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APEX_PLSQL_JOB.SUBMIT_PROCESS' must be declared

Mohammed Rahim4 days ago

Hi,

I have upgraded Oracle APEX from 4.0 to 22.2 and when I run the application I am getting the error as show below

But the same application works fine in 4.2 version but not in 22.2

Below is the code

DECLARE
l_Sql VARCHAR2(400);
l_Job NUMBER;
BEGIN
IF NVL( :P1_OTHER_FILE_FLAG, 'N' ) = 'Y'
THEN
--
NULL ;
--
ELSE
--
l_Sql := 'BEGIN XXCUST.XXEMP_IFACE_PROCESS.RUN_PROCESS'
||'( ProcessName => ''INITIAL_LOAD'' '
||', FileID => &P1_FILEID. ); END;' ;
:P1_ERROR_MSG := l_Sql;
l_Job := APEX_PLSQL_JOB.SUBMIT_PROCESS( p_sql => l_Sql, p_status => 'Background process submitted' );
--
END IF;

EXCEPTION

WHEN OTHERS
THEN

:P1_ERROR_MSG := 'Failed to run process: '||SQLERRM ;
--
XXCUST.XXEMP_APEX_UTILS.Error_Redirect
( p_CUSTOM_MSG => :P1_ERROR_MSG
, p_GOTO_PAGE => v ('APP_PAGE_ID') ) ;
--
END;

Kindly please provide your support as I am struggling

Thanks

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