Error faced while creating Oracle Job

Question

Hi,
I have the following script to create a job.
It executed successfully on my database. When we tried to execute the same at our clients place we got some error,
===========================================
SQL> DECLARE
2 job NUMBER;
3 m_sleeptimegap NUMBER;
4 job_interval VARCHAR2(100);
5
6
7 BEGIN
8 SELECT to_number(tdvalue)
9 INTO m_sleeptimegap
10 FROM tabledetails
11 WHERE tdidcode = 'LICENSE_PULSE'
12 AND tdkey = 'STATUS_DELAY';
13
14 job_interval := 'SYSDATE +'|| m_sleeptimegap / 86400;
15
16 DBMS_JOB.SUBMIT(job, 'License_Pulse_Check;', SYSDATE, job_interval, 'FALSE');
17 UPDATE tabledetails
18 SET tdvalue = LTRIM( RTRIM( TO_CHAR(job) ) )
19 WHERE tdidcode = 'LICENSE_PULSE'
20 AND tdkey = 'ORACLE-JOB-ID';
21 COMMIT;
22 END;
23 /
DECLARE
ERROR at line 1:
ORA-23319: parameter value "SYSDATE
+,000347222222222222222222222222222222222222" is not appropriate
ORA-06512: at "SYS.DBMS_JOB", line 60
ORA-06512: at "SYS.DBMS_JOB", line 138
ORA-06512: at line 16
===========================================
'm_sleeptimegap' is = 30 , which we have configured in a table called "tabledetails".
This is present at the client's place.
What could be the reason of this script failure ?
Could it be grants issue or something else ?
Thanks.

Answer 1

Hi,
But that parameter in the dbms_job.submit procedure is of VARCHAR2 type,
dbms_job.submit(
JOB OUT BINARY_INTEGER,
WHAT IN VARCHAR2,
NEXT_DATE IN DATE DEFAULT SYSDATE,
INTERVAL IN VARCHAR2 DEFAULT 'NULL',
NO_PARSE IN BOOLEAN DEFAULT FALSE,
INSTANCE IN BINARY_INTEGER DEFAULT 0,
FORCE IN BOOLEAN DEFAULT FALSE);
[http://www.psoug.org/reference/dbms_job.html?PHPSESSID=5a94d055d8b65bd4b4661d1d24dedbf8]
Thanks.