Remove all 'at' Jobs except first five jobs

Question

I want to keep only first five scheduled jobs (as in the lowest 5 job ID numbers) and remove the rest of the scheduled atq jobs. How to can I do this?

what are you calling 'first' ? next to be run ? oldest/latest to have been submitted ? top 5 in at -l | head -5 (which is more or less random) — Archemar, Apr 18 '15 at 06:58
at command schedules job by job id. I want to run only first 5 schedule job Ids and remove others. — KK Patel, Apr 18 '15 at 07:39

score 0 · Answer 1 · answered Apr 18 '15 at 12:30

On my Debian system, at sorts jobs by the time they are scheduled to start and not the order they were given to at in:

$ for i in 10 20 30 40 50 60 70; do 
    at now + "$i" min < scripts/foo.sh; sleep 1; 
done
warning: commands will be executed using /bin/sh
job 8 at Sat Apr 18 15:31:00 2015
warning: commands will be executed using /bin/sh
job 9 at Sat Apr 18 15:41:00 2015
warning: commands will be executed using /bin/sh
job 10 at Sat Apr 18 15:51:00 2015
warning: commands will be executed using /bin/sh
job 11 at Sat Apr 18 16:01:00 2015
warning: commands will be executed using /bin/sh
job 12 at Sat Apr 18 16:12:00 2015
warning: commands will be executed using /bin/sh
job 13 at Sat Apr 18 16:22:00 2015
warning: commands will be executed using /bin/sh
job 14 at Sat Apr 18 16:32:00 2015

$ atq
9   Sat Apr 18 15:41:00 2015 a terdon
11  Sat Apr 18 16:01:00 2015 a terdon
10  Sat Apr 18 15:51:00 2015 a terdon
12  Sat Apr 18 16:12:00 2015 a terdon
8   Sat Apr 18 15:31:00 2015 a terdon
14  Sat Apr 18 16:32:00 2015 a terdon
13  Sat Apr 18 16:22:00 2015 a terdon

As you can see, at will number the jobs in the order in which they will be run, but atq lists them in an apparently random order.

To remove the first 5 jobs as listed by atq, you can do:
```
atrm $(atq | head -5 | cut -f 1)
```
To delete the first 5 jobs based on the order they will be launched in, do:
```
atrm $(atq | sort -n | head -5 | cut -f 1)
```

ctrl-alt-delor · Answer 2 · 2015-04-18T17:28:39.340

0

This removes the first 5, so is wrong, if you can findout how to do an inverted head (remove head), then you will have the answer. A combination of wc and tail may do it.

atq | sort -g  | head -5 | cut -f1 | xargs atrm

Correct answer

atq | sort -g  | tail -n +6 | cut -f1 | xargs atrm

edited Apr 18 '15 at 17:28

answered Apr 18 '15 at 12:31

ctrl-alt-delor

27,993

1

tail -n +6 will start at line 6. (Yes, bizarre syntax.) Or, untested, sed -ne '6,$p' – Ulrich Schwarz Apr 18 '15 at 17:24
@UlrichSchwarz cool I experimented and head -n -5 does all but last 5. – ctrl-alt-delor Apr 18 '15 at 17:29
@Richard but that's not what the OP wants either. They want to keep the first five and cancel all the others. – Chris Davies Apr 18 '15 at 22:11
@roaima are you commenting my comment or answer, I think the answer is correct, at least the bit that says it is correct. The comment is just commenting about the previous comment (an aside, may be not too relevent). – ctrl-alt-delor Apr 18 '15 at 22:20
@Richard I was responding to your comment. – Chris Davies Apr 19 '15 at 12:14

Remove all 'at' Jobs except first five jobs

2 Answers2