Getting totals in a CFC

I am trying to use two SQL calls and set a couple variables
to gather the
balance of projects left for a member.
I first find out how many projects they have total for their
account.
(available)
Then I use the recordcount to find out how many projects they
current have
active. (used)
Then I subtract the active projects from the total and get
the
'projectsleft'
What did I do wrong in this component and how can I output
the final number
(projectsleft)
I want to create a component on another page and use the
'projectsleft'.
<cfcomponent>
<cffunction name="Balance" access="remote"
returntype="numeric">
<cfquery name="UsedProjects"
datasource="#Request.MainDSN#">
Select title
FROM projects
Where dirname = '#SESSION.MM_username#'
</cfquery>
<cfset used = #UsedProjects.RecordSet#>
<cfquery name="AvailProjects"
datasource="#Request.MainDSN#">
Select projects
FROM Members
Where username = '#SESSION.MM_username#'
</cfquery>
<cfset available = #AvailProjects.project#>
<cfset projectsleft = available - used>
<cfreturn projectsleft>
</cffunction>
</cfcomponent>
Thanks!
Wally Kolcz
Developer / Support

I appreciate your submission to my question, but I am asking
cause I don't
know. I apologize for my lack of knowledge. If I knew how to
do it
efficiently I would not be asking.
Also, I know the item is wrong. That is why I am asking how
make it right.
"Dan Bracuk" <[email protected]> wrote in
message
news:e9o4fe$aqf$[email protected]..
> What did you do wrong? This
> <cfset used = #UsedProjects.RecordSet#>
> does not give a recordcount.
>
> and this
> <cfset available = #AvailProjects.project#>
> does not give you a number.
>
> Plus if I remember correctly, cfc's do not recognize
session variables.
>
> That's the stuff that is definitely wrong. You are also
being
> inefficient.
> You should be able to get the answer in one query.
>

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          i_datum_von                   = p_lv_date2
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    Regards,
    Sunil kairam.

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    user11100190 wrote:
    Hi,
    Thanks for suggesting a soultion, it works well.
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    Since the enqueueing and dequeueing may be interleaved from multiple threads, it may be possible that the partial total count returned in a result does not correspond to the data in the partial result, so you should not base anything on that assumption.
    Once again, that approach may leak memory based on how Coherence is internally implemented, so I can't recommend this approach but it may work.
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    Best regards,
    Robert

  • HOW TO GET TOTAL BEFORE DISCOUNT AMOUNT IN MARKETING DOCUMNET

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  • How to get total of any field in sapscript?

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    LIKE:
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  • Query to get total counts in the given scenario

    Hi,
    I am using Oracle 10g
    I have a table A with following data
    AgId     Trm     CD     S
    1000     100010     12-JAN     A
    1000     100019     20-MAR     A
    1000     100019     20-JUL     D
    1001     100011     25-JAN     A
    1001     100011     20-FEB     D
    1001     100011     23-MAR     A
    1001     100012     31-JAN     A
    1002     100013     14-FEB     A
    1002     100013     05-APR     D
    1002     100015     02-MAY     A
    1003     100014     03-MAR     A
    1003     100014     25-MAR     D
    1004     100016     22-MAY     A
    1004     100017     21-JUN     A
    1004     100018     01-JUL     A
    1005     100020     21-MAY     D
    1005     100020     21-JUL     A
    1005     100020     11-AUG     D
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    status of AgId '1002' is A
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    Hi,
    when you put some code please enclose it between two lines starting with {noformat}{noformat}
    i.e.:
    {noformat}{noformat}
    CREATE TABLE ... etc.
    {noformat}{noformat}
    Also do not forget to put correct statements. Your insert were missing semicolon at the end.
    Coming back to your problem, I assum that the current status of any trm is the latest status so I have done what follows:
    {code:sql}
    SELECT agid
         , trm
         , cd
         , s
         , ROW_NUMBER () OVER (PARTITION BY agid, trm ORDER BY cd) rn
         , COUNT (*) OVER (PARTITION BY agid, trm) rn_tot
    FROM a
    ORDER BY agid, trm, cd;
    Output:
          AGID        TRM CD        S         RN     RN_TOT
          1000     100010 12-JAN-12 A          1          1
          1000     100019 20-MAR-12 A          1          2
          1000     100019 20-JUL-12 D          2          2
          1001     100011 25-JAN-12 A          1          3
          1001     100011 20-FEB-12 D          2          3
          1001     100011 23-MAR-12 A          3          3
          1001     100012 31-JAN-12 A          1          1
          1002     100013 14-FEB-12 A          1          2
          1002     100013 05-APR-12 D          2          2
          1002     100015 02-MAY-12 A          1          1
          1003     100014 03-MAR-12 A          1          2
          1003     100014 25-MAR-12 D          2          2
          1004     100016 22-MAY-12 A          1          1
          1004     100017 21-JUN-12 A          1          1
          1004     100018 01-JUL-12 A          1          1
          1005     100020 21-MAY-12 D          1          3
          1005     100020 21-JUL-12 A          2          3
          1005     100020 11-AUG-12 D          3          3I have used to columns rn and rn_tot to find the latest entry for a specific agid, trm
    Using the query above I have done the following:
    WITH mydata AS
       SELECT agid
            , trm
            , cd
            , s
            , ROW_NUMBER () OVER (PARTITION BY agid, trm ORDER BY cd) rn
            , COUNT (*) OVER (PARTITION BY agid, trm) rn_tot
        FROM a
    SELECT agid
         , CASE WHEN SUM (CASE s WHEN 'A' THEN 1 END) > 0 THEN 'A' ELSE 'D' END s
      FROM mydata
    WHERE rn = rn_tot
    GROUP BY agid;
    Output:
          AGID S
          1000 A
          1001 A
          1002 A
          1003 D
          1004 A
          1005 D
    {code}
    Which is listing the agid and its status if at least there is one trm with final status 'A' for that agid.
    To sum up everything you can do simply like this:
    {code:sql}
    WITH mydata AS
       SELECT agid
            , trm
            , cd
            , s
            , ROW_NUMBER () OVER (PARTITION BY agid, trm ORDER BY cd) rn
            , COUNT (*) OVER (PARTITION BY agid, trm) rn_tot
        FROM a
    totdata AS
       SELECT agid
            , CASE WHEN SUM (CASE s WHEN 'A' THEN 1 END) > 0 THEN 'A' ELSE 'D' END s
         FROM mydata
        WHERE rn = rn_tot
        GROUP BY agid
    SELECT S, COUNT(*) cnt
      FROM totdata
    GROUP BY S;
    Output:
    S        CNT
    D          2
    A          4
    {code}
    Regards.
    Al                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      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    Does anybody know how to get total at each level.
    I mean, if I have the following structure...
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    Level 3
    Level 4
    Level 5
    At level 5 we have the following calculation
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    6                     3                 12,00
    7                     4                  28,00
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    My question is how to get a sum of (10,00 + 12,00 +28,00 = 50,00) at level 4?
    Thanks.
    With regards,
    Anand

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