Re: dbms_jobs

Less keystrokes, yes, but the long version is less likely to generate questions like 'What do those numbers mean?' :)

Other jobs running on the database is not really a factor in the timing of my test, as there are no other jobs in the database, and I am the only user of the entire box.

Jared

On Friday 07 September 2001 10:21, MacGregor, Ian A. wrote:
> I find it easier, takes less keystrokes, I am the world's worst typist, to
> SCHEDULE A JOB TO RUN AT 9:45 pm and run every five minutes thereafter
>
> exec dbms_job.submit(:<jobno>,'<procedure>;', trunc(sysdate) + 21.75/24,
> 'trunc(sysdate,''MI'') + 5/1440')
>
> Also bear in mind the job_queue_interval parameter and the number of job
> queue procceses running. There's no guarantee the job will start precisely
> when it is scheduled.
>
> Ian MacGregor
> Stanford Linear Accelerator Center
> ian_at_slac.stanford.edu
>
> -----Original Message-----
> Sent: Thursday, September 06, 2001 11:00 PM
> To: Multiple recipients of list ORACLE-L
>
>
>
> Any thoughts on how this was scheduled this way?
>
> Schedule a job that takes 10 minutes, set the interval
> to run 5 minutes after the first job starts.
>
> Here's the job:
>
> create or replace procedure dummy
> is
> begin
> -- sleep for 10 minutes
> -- envy the computer
> -- waiting for interruption
> dbms_lock.sleep(10*60);
> end;
> /
>
> Here's the submission:
>
> declare
> jobno integer;
> begin
> dbms_job.submit(
> job => jobno
> , what => 'dummy;'
> -- provide resolution to the second
> -- midnight hour of minute of units
> per day -- of current day day to run hr to run ( 1
> second ) , next_date => trunc(sysdate) + ((21 * (60*60) + ( 60*45)) * (
> 1/(60*60*24))) , interval => 'trunc(sysdate) + ((21 * (60*60) + ( 60*50)) *
> ( 1/(60*60*24)))' );
> commit;
> end;
> /
>
> Here's before it ran:
> FAIL
> SCHEMA_USE PRIV_USER LOG_USER JOB LAST_DATE LAST_SEC
> NEXT_DATE NEXT_SEC TOTAL_TIME B INTERVAL URES WHAT
> ---------- ---------- ---------- ------ ------------------- --------
> ------------------- -------- ---------- - -------------------- ----
> --------------------
> JKSTILL JKSTILL JKSTILL 22
> 09/06/2001 21:45:00 21:45:00 244 N trunc(sysdate) + ((2 dummy;
>
> 1 * (60*60) + ( 60*5
>
> 0)) * ( 1/(60*60*24)
>
> ))
>
>
> Here's after it ran:
>
> FAIL
> SCHEMA_USE PRIV_USER LOG_USER JOB LAST_DATE LAST_SEC
> NEXT_DATE NEXT_SEC TOTAL_TIME B INTERVAL URES WHAT
> ---------- ---------- ---------- ------ ------------------- --------
> ------------------- -------- ---------- - -------------------- ----
> --------------------
> JKSTILL JKSTILL JKSTILL 22 09/06/2001 21:45:14 21:45:14
> 09/06/2001 21:57:28 21:57:28 614 N trunc(sysdate) + ((2 1 dummy;
>
> 1 * (60*60) + ( 60*5
>
> 0)) * ( 1/(60*60*24)
>
> ))
>
> Notice the next run time is 00:02:28 after completion of the first job.
>
> Jared
>
> On Thursday 06 September 2001 00:50, nlzanen1_at_EY.NL wrote:
> > Hi
> >
> >
> > As far as I know the job will be rescheduled after the job completes. So
> > in your examples the job will start one hour after the two hour job
> > finishes.
> >
> >
> > Jack
> >
> >
> >
> >
> > David Turner <turner_at_tellme.com>@fatcity.com on 06-09-2001 05:35:32
> >
> > Please respond to ORACLE-L_at_fatcity.com
> >
> > Sent by: root_at_fatcity.com
> >
> >
> > To: Multiple recipients of list ORACLE-L <ORACLE-L_at_fatcity.com>
> > cc: (bcc: Jack van Zanen/nlzanen1/External/MEY/NL)
> >
> > I was wondering if you schedule a job to run every hour and say
> > the job takes 2 hours to run. Will the next run of the job queue up
> > or will it run in parallel with the current job? I'll be testing this
> > but if anyone knows I would appreciate it?
> >
> > Also if the second job waits for the first job to finish how can you see
> > how many jobs have queued up?
> >
> > Thanks, Dave
> > --
> > Please see the official ORACLE-L FAQ: http://www.orafaq.com
> > --
> > Author: David Turner
> > INET: turner_at_tellme.com
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-- 
Please see the official ORACLE-L FAQ: http://www.orafaq.com
-- 
Author: Jared Still
  INET: jkstill_at_cybcon.com

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