BPM: How to return the total number of lines in Multiline Container Element

Hi all:
I am trying to follow the following blog to do message splitting:
/people/narendra.jain/blog/2005/12/30/various-multi-mappings-and-optimizing-their-implementation-in-integration-processes-bpm-in-xi
In the blog, total number of messages for Message1 was returned to Count in Message 2.
If I do not want to create message 2, if there any way to return total number lines in multiline elements ? Because this number need to be used to specify in loop condition:
While Indxe != TotalNumberOfLines
Thanks
Liang

Hi, Matias:
For the loop block option, your answer is passing message count in different message type is best option !
I really do not know why SAP does not deliver a build-in function to return the number of items in multiline element at run time ?
Now the question for the second option: ForEach block:
When I leave current line of block properties empty, the compile does not pass
I have to input a another element (Var_Cur) in current line
Inside my block, I have one send step, it will send Var_Dest[Index], followed by a container operation step to increment Index by 1.
With above setup, I compile it, compile passed, but it says:
"Container Element Var_Cur in Step Block1 is initialized but not used"
then I added another containter operation step in block:
Var_Cur = Var_Dest[Index]
I modified send step to send Var_Cur rather than Var_Dest.
Up to now, I can compile it without any warnings.
But when I send message to IE, the message pass pipe line, but it seems does not pass the BPE.
When I go to SXI_CACHE, to check the return code for my Integration process is 0, means no error.
But my message does not reach my target system.
Where do you think I made wrong ?
Thanks
Liang

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How to find the total number of pair under particular parent according to pattern 1:2or2:1,1:1 day to day maximum up to 5 pair caping daily

Dear friends,
I provide full details here ,my database table structure ,data and stored procedure and my problem ,
please review it and provide the solution or any idea to solve my problem.
I am working on a project in which members are added in a tree pattern, and get the payment accordingly.
below is my table structure ,data and stored procedure
CREATE TABLE Associate_Income
ID varchar(30) NOT NULL,
ParentID varchar(30) NULL,
IsLeft tinyint NULL,
IsRight tinyint NULL,
joingdate datetime NOT NULL
go
INSERT Associate_Income
(ID, ParentID, IsLeft, IsRight, joingdate)
SELECT 'Ramesh123', NULL, NULL, NULL '2014-01-03 16:31:15.000' UNION ALL
SELECT 'Sonu', 'Ramesh123', 1, NULL, '2014-01-03 16:45:21.000' UNION ALL
SELECT 'Pawan kumar', 'Ramesh123', NULL, 1, '2014-01-04 16:50:23.000' UNION ALL
SELECT 'Ravi123', 'Sonu', 1, NULL, '2014-01-04 17:03:22.000' UNION ALL
SELECT 'Vineet123', 'Sonu', NULL, 1, '2014-01-04 17:26:01.000' UNION ALL
SELECT 'dev123', 'Ravi123', 1, NULL, '2014-01-05 19:35:16.000' UNION ALL
SELECT 'Mukesh123', 'Ravi123', NULL, 1, '2014-01-05 19:40:41.000' UNION ALL
SELECT 'poonam123', 'Vineet123', 1, NULL, '2014-01-05 19:49:49.000' UNION ALL
SELECT 'monu', 'Pawan kumar', 1, NULL, '2014-01-05 17:32:58.000' UNION ALL
SELECT 'Arti123', 'Pawan kumar', NULL, 1, '2014-01-05 19:54:35.000' UNION ALL
My database table Associate_Income structure and data is as follow:
ID ParentID IsLeft IsRight joingdate
Ramesh123 NULL NULL NULL 2014-01-03 16:31:15.000
Sonu Ramesh123 1 NULL 2014-01-03 16:45:21.000
Pawan kumar Ramesh123 NULL 1 2014-01-04 16:50:23.000
Ravi123 Sonu 1 NULL 2014-01-04 17:03:22.000
Vineet123 Sonu NULL 1 2014-01-04 17:26:01.000
dev123 Ravi123 1 NULL 2014-01-05 19:35:16.000
Mukesh123 Ravi123 NULL 1 2014-01-05 19:40:41.000
poonam123 Vineet123 1 NULL 2014-01-05 19:49:49.000
monu Pawan kumar 1 NULL 2014-01-05 17:32:58.000
Arti123 Pawan kumar NULL 1 2014-01-05 19:54:35.000
by using below stored procedure i can count the total number of pairs under particular node in 2:1,1:1 ratio means first pair is completed when two node added to the left side of given parent node and one node added
right side of given parent node after that all pairs are completed when one node added left side and one node added right side of parent node (1:1 ratio)
example if i execute my stored procedure as follows it would return following.
EXEC count_pairs 'Ramesh123'
3
so there is 3 pairs as shown in my figure.
when we execute my stored procedure for ParentID 'sonu' it would return following.
EXEC count_pairs 'sonu'
2
so there is 2 pairs as shown in my figure.
My problem is to find the query which can return the total number of pair under particular node any given parent node. day to
day maximum 5 pairs in a day please any one can suggest us
CREATE proc [dbo].[count_pairs]
@ParentID nvarchar(50)
as
begin
Declare @ParentSUM SMALLINT = 0
Declare @SubLeftID nvarchar(50)
Declare @SubRightID nvarchar(50)
SELECT @SubLeftID = CASE WHEN [IsLeft] = 1 THEN [ID] ELSE @SubLeftID END
,@SubRightID = CASE WHEN [IsRight] = 1 THEN [ID] ELSE @SubRightID END
FROM Associate_Income
WHERE ParentID = @ParentID
IF @SubLeftID IS NOT NULL AND @SubRightID IS NOT NULL AND EXISTS(SELECT 1 FROM Associate_Income WHERE [IsLeft] = 1 AND ParentID = @SubLeftID)
BEGIN
SET @ParentSUM = 1
;WITH Associate_Income_CTE AS
SELECT [ID], [ParentID], [IsLeft], [IsRight], 0 AS [Level]
FROM Associate_Income
WHERE [ParentID] = @ParentID
UNION ALL
SELECT RecursiveMember.[ID], RecursiveMember.[ParentID], RecursiveMember.[IsLeft], RecursiveMember.[IsRight], Level + 1
FROM Associate_Income RecursiveMember
INNER JOIN Associate_Income_CTE AnchorMember
ON RecursiveMember.[ParentID] = AnchorMember.[ID]
SELECT @ParentSUM = @ParentSUM + COUNT([ParentID])
FROM
SELECT [ParentID]
,'IsLeft' AS [Direction]
,1 AS [Value]
FROM Associate_Income
WHERE [IsLeft] = 1
AND [ID] <> @ParentID --AND [ID] NOT IN (@SubLeftID, @ParentID)
AND [ParentID] NOT IN (@ParentID, @SubLeftID)
UNION ALL
SELECT [ParentID]
,'IsRight' AS [Direction]
,1 AS [Value]
FROM Associate_Income
WHERE [IsRight] = 1
AND [ParentID] <> @ParentID
) AS Associate_Income
PIVOT
MAX([Value]) FOR [Direction] IN ([IsLeft], [IsRight])
) PVT
WHERE [IsLeft] IS NOT NULL AND [IsRight] IS NOT NULL
END
SELECT @ParentSUM
Jitendra Kumar Sr. Software Developer at Ruvixo Technologies 7895253402

I don't think this is homework, I am not sure how helpful it was by Kalman to merge the two threads. It appears that two different persons posted the questions. It could be though, that it is the same problem and Jitendra has taken over Chandra's
task.
However, I was not able to understand the problem nor the figure. And nor the definition of pairs in this context. My assumption is that what Jitendra posted is an abstraction of the actual problem in order to not reveal intellecutal property. Possibly this
makes the problem more difficult to understand for us outsiders.
I've come so far that I worked out a table definition and INSERT statements with sample data, as well as a call to the procedure which returns 3 and this appears to map to the figure, but I don't know what it means. Since I don't know what this
is about, I have not made any attepmts to understand the code, but I would appreciate clarification about the underlying business rules as well as the expected results for various test cases.
CREATE TABLE Associate_Income(ID varchar(30) NOT NULL,
ParentID varchar(30) NULL,
IsLeft tinyint NULL,
IsRight tinyint NULL,
joingdate datetime NOT NULL)
go
INSERT Associate_Income (ID, ParentID, IsLeft, IsRight, joingdate)
SELECT 'Ramesh123', NULL, NULL, NULL, '2014-01-03 16:31:15.000' UNION ALL
SELECT 'Sonu', 'Ramesh123', 1, NULL, '2014-01-03 16:45:21.000' UNION ALL
SELECT 'Pawan kumar', 'Ramesh123', NULL, 1, '2014-01-04 16:50:23.000' UNION ALL
SELECT 'Ravi123', 'Sonu', 1, NULL, '2014-01-04 17:03:22.000' UNION ALL
SELECT 'Vineet123', 'Sonu', NULL, 1, '2014-01-04 17:26:01.000' UNION ALL
SELECT 'dev123', 'Ravi123', 1, NULL, '2014-01-05 19:35:16.000' UNION ALL
SELECT 'Mukesh123', 'Ravi123', NULL, 1, '2014-01-05 19:40:41.000' UNION ALL
SELECT 'poonam123', 'Vineet123', 1, NULL, '2014-01-05 19:49:49.000' UNION ALL
SELECT 'monu', 'Pawan kumar', 1, NULL, '2014-01-05 17:32:58.000' UNION ALL
SELECT 'Arti123', 'Pawan kumar', NULL, 1, '2014-01-05 19:54:35.000'
go
CREATE proc [dbo].[count_pairs]
@ParentID nvarchar(50)
as
begin
Declare @ParentSUM SMALLINT = 0
Declare @SubLeftID nvarchar(50)
Declare @SubRightID nvarchar(50)
SELECT @SubLeftID = CASE WHEN [IsLeft] = 1 THEN [ID] ELSE @SubLeftID END
,@SubRightID = CASE WHEN [IsRight] = 1 THEN [ID] ELSE @SubRightID END
FROM Associate_Income
WHERE ParentID = @ParentID
IF @SubLeftID IS NOT NULL AND @SubRightID IS NOT NULL AND EXISTS(SELECT 1 FROM Associate_Income WHERE [IsLeft] = 1 AND ParentID = @SubLeftID)
BEGIN
SET @ParentSUM = 1
;WITH Associate_Income_CTE AS
SELECT [ID], [ParentID], [IsLeft], [IsRight], 0 AS [Level]
FROM Associate_Income
WHERE [ParentID] = @ParentID
UNION ALL
SELECT RecursiveMember.[ID], RecursiveMember.[ParentID], RecursiveMember.[IsLeft], RecursiveMember.[IsRight], Level + 1
FROM Associate_Income RecursiveMember
INNER JOIN Associate_Income_CTE AnchorMember
ON RecursiveMember.[ParentID] = AnchorMember.[ID]
SELECT @ParentSUM = @ParentSUM + COUNT([ParentID])
FROM
SELECT [ParentID]
,'IsLeft' AS [Direction]
,1 AS [Value]
FROM Associate_Income
WHERE [IsLeft] = 1
AND [ID] <> @ParentID --AND [ID] NOT IN (@SubLeftID, @ParentID)
AND [ParentID] NOT IN (@ParentID, @SubLeftID)
UNION ALL
SELECT [ParentID]
,'IsRight' AS [Direction]
,1 AS [Value]
FROM Associate_Income
WHERE [IsRight] = 1
AND [ParentID] <> @ParentID
) AS Associate_Income
PIVOT
MAX([Value]) FOR [Direction] IN ([IsLeft], [IsRight])
) PVT
WHERE [IsLeft] IS NOT NULL AND [IsRight] IS NOT NULL
END
SELECT @ParentSUM
END
go
EXEC count_pairs 'Ramesh123'
go
DROP PROCEDURE count_pairs
DROP TABLE Associate_Income
Erland Sommarskog, SQL Server MVP, [email protected]

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How to get the total number of occurrences based on the value of a column.

Hello everyone,
This is the first time that I will ask question here on your forum but has been following several threads ever since. I guess that now is my turn to ask a question. So anyway here is the thing, I have a query that should return count the number of rows depending on the value of SLOT. Something like this:
WIPDATAVALUE          SLOT             N            M
1-2                   TRALTEST43S1     1            3
1-2                   TRALTEST43S1     2            3
3                     TRALTEST43S1     3            3
4-6                   TRALTEST43S2     1            4
4-6                   TRALTEST43S2     2            4
4-6                   TRALTEST43S2     3            4
7                     TRALTEST43S2     4            4-----
As you can see above, on the SLOT TRALTEST43S1, there are three occurrences so M (Total number of occurrences) should be three and that column N should count it. Same goes with the SLOT TRALTEST43S2. This is the query that I have so far:
SELECT DISTINCT
WIPDATAVALUE, SLOT
, LEVEL AS n
, m
FROM
SELECT
    WIPDATAVALUE
    , SLOT
    , (dulo - una) + 1 AS m
FROM
    SELECT
      WIPDATAVALUE
      , SLOT
      , CASE WHEN INSTR(wipdatavalue, '-') = 0 THEN wipdatavalue ELSE SUBSTR(wipdatavalue, 1, INSTR(wipdatavalue, '-')-1) END AS una
      , CASE WHEN INSTR(wipdatavalue, '-') = 0 THEN wipdatavalue ELSE SUBSTR(wipdatavalue, INSTR(wipdatavalue, '-') + 1) END AS dulo
    FROM trprinting
    WHERE (containername = :lotID OR SLOT= :lotID) AND WIPDATAVALUE LIKE :wip
) CONNECT BY LEVEL <= m
ORDER BY wipdatavalue;And that it results to something like this:
WIPDATAVALUE          SLOT             N            M
1-2                   TRALTEST43S1     1            2
1-2                   TRALTEST43S1     2            2
3                     TRALTEST43S1     1            1
4-6                   TRALTEST43S2     1            3
4-6                   TRALTEST43S2     2            3
4-6                   TRALTEST43S2     3            3
7                     TRALTEST43S2     1            1-----
I think that my current query is basing its M and N results on WIPDATAVALUE and not the SLOT that is why I get the wrong output. I have also tried to use the WITH Statement and it works well but unfortunately, our system cant accept subquery factoring.
I know you guys will be helping out because you are all awesome. Thanks everyone
Edited by: 1001275 on Apr 19, 2013 8:07 PM
Edited by: 1001275 on Apr 19, 2013 8:18 PM

Hi,
Sorry, it's still not clear what you want.
Are you saying that, given this table:
CREATE TABLE trprinting
WIPDATAVALUE       VARCHAR2(255)
, SLOT               VARCHAR2(255)
INSERT INTO trprinting (wipdatavalue, slot) VALUES ('1-2', 'TRALTEST43S1');
INSERT INTO trprinting (wipdatavalue, slot) VALUES ('3',    'TRALTEST43S1');
INSERT INTO trprinting (wipdatavalue, slot) VALUES ('4-6', 'TRALTEST43S2');
INSERT INTO trprinting (wipdatavalue, slot) VALUES ('7',    'TRALTEST43S2');you want to produce this output:
WIPDATAVALUE SLOT                     N          M
1-2          TRALTEST43S1             1          3
1-2          TRALTEST43S1             2          3
3            TRALTEST43S1             3          3
4-6          TRALTEST43S2             1          4
4-6          TRALTEST43S2             2          4
4-6          TRALTEST43S2             3          4
7            TRALTEST43S2             4          4? If so, here's one way:
WITH     got_numbers     AS
     SELECT     wipdatavalue
     ,     slot
     ,     TO_NUMBER ( SUBSTR ( wipdatavalue
                           , 1
                       , INSTR ( wipdatavalue || '-'
                            ) - 1
                 )          AS low_number
     ,     TO_NUMBER ( SUBSTR ( wipdatavalue
                           , 1 + INSTR ( wipdatavalue
                 )          AS high_number
     FROM     trprinting
SELECT       wipdatavalue
,       slot
,       ROW_NUMBER () OVER ( PARTITION BY slot
                             ORDER BY          low_number
                    )                    AS n
,       COUNT (*)     OVER ( PARTITION BY slot )     AS m
FROM       got_numbers
CONNECT BY     LEVEL               <= high_number + 1 - low_number
     AND     low_number          = PRIOR low_number
     AND     PRIOR SYS_GUID ()      IS NOT NULL
ORDER BY low_number
,            n
;Much of the complexity here is caused by storing 2 numbers in 1 VARCHAR2 column, wipdatavalue. Relational databases work best when there is no more than 1 item in any given column of any given row. This is so basic to datbase design that it is called First Normal Form. Also, numbers belong in NUMBER columns, not VARCHAR2. If you stored your data like that in the fist place, then you wouldn't need the sub-query I called got_numbers, which is about 60% of the code above. (That could be reduced by replacing SUSTR and INSTR with the less efficient REGEGP_SUBSTR.)

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L   U                                  0 UL            0.00 USD       1 EA
00002                    Chief Inspector                          R3VNI
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How can I see the total number of books (size of library) on iBooks on my iMac and IPad Mini?

I have installed iBooks on both my iMac (main library) as well as on my iPad Mini Retina. I can't seem to find where to see the total number of books in each library? Can anyone enlighten me please? I can see the covers, lists etc. but nowhere how many books I have total in each library (Total: X number of books stored) or something like that. Can someone point me where this information is viewable?

Did you have 'find my iPhone' activated on it? Because if you go to iCloud.com you can log in there and you can remotely lock or wipe your device. From what I hear, the police won't do anything about going and fetching stolen iPads and iPhones, but if you have find my iphone/ipad/ipod/mac turned on then it will give you a map of where it is.
Just keep in mind that sometimes iTunes does ask you for your password regardless, and only a few weeks ago apple brought in security questions, so everyone had to fill in security questions for their iTunes account to make it more secure If all else fails, ring Apple cusomer support and they might be able to bar your account or something, however you may have to make a new account (the might be able to set it up so you have all your previous information on a new account)
Hope this helps!

How to know the total page number of the standard report

How to know the total page number of the standard report?
If I insert a graph, a table, or text into the end of the standard report, how to know how much space is left for the current page?

Hello Net,
Unfortunately, we do not have any VIs in LabVIEW that can figure out the number of pages of a standard report. However, with the use of ActiveX calls you can achieve this task. Here are a couple of discussion forums which discuss this:
Excel Page Numbering with ActiveX:
http://forums.ni.com/ni/board/message?board.id=170&message.id=194479&requireLogin=False
Number Pages in Word Report:
http://forums.ni.com/ni/board/message?board.id=170&message.id=162972&requireLogin=False
There is also great documentation on using ActiveX on msdn. You can search this document for page number objects. Here is a link to this:
http://msdn2.microsoft.com/en-us/library/aa223048(office.11).aspx
Please refer to these links. If you still have questions please feel free to let us know so we can assist you.
Thank you
Best Regards,
Dominic L.

BPM: How to return the total number of lines in Multiline Container Element

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