Reg: Resolve the dependencies of employment-
Hi Experts,
Need an idea about this scenario-
Every day, I have scheduled tasks, at 3:30, and another expected at 06:00. For simplicity, we called first job as 'X' and 'Y '.
Ideally X gets completed prior to the start of it, and all is well. But, the days where X is not complete from 6:00, I must turn off there and wait times first complete first. Once X is complete, I trigger Y.
So, I need to create a sort of chain of employment, where the behavior can vary every day.
In short:
1 - X Job starts at 03:00, jobs started at 06:00.
2. If employment X is not complete before 06:00, stop work until the end of X. When X is complete, start Y immediately.
Please notify. Entries are appreciated.
Thank you and best regards,
-Nordine
(on Oracle 10.2.0.1.0)
Nordine salvation,
You can do this easily by playing with the procedure TURN.
Here's how I would do:
-Create jobs, X and Y with their respective calendar
-At the beginning of the logical x, disable Y (DBMS_SCHEDULER.disable('Y'));
-At the end of the logical x, check if the time is before 06:00. If so, then simply select Y (DBMS_SCHEDULER.enable('Y')); if not, run Y using DBMS_SCHEDULER. RUN_JOB (it must be enabled).
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