My Oracle DBA have setup a task with following repeat_interval:
Start Date :"30/JAN/20 08:00AM"
Repeat_interval: "FREQ=DAILY; INTERVAL=0; BYMINUTE=15"
Can I ask what is "Interval=0" means?
Does it means this task will run daily from 8AM, and will repeat every 15 mins until success?
I tried to get the answer from Google, but what I find is what is Interval=1, but nothing for 0.
So would be great if anyone can share me some light here.
Thanks in advance!
INTERVAL is the number of increments of the FREQ value between executions. I believe in this case that a value of 0 or 1 would be the same. The schedule as shown would execute once per day (FREQ=DAILY), at approximately 15 minutes past a random hour (BYMINUTE=15, but BYHOUR and BYSECOND are not set).
Schedule has nothing to do with whether or not the previous execution succeeded or not. Start Date is only the date at which the job was enabled, not when it actually starts processing.
If you want it to run every 15 minutes from the moment you enable it, you should set as follows:
FREQ=MINUTELY; INTERVAL=15
If you want it to run exactly on the quarter hour, then this:
FREQ=MINUTELY; BYMINUTE=0,15,30,45; BYSECOND=0
If you want it to run every day at 8am, then this:
FREQ=DAILY; BYHOUR=8; BYMINUTE=0; BYSECOND=0
Related
I am looking at https://github.com/jmettraux/rufus-scheduler which nicely allows me to schedule and chain events.
But I want the events to stop after, say, the 10th occurrence or after the 24 days.
How do I do this?
My case would be:
run a script which creates the recurring jobs based on intervals and then stops after a given date or occurrence.
This is what I have done.
def run_schedule(url, count, method, interval)
puts "running scheduler"
scheduler = Rufus::Scheduler.new
scheduler.every interval do
binding.pry
attack_loop(url, count, method)
end
end
I am testing my site and want the attack_loop to be scheduled in memory to run against the interval.
But it appears it never hits the binding.pry line.
Normally these schedulers are running via cron jobs. Then the problem with your requirement is cron job doesn't know whether you hit the 10th occurrence or the 24 days as it doesnt keep a track. One possible solution would be to create a separate table to update the cron job details.
I'm thinking a table like,
scheduler_details
- id
- occurrence_count
- created_date
- updated_date
_ scheduler_type
So, now when you run a script, you can create or update the details. (You can search the script by scheduler_type, that way
you can check the number of occurrences
with created date, you can calculate the 24 days
HTH
Good day, you can specify the number of times a job should run:
https://github.com/jmettraux/rufus-scheduler/#times--nb-of-times-before-auto-unscheduling
If you need a more elaborate stop condition, you could do:
scheduler.every interval do |job|
if Time.now > some_max_date
job.unschedule
else
attack_loop(url, count, method)
end
end
But it appears it never hits the binding.pry line.
What has it to do with your issue title? Are you mixing issues?
I was wanting to write a program in C that I can simply type in the hours that I worked for each day of the week, including time on break, that will take my input and return the total number of hours I have worked for that week. It's dumb, I know, but I am not sure how to do the math for this regarding time on the clock.
Thank you
At beginning of work: get the current date, make it into seconds.
At end of work: get the current date, make it into seconds.
So working seconds = end seconds - beginning seconds
Then you'll just have to make those into hours.
I have a DBMS_jobs which is scheduled to run a procedure FINDING_PROCEDURE at 6 am evey day. Can anyone tell me how can i change the start time so that it is scheduled to run at 9 am from tomorrow. Thanks in advance.
As I already mentioned in my comment - your job doesn't run at 6 am every day, it runs every 21 hours.
As a second remark, you should seriously consider switching to DBMS_SCHEDULER - it's so much nicer than DBMS_JOB.
Anyway, to let this job run at 9am every day, this should do the trick:
DBMS_JOB.CHANGE (
job => your_job_id,
interval => 'trunc(sysdate) + 1 + 9/24');
you can use DBMS_JOB.CHANGE() to Alter your job schedule.
Click on this link for complete reference from
Oracle Documentation:DBMS_JOB
and find DBMS_JOB.CHANGE()
I have a spring batch application and i am using CRON to set how often this application runs. But the problem i am running into is that i want to run the job on specific hours
3 am
7 am
11 am
3 pm
7 pm
11 pm
As you can see it is every 4 hours but starts at 3 am so i cannot use */4 in the hours section of the timing format as this would start the job at 4am
I have also tried '3,7,11,15,19,23' in the hours section but this does not work either (guessing it only works in the minutes section). Does someone know how i can do this?
Use
#Scedule(cron="0 0 3/4 * * ?")
The Pattern x/y means: where <timepart> mod y = x
or
#Scedule(cron="0 0 3,7,11,15,19,21 * * ?")
According to the Quartz Cron Trigger Tutorial:
The '/' character can be used to specify increments to values. For
example, if you put '0/15' in the Minutes field, it means 'every 15th
minute of the hour, starting at minute zero'. If you used '3/20' in
the Minutes field, it would mean 'every 20th minute of the hour,
starting at minute three' - or in other words it is the same as
specifying '3,23,43' in the Minutes field. Note the subtlety that
"/35" does *not mean "every 35 minutes" - it mean "every 35th minute
of the hour, starting at minute zero" - or in other words the same as
specifying '0,35'.
0 0 3,7,11,15,19,23 * * ?
Fires for 0 minute starting at 3am and ending at 23:00 pm every day.
judging by the two answers above the error i was making was i was keeping the apostrophe at the start and end of my hours... very silly
i managed to solve this by using 3-23/4 for the hour as this starts from 3am and then every other fourth hour (just a different way of doing it to the other answers)
I need to run a vbs script at login that will run a batch script based on a time range of between 22:00 and 06:00 the following day.
I have the current script as follows
If Hour(Now()) >= 20 AND hour(Now()) < 6 Then
//RUN SCRIPT
Else
//RUN OTHER SCRIPT
End If
Now the script runs fine when I use pre noon times e.g 6 and 11 but the about it does not. I can see the issue in that is is not factoring in the following days time and actually going back in time. What I need is the following
if time is 20:00 on day 1 but less than 06:00 on day 2 run the script else run the other script
This needs to run continuously between these times for every day of the week.
Please can you help?
Why not just change AND to OR? In that case when the hour of the day is greater than 20 then it will fire. if the hour of the day is less than 6 it will also fire. Look at it less of the time frame when it needs to fire and more of a time frame excluding those hours when it does not need to fire.
If Hour(Now()) >= 20 OR hour(Now()) < 6 Then
//RUN SCRIPT
Else
//RUN OTHER SCRIPT
End If
VB Script Date Functions