Remove all `at` jobs

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I know that to remove a scheduled at job I have to use atrm "numjob1 numjob2", but is there an easy way to do that for all the jobs?

edited Aug 25 '16 at 11:09

Jeff Schaller

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asked Oct 28 '12 at 20:20

robob

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up vote
26
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I know that to remove a scheduled at job I have to use atrm "numjob1 numjob2", but is there an easy way to do that for all the jobs?

edited Aug 25 '16 at 11:09

Jeff Schaller

34.2k951113

asked Oct 28 '12 at 20:20

robob

233135

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up vote
26
down vote

favorite

I know that to remove a scheduled at job I have to use atrm "numjob1 numjob2", but is there an easy way to do that for all the jobs?

edited Aug 25 '16 at 11:09

Jeff Schaller

34.2k951113

asked Oct 28 '12 at 20:20

robob

233135

I know that to remove a scheduled at job I have to use atrm "numjob1 numjob2", but is there an easy way to do that for all the jobs?

command-line at

edited Aug 25 '16 at 11:09

Jeff Schaller

34.2k951113

asked Oct 28 '12 at 20:20

robob

233135

edited Aug 25 '16 at 11:09

Jeff Schaller

34.2k951113

asked Oct 28 '12 at 20:20

robob

233135

edited Aug 25 '16 at 11:09

Jeff Schaller

34.2k951113

edited Aug 25 '16 at 11:09

Jeff Schaller

34.2k951113

edited Aug 25 '16 at 11:09

Jeff Schaller

34.2k951113

asked Oct 28 '12 at 20:20

robob

233135

asked Oct 28 '12 at 20:20

robob

233135

asked Oct 28 '12 at 20:20

robob

233135

add a commentÂ |Â

5 Answers
5

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up vote
27
down vote

accepted

You can run this command to remove all the jobs at the atq

 for i in `atq | awk 'print $1'`;do atrm $i;done

answered Oct 28 '12 at 20:34

pradeepchhetri

6,11583254

1

variation on this answer at -l | awk 'printf "%s ", $1' | xargs atrm
â€“Â Sergiy Kolodyazhnyy
Apr 24 '15 at 10:11

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up vote
10
down vote

You could do something like this:

for i in $(atq | cut -f 1); do atrm $i; done

answered Oct 28 '12 at 20:29

DaveEmme

346125

In FreeBSD it's cut -f3 First column is date
â€“Â David Jashi
Oct 26 '14 at 9:51

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up vote
5
down vote

This seems to me a short line:

atrm $(atq | cut -f1)

edited Jul 28 '14 at 19:50

Networker

5,83993867

answered Jul 28 '14 at 19:19

user78969

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up vote
2
down vote

For more AIX 6 systems you can simply do:

atrm -

Ref: http://pic.dhe.ibm.com/infocenter/aix/v6r1/index.jsp?topic=%2Fcom.ibm.aix.cmds%2Fdoc%2Faixcmds1%2Fatrm.htm

answered Sep 3 '13 at 2:35

Craig

1213

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up vote
0
down vote

Here is my xargs version that avoids braces and is hopefully intuitive:

atq | cut -f 1 | xargs atrm

You can also grep specific jobs by timestamp/userid and then remove them:

atq | grep "2018-10-22 16:" | cut -f 1 | xargs atrm

answered 7 mins ago

Thyag

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
27
down vote

accepted

You can run this command to remove all the jobs at the atq

 for i in `atq | awk 'print $1'`;do atrm $i;done

answered Oct 28 '12 at 20:34

pradeepchhetri

6,11583254

1

variation on this answer at -l | awk 'printf "%s ", $1' | xargs atrm
â€“Â Sergiy Kolodyazhnyy
Apr 24 '15 at 10:11

add a commentÂ |Â

up vote
27
down vote

accepted

You can run this command to remove all the jobs at the atq

 for i in `atq | awk 'print $1'`;do atrm $i;done

answered Oct 28 '12 at 20:34

pradeepchhetri

6,11583254

1

variation on this answer at -l | awk 'printf "%s ", $1' | xargs atrm
â€“Â Sergiy Kolodyazhnyy
Apr 24 '15 at 10:11

add a commentÂ |Â

up vote
27
down vote

accepted

You can run this command to remove all the jobs at the atq

 for i in `atq | awk 'print $1'`;do atrm $i;done

answered Oct 28 '12 at 20:34

pradeepchhetri

6,11583254

You can run this command to remove all the jobs at the atq

 for i in `atq | awk 'print $1'`;do atrm $i;done

answered Oct 28 '12 at 20:34

pradeepchhetri

6,11583254

answered Oct 28 '12 at 20:34

pradeepchhetri

6,11583254

answered Oct 28 '12 at 20:34

pradeepchhetri

6,11583254

answered Oct 28 '12 at 20:34

pradeepchhetri

6,11583254

1

variation on this answer at -l | awk 'printf "%s ", $1' | xargs atrm
â€“Â Sergiy Kolodyazhnyy
Apr 24 '15 at 10:11

add a commentÂ |Â

1

variation on this answer at -l | awk 'printf "%s ", $1' | xargs atrm
â€“Â Sergiy Kolodyazhnyy
Apr 24 '15 at 10:11

variation on this answer at -l | awk 'printf "%s ", $1' | xargs atrm
â€“Â Sergiy Kolodyazhnyy
Apr 24 '15 at 10:11

add a commentÂ |Â

up vote
10
down vote

You could do something like this:

for i in $(atq | cut -f 1); do atrm $i; done

answered Oct 28 '12 at 20:29

DaveEmme

346125

In FreeBSD it's cut -f3 First column is date
â€“Â David Jashi
Oct 26 '14 at 9:51

add a commentÂ |Â

up vote
10
down vote

You could do something like this:

for i in $(atq | cut -f 1); do atrm $i; done

answered Oct 28 '12 at 20:29

DaveEmme

346125

In FreeBSD it's cut -f3 First column is date
â€“Â David Jashi
Oct 26 '14 at 9:51

add a commentÂ |Â

up vote
10
down vote

You could do something like this:

for i in $(atq | cut -f 1); do atrm $i; done

answered Oct 28 '12 at 20:29

DaveEmme

346125

You could do something like this:

for i in $(atq | cut -f 1); do atrm $i; done

answered Oct 28 '12 at 20:29

DaveEmme

346125

answered Oct 28 '12 at 20:29

DaveEmme

346125

answered Oct 28 '12 at 20:29

DaveEmme

346125

answered Oct 28 '12 at 20:29

DaveEmme

346125

In FreeBSD it's cut -f3 First column is date
â€“Â David Jashi
Oct 26 '14 at 9:51

add a commentÂ |Â

In FreeBSD it's cut -f3 First column is date
â€“Â David Jashi
Oct 26 '14 at 9:51

In FreeBSD it's cut -f3 First column is date
â€“Â David Jashi
Oct 26 '14 at 9:51

add a commentÂ |Â

up vote
5
down vote

This seems to me a short line:

atrm $(atq | cut -f1)

edited Jul 28 '14 at 19:50

Networker

5,83993867

answered Jul 28 '14 at 19:19

user78969

5111

add a commentÂ |Â

up vote
5
down vote

This seems to me a short line:

atrm $(atq | cut -f1)

edited Jul 28 '14 at 19:50

Networker

5,83993867

answered Jul 28 '14 at 19:19

user78969

5111

add a commentÂ |Â

up vote
5
down vote

This seems to me a short line:

atrm $(atq | cut -f1)

edited Jul 28 '14 at 19:50

Networker

5,83993867

answered Jul 28 '14 at 19:19

user78969

5111

This seems to me a short line:

atrm $(atq | cut -f1)

edited Jul 28 '14 at 19:50

Networker

5,83993867

answered Jul 28 '14 at 19:19

user78969

5111

edited Jul 28 '14 at 19:50

Networker

5,83993867

edited Jul 28 '14 at 19:50

Networker

5,83993867

edited Jul 28 '14 at 19:50

Networker

5,83993867

answered Jul 28 '14 at 19:19

user78969

5111

answered Jul 28 '14 at 19:19

user78969

5111

answered Jul 28 '14 at 19:19

user78969

5111

add a commentÂ |Â

up vote
2
down vote

For more AIX 6 systems you can simply do:

atrm -

Ref: http://pic.dhe.ibm.com/infocenter/aix/v6r1/index.jsp?topic=%2Fcom.ibm.aix.cmds%2Fdoc%2Faixcmds1%2Fatrm.htm

answered Sep 3 '13 at 2:35

Craig

1213

add a commentÂ |Â

up vote
2
down vote

For more AIX 6 systems you can simply do:

atrm -

Ref: http://pic.dhe.ibm.com/infocenter/aix/v6r1/index.jsp?topic=%2Fcom.ibm.aix.cmds%2Fdoc%2Faixcmds1%2Fatrm.htm

answered Sep 3 '13 at 2:35

Craig

1213

add a commentÂ |Â

up vote
2
down vote

For more AIX 6 systems you can simply do:

atrm -

Ref: http://pic.dhe.ibm.com/infocenter/aix/v6r1/index.jsp?topic=%2Fcom.ibm.aix.cmds%2Fdoc%2Faixcmds1%2Fatrm.htm

answered Sep 3 '13 at 2:35

Craig

1213

For more AIX 6 systems you can simply do:

atrm -

Ref: http://pic.dhe.ibm.com/infocenter/aix/v6r1/index.jsp?topic=%2Fcom.ibm.aix.cmds%2Fdoc%2Faixcmds1%2Fatrm.htm

answered Sep 3 '13 at 2:35

Craig

1213

answered Sep 3 '13 at 2:35

Craig

1213

answered Sep 3 '13 at 2:35

Craig

1213

answered Sep 3 '13 at 2:35

Craig

1213

add a commentÂ |Â

up vote
0
down vote

Here is my xargs version that avoids braces and is hopefully intuitive:

atq | cut -f 1 | xargs atrm

You can also grep specific jobs by timestamp/userid and then remove them:

atq | grep "2018-10-22 16:" | cut -f 1 | xargs atrm

answered 7 mins ago

Thyag

1012

New contributor

add a commentÂ |Â

up vote
0
down vote

Here is my xargs version that avoids braces and is hopefully intuitive:

atq | cut -f 1 | xargs atrm

You can also grep specific jobs by timestamp/userid and then remove them:

atq | grep "2018-10-22 16:" | cut -f 1 | xargs atrm

answered 7 mins ago

Thyag

1012

New contributor

add a commentÂ |Â

up vote
0
down vote

Here is my xargs version that avoids braces and is hopefully intuitive:

atq | cut -f 1 | xargs atrm

You can also grep specific jobs by timestamp/userid and then remove them:

atq | grep "2018-10-22 16:" | cut -f 1 | xargs atrm

answered 7 mins ago

Thyag

1012

New contributor

Here is my xargs version that avoids braces and is hopefully intuitive:

atq | cut -f 1 | xargs atrm

You can also grep specific jobs by timestamp/userid and then remove them:

atq | grep "2018-10-22 16:" | cut -f 1 | xargs atrm

answered 7 mins ago

Thyag

1012

New contributor

answered 7 mins ago

Thyag

1012

New contributor

answered 7 mins ago

Thyag

1012

answered 7 mins ago

Thyag

1012

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