt-hienm.
DATA:
l_s_dynpfields TYPE dynpread,
l_t_dynpfields TYPE STANDARD TABLE OF dynpread.
CLEAR: l_t_dynpfields, l_s_dynpfields.
l_s_dynpfields-fieldname = 'P_VERS'.
WRITE l_s_hiertxt-version TO l_s_dynpfields-fieldvalue.
APPEND l_s_dynpfields TO l_t_dynpfields.
l_s_dynpfields-fieldname = 'P_DATETO'.
WRITE l_s_hiertxt-dateto TO l_s_dynpfields-fieldvalue.
APPEND l_s_dynpfields TO l_t_dynpfields.
CALL FUNCTION 'DYNP_VALUES_UPDATE'
    EXPORTING
      dyname     = sy-repid
      dynumb     = sy-dynnr
    TABLES
      dynpfields = l_t_dynpfields.
AT SELECTION-SCREEN ON VALUE-REQUEST FOR p_fname.
DATA:
l_filename1 TYPE string,
l_filename2 TYPE string,
l_path TYPE string,
l_fullpath TYPE string,
l_action TYPE i.
l_filename1 = p_fname.
CALL METHOD cl_gui_frontend_services=>file_save_dialog
EXPORTING
window_title = 'Select Download File'
default_extension = 'txt'
default_file_name = l_filename1
file_filter = 'All Files (.)|.|Text files (.txt)|.txt'
CHANGING
filename = l_filename2
path = l_path
fullpath = l_fullpath
user_action = l_action
EXCEPTIONS
cntl_error = 1
OTHERS = 2.                                               "#EC NOTEXT
CHECK sy-subrc = 0.
CALL METHOD cl_gui_cfw=>flush.
IF l_action = 0.
    p_fname = l_fullpath.
ENDIF.
INITIALIZATION.
GET PARAMETER ID 'RSC' FIELD p_iobjnm.
START-OF-SELECTION.
Check input
CHECK NOT p_fname IS INITIAL.
IF p_langu IS INITIAL.
    p_langu = sy-langu.
ENDIF.
IF p_dateto IS INITIAL.
    p_dateto = '99991231'.
ENDIF.
Read hierarchy
CLEAR g_s_hiesel.
g_s_hiesel-objvers = rs_c_objvers-active.
g_s_hiesel-hienm = p_hienm.
g_s_hiesel-version = p_vers.
g_s_hiesel-iobjnm = p_iobjnm.
g_s_hiesel-dateto = p_dateto.
CALL FUNCTION 'RSNDI_SHIE_STRUCTURE_GET'
    EXPORTING
      i_s_hiesel        = g_s_hiesel
      i_no_nodenm_table = rs_c_true
    IMPORTING
      e_s_hiedir        = g_s_hiedir
      e_subrc           = g_subrc
    TABLES
      e_t_hiedirt       = g_t_hiedirt
      e_t_hierstruc     = g_t_hierstruc
      e_t_thiernode     = g_t_thiernode
      e_t_hierintvl     = g_t_hierintvl
      e_t_message       = g_t_message.
IF g_subrc <> 0.
    READ TABLE g_t_message INTO g_s_message INDEX 1.
    IF sy-subrc = 0.
      MESSAGE ID g_s_message-msgid TYPE 'I' NUMBER g_s_message-msgno
      WITH g_s_message-msgv1 g_s_message-msgv2
      g_s_message-msgv3 g_s_message-msgv4.
    ELSE.
      MESSAGE ID 'RSBO' TYPE 'I' NUMBER 899
      WITH 'Hierarchy read error'.
    ENDIF.
    EXIT.
ENDIF.
Defined output structures
IF p_dates IS INITIAL AND p_inter IS INITIAL.
    g_struct_s = 'Y_S_HIERFILE_1'.
    g_struct_t = 'Y_T_HIERFILE_1'.
ELSEIF p_dates = 'X' AND p_inter IS INITIAL.
    g_struct_s = 'Y_S_HIERFILE_2'.
    g_struct_t = 'Y_T_HIERFILE_2'.
ELSEIF p_dates IS INITIAL AND p_inter = 'X'.
    g_struct_s = 'Y_S_HIERFILE_3'.
    g_struct_t = 'Y_T_HIERFILE_3'.
ELSE.
    g_struct_s = 'Y_S_HIERFILE_4'.
    g_struct_t = 'Y_T_HIERFILE_4'.
ENDIF.
CREATE DATA gr_s_file TYPE (g_struct_s).
ASSIGN gr_s_file->* TO <g_s_file>.
CREATE DATA gr_t_file TYPE (g_struct_t).
ASSIGN gr_t_file->* TO <g_t_file>.
Nodes
REFRESH <g_t_file>.
LOOP AT g_t_hierstruc INTO g_s_hierstruc.
    CLEAR <g_s_file>.
    MOVE-CORRESPONDING g_s_hierstruc TO <g_s_file>.
Texts for nodes
    READ TABLE g_t_thiernode INTO g_s_thiernode WITH TABLE KEY
    langu = p_langu
    hieid = g_s_hierstruc-hieid
    objvers = rs_c_objvers-active
    nodename = g_s_hierstruc-nodename.
    IF sy-subrc = 0.
      MOVE-CORRESPONDING g_s_thiernode TO <g_s_file>.
    ELSE.
Texts for characteristic values
      REFRESH g_t_chavlinfo.
      CLEAR g_s_chavlinfo.
      g_s_chavlinfo-c_chavl = g_s_hierstruc-nodename.
      APPEND g_s_chavlinfo TO g_t_chavlinfo.
      CALL FUNCTION 'RSD_CHAVL_READ_ALL'
        EXPORTING
          i_iobjnm                  = g_s_hierstruc-iobjnm
          i_langu                   = p_langu
          i_dateto                  = p_dateto
          i_check_value             = space
          i_sid_in                  = space
          i_hieid                   = g_s_hiedir-hieid
          i_objvers                 = g_s_hiedir-objvers
        CHANGING
          c_t_chavlinfo             = g_t_chavlinfo
        EXCEPTIONS
          info_object_not_found     = 1
          routines_generation_error = 2
          check_table_not_existing = 3
          text_table_not_existing   = 4
          OTHERS                    = 5.
      IF sy-subrc = 0.
        READ TABLE g_t_chavlinfo INTO g_s_chavlinfo INDEX 1.
        IF sy-subrc = 0.
          MOVE-CORRESPONDING g_s_chavlinfo-e_chatexts TO <g_s_file>.
          ASSIGN COMPONENT 'LANGU' OF STRUCTURE <g_s_file> TO <g_langu>.
          IF sy-subrc = 0.
            <g_langu> = p_langu.
          ENDIF.
        ENDIF.
      ENDIF.
    ENDIF.
Intervals
    IF g_s_hierstruc-intervl = 'X' AND p_inter = 'X'.
      READ TABLE g_t_hierintvl INTO g_s_hierintvl WITH TABLE KEY
      hieid = g_s_hierstruc-hieid
      objvers = rs_c_objvers-active
      nodeid = g_s_hierstruc-nodeid.
      IF sy-subrc = 0.
        MOVE-CORRESPONDING g_s_hierintvl TO <g_s_file>.
      ENDIF.
    ENDIF.
    APPEND <g_s_file> TO <g_t_file>.
ENDLOOP.
Download output table
g_fname = p_fname.
CALL FUNCTION 'GUI_DOWNLOAD'
    EXPORTING
      filename                = g_fname
      write_field_separator   = space
    TABLES
      data_tab                = <g_t_file>
    EXCEPTIONS
      file_write_error        = 1
      no_batch                = 2
      gui_refuse_filetransfer = 3
      invalid_type            = 4
      no_authority            = 5
      unknown_error           = 6
      header_not_allowed      = 7
      separator_not_allowed   = 8
      filesize_not_allowed    = 9
      header_too_long         = 10
      dp_error_create         = 11
      dp_error_send           = 12
      dp_error_write          = 13
      unknown_dp_error        = 14
      access_denied           = 15
      dp_out_of_memory        = 16
      disk_full               = 17
      dp_timeout              = 18
      file_not_found          = 19
      dataprovider_exception = 20
      control_flush_error     = 21
      OTHERS                  = 22.
IF sy-subrc = 0.
    MESSAGE ID 'RSBO' TYPE 'I' NUMBER 899
    WITH 'Hierarchy download successful!'.
ELSE.
    MESSAGE ID sy-msgid TYPE 'I' NUMBER sy-msgno
    WITH sy-msgv1 sy-msgv2 sy-msgv3 sy-msgv4.
ENDIF.
end

Locking Periods using the Second Hierarchy in the Time Dimension

Hi,
They are not usiing work status for this...
I am at a client, and they used the second hierarchy in the time dimension to lock periods. If they left the second hierarchy blank, no data can be written to it. To open the open the period, they code the Second Heirarchy with CONS_OPEN. Just wondering if anyone has information on this?
It was working fine, then we needed to run a transport to move BPC to a new machine, and it stopped working. Not sure if anyone has done it this way before?
Thanks,
Chandra

Dear Kacper,
the way I understood, there are two separate questions in this post, namely:
I'm using a shared resource in multiple actors, and the code accessing the resource can take quite some time to execute. How can can I ensure this does not mess up the timing of one or all accessors?
I'm using a shared resource in multiple actors, and the code accessing the resource can generate errors if I try to access it from multiple places. How can I ensure synchronization between all accessors?
As for the first question, the optimal solution to separate the resource prone to timing issues in a different thread/actor. Put anything that is not strictly timed (file access, network comunication) in a separate loop. For each actor, you will get queue references to communicate with, so instead of logging the measurements in the actors where they are made, just queue them up for a different loop to process.
For the second part, the same logic still applies. Ideally, every singular resource should be handled by its own thread and nowhere else, so if anything else needs data from/to said resource, it can send a request to the dedicated thread or actor.
If, for some reason, this is not sufficient, you have to handle synchronization in some other way. There are a lot of techniques here, for example:
Create a named semaphore or lock as a part of the class. Have class functions use the semaphore before accessing the resource.
Use actor messages. Have a "resource in use" or "resource released" messages sent to all users whenever obtaining or releasing said resource.
Have a separate actor handle all resources, awarding them to threads needing them. This method also allows setting priorities between requests.
These are just a few examples, there are many other options.
Please let me know if this was helpful.
Kind regards:
Andrew Valko
National Instruments Hungary

One of my friends is getting an error "Can not buy this item. You must buy the application to which this item is intended to buy the item" when they try to make an In app Purchase

More Details: http://feedback.photoshop.com/photoshop_family/topics/cannot_download_camera_pac k_inside_photoshop_app
My friend is trying to purchase an IAP on Adobe Photoshop Express, however they get an error saying "Can not buy this item. You must buy the application to which this item is intended to buy the item" when they try to make an In app Purchase". Everything other app is working fine.
Any thoughts/suggestion to make this error go away?

I suggest trying the update again after an hour or two. Sometimes, the notice that an update is available comes an hour or two before the developers actually sends the revised app out to be downloaded.

Intend hierarchy in the colomn

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