AW: Embedded SQL does not work

[Sharma Deepak <deepak.sharma@xxxxxxxx>]

Please check the Compile time, COMMIT, option. May be helpful .

Deepak
-----Ursprungliche Nachricht-----
Von:    Silvio Santos [SMTP:Silvio.Santos@biw-ag.de]
Gesendet am:    Mittwoch, 13. Januar 1999 13:50
An:     'RPG400'
Betreff:        Embedded SQL does not work

Hello!
I am trying to do an update in a table B with data from table A, for that I 
use a cursor. Here is the code:

     IDSA         DS
     I                                        1  10 FCTNA
     I                                       11  11 CASSA
     I                                       12  15 POSNA
     I                                       16  65 FCDS
     I                                       66  90 FCSD
     C*
     C/EXEC SQL
     C+         DECLARE CA CURSOR FOR
     C+         SELECT A.FCTN0O, A.CASS0O, A.POSN0O, A.FCDS0O, A.FCSD0O
     C+           FROM PFA A, PFB B
     C+           WHERE A.FCTN0O = B.FCTN0O  AND
     C+                 A.CASS0O = B.CASS0O  AND
     C+                 A.POSN0O = B.POSN0O
     C+         FOR READ ONLY
     C/END-EXEC
     C*
     C/EXEC SQL
     C+         OPEN CA
     C/END-EXEC
     C*
     C/EXEC SQL
     C+         WHENEVER NOT FOUND
     C+         GO TO CLOSEA
     C/END-EXEC
     C*
     C           CICLO     TAG
     C/EXEC SQL
     C+         FETCH CA
     C+           INTO  :FCTNA, :CASSA, :POSNA, :FCDS, :FCSD
     C/END-EXEC
     C*
     C           SQLCOD    IFNE 0
     C                     GOTO CLOSEA
     C                     ENDIF
     C*
     C/EXEC SQL
     C+         UPDATE TESTLIB/PFB
     C+           SET FCDS0O = :FCDS,
     C+               FCSD0O = :FCSD
     C+           WHERE FCTN0O = :FCTNA  AND
     C+                 CASS0O = :CASSA  AND
     C+                 POSN0O = :POSNA
     C/END-EXEC
     C                     GOTO CICLO
     C           CLOSEA    TAG
     C/EXEC SQL
     C+         CLOSE CA
     C/END-EXEC
     C*
     C                     SETON                     LR
     C*

But the update is not working. I also tried to run the debug an I saw that 
the sqlcod is with the value = 0 before the update, but after it contains 
-7008,
I also tried to see the sqlerm but it appears strange things.
I tested an equivalent update in the command line of the STRSQL and it 
works, so what can be wrong ?

Silvio.


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