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Why can't I get pgrep output right to variable on bash script?
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I'm trying to make a script to either quit compton if it's running or start it if it's not running. I've read from man that it should exit 1 if process is found, so I've tried to make a script that uses that... However this just doesn't work, It starts if it's closed but doesn't close it. what am I doing wrong ??
#!/bin/bash
status=$(pgrep compton 2>&1)
if [[ $status == 1 ]];
then
killall compton
else
exec compton -b
fi
echo $status
bash stdout stderr exit-status pgrep
add a comment |
I'm trying to make a script to either quit compton if it's running or start it if it's not running. I've read from man that it should exit 1 if process is found, so I've tried to make a script that uses that... However this just doesn't work, It starts if it's closed but doesn't close it. what am I doing wrong ??
#!/bin/bash
status=$(pgrep compton 2>&1)
if [[ $status == 1 ]];
then
killall compton
else
exec compton -b
fi
echo $status
bash stdout stderr exit-status pgrep
1
0
or1
is exit status, but$(...)
captures output.
– Charles Duffy
Mar 19 at 23:41
grab$?
. doman bash
.
– user2497
Mar 20 at 19:09
add a comment |
I'm trying to make a script to either quit compton if it's running or start it if it's not running. I've read from man that it should exit 1 if process is found, so I've tried to make a script that uses that... However this just doesn't work, It starts if it's closed but doesn't close it. what am I doing wrong ??
#!/bin/bash
status=$(pgrep compton 2>&1)
if [[ $status == 1 ]];
then
killall compton
else
exec compton -b
fi
echo $status
bash stdout stderr exit-status pgrep
I'm trying to make a script to either quit compton if it's running or start it if it's not running. I've read from man that it should exit 1 if process is found, so I've tried to make a script that uses that... However this just doesn't work, It starts if it's closed but doesn't close it. what am I doing wrong ??
#!/bin/bash
status=$(pgrep compton 2>&1)
if [[ $status == 1 ]];
then
killall compton
else
exec compton -b
fi
echo $status
bash stdout stderr exit-status pgrep
bash stdout stderr exit-status pgrep
edited Mar 19 at 23:19
Kusalananda♦
138k17258426
138k17258426
asked Mar 19 at 21:04
TubeTube
161
161
1
0
or1
is exit status, but$(...)
captures output.
– Charles Duffy
Mar 19 at 23:41
grab$?
. doman bash
.
– user2497
Mar 20 at 19:09
add a comment |
1
0
or1
is exit status, but$(...)
captures output.
– Charles Duffy
Mar 19 at 23:41
grab$?
. doman bash
.
– user2497
Mar 20 at 19:09
1
1
0
or 1
is exit status, but $(...)
captures output.– Charles Duffy
Mar 19 at 23:41
0
or 1
is exit status, but $(...)
captures output.– Charles Duffy
Mar 19 at 23:41
grab
$?
. do man bash
.– user2497
Mar 20 at 19:09
grab
$?
. do man bash
.– user2497
Mar 20 at 19:09
add a comment |
2 Answers
2
active
oldest
votes
You are getting the pgrep
output in your status
variable. It's just not the output that you expect it to be.
pgrep
outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton
, then $status
would be the PID of that process, or of those processes. pgrep
also returns an exit status, but an exit status is not captured by a command substitution as a string.
In your test, you compare $status
against 1
. It is unlikely that compton
has PID 1.
If you want to kill any compton
process if they exist, and start compton -b
if no compton
process exists, you may do that with
#!/bin/sh
if ! pkill compton; then
exec compton -b
fi
This uses the exit status of pkill
. The pkill
tool works in an equivalent way to pgrep
(they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep
would do, pkill
sends the TERM
signal (by default) to the matching processes.
The if
keyword uses the exit status of the command that you use with it.
The !
inverts the sense of the test so that
If
pkill compton
succeeds, it means that there was one or severalcompton
processes that have now been killed, or at least signalled, andexec compton -b
will not be executed.If
pkill compton
fails (no process matched the name, or there was some internal error inpkill
), the body of theif
statement would call yourexec compton -b
, which would replace the shell process with the process resulting from runningcompton -b
.
add a comment |
You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"
so
#!/bin/bash
pgrep compton >/dev/null
if [[ $? -eq 0 ]]
then
killall compton
else
exec compton -b
fi
or
#!/bin/bash
status=$(pgrep compton 2>&1)
if [[ -n "$status" ]]
then
killall compton
else
exec compton -b
fi
3
if pgrep compton >/dev/null; then
is the better practice, so one doesn't directly inspect$?
at all. In addition to avoiding bugs where$?
refers to a different command than you think it does when editing code to add logging or such, this also means thatset -e
will no longer treatpgrep
returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").
– Charles Duffy
Mar 19 at 23:41
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are getting the pgrep
output in your status
variable. It's just not the output that you expect it to be.
pgrep
outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton
, then $status
would be the PID of that process, or of those processes. pgrep
also returns an exit status, but an exit status is not captured by a command substitution as a string.
In your test, you compare $status
against 1
. It is unlikely that compton
has PID 1.
If you want to kill any compton
process if they exist, and start compton -b
if no compton
process exists, you may do that with
#!/bin/sh
if ! pkill compton; then
exec compton -b
fi
This uses the exit status of pkill
. The pkill
tool works in an equivalent way to pgrep
(they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep
would do, pkill
sends the TERM
signal (by default) to the matching processes.
The if
keyword uses the exit status of the command that you use with it.
The !
inverts the sense of the test so that
If
pkill compton
succeeds, it means that there was one or severalcompton
processes that have now been killed, or at least signalled, andexec compton -b
will not be executed.If
pkill compton
fails (no process matched the name, or there was some internal error inpkill
), the body of theif
statement would call yourexec compton -b
, which would replace the shell process with the process resulting from runningcompton -b
.
add a comment |
You are getting the pgrep
output in your status
variable. It's just not the output that you expect it to be.
pgrep
outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton
, then $status
would be the PID of that process, or of those processes. pgrep
also returns an exit status, but an exit status is not captured by a command substitution as a string.
In your test, you compare $status
against 1
. It is unlikely that compton
has PID 1.
If you want to kill any compton
process if they exist, and start compton -b
if no compton
process exists, you may do that with
#!/bin/sh
if ! pkill compton; then
exec compton -b
fi
This uses the exit status of pkill
. The pkill
tool works in an equivalent way to pgrep
(they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep
would do, pkill
sends the TERM
signal (by default) to the matching processes.
The if
keyword uses the exit status of the command that you use with it.
The !
inverts the sense of the test so that
If
pkill compton
succeeds, it means that there was one or severalcompton
processes that have now been killed, or at least signalled, andexec compton -b
will not be executed.If
pkill compton
fails (no process matched the name, or there was some internal error inpkill
), the body of theif
statement would call yourexec compton -b
, which would replace the shell process with the process resulting from runningcompton -b
.
add a comment |
You are getting the pgrep
output in your status
variable. It's just not the output that you expect it to be.
pgrep
outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton
, then $status
would be the PID of that process, or of those processes. pgrep
also returns an exit status, but an exit status is not captured by a command substitution as a string.
In your test, you compare $status
against 1
. It is unlikely that compton
has PID 1.
If you want to kill any compton
process if they exist, and start compton -b
if no compton
process exists, you may do that with
#!/bin/sh
if ! pkill compton; then
exec compton -b
fi
This uses the exit status of pkill
. The pkill
tool works in an equivalent way to pgrep
(they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep
would do, pkill
sends the TERM
signal (by default) to the matching processes.
The if
keyword uses the exit status of the command that you use with it.
The !
inverts the sense of the test so that
If
pkill compton
succeeds, it means that there was one or severalcompton
processes that have now been killed, or at least signalled, andexec compton -b
will not be executed.If
pkill compton
fails (no process matched the name, or there was some internal error inpkill
), the body of theif
statement would call yourexec compton -b
, which would replace the shell process with the process resulting from runningcompton -b
.
You are getting the pgrep
output in your status
variable. It's just not the output that you expect it to be.
pgrep
outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton
, then $status
would be the PID of that process, or of those processes. pgrep
also returns an exit status, but an exit status is not captured by a command substitution as a string.
In your test, you compare $status
against 1
. It is unlikely that compton
has PID 1.
If you want to kill any compton
process if they exist, and start compton -b
if no compton
process exists, you may do that with
#!/bin/sh
if ! pkill compton; then
exec compton -b
fi
This uses the exit status of pkill
. The pkill
tool works in an equivalent way to pgrep
(they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep
would do, pkill
sends the TERM
signal (by default) to the matching processes.
The if
keyword uses the exit status of the command that you use with it.
The !
inverts the sense of the test so that
If
pkill compton
succeeds, it means that there was one or severalcompton
processes that have now been killed, or at least signalled, andexec compton -b
will not be executed.If
pkill compton
fails (no process matched the name, or there was some internal error inpkill
), the body of theif
statement would call yourexec compton -b
, which would replace the shell process with the process resulting from runningcompton -b
.
edited yesterday
answered Mar 19 at 22:00
Kusalananda♦Kusalananda
138k17258426
138k17258426
add a comment |
add a comment |
You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"
so
#!/bin/bash
pgrep compton >/dev/null
if [[ $? -eq 0 ]]
then
killall compton
else
exec compton -b
fi
or
#!/bin/bash
status=$(pgrep compton 2>&1)
if [[ -n "$status" ]]
then
killall compton
else
exec compton -b
fi
3
if pgrep compton >/dev/null; then
is the better practice, so one doesn't directly inspect$?
at all. In addition to avoiding bugs where$?
refers to a different command than you think it does when editing code to add logging or such, this also means thatset -e
will no longer treatpgrep
returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").
– Charles Duffy
Mar 19 at 23:41
add a comment |
You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"
so
#!/bin/bash
pgrep compton >/dev/null
if [[ $? -eq 0 ]]
then
killall compton
else
exec compton -b
fi
or
#!/bin/bash
status=$(pgrep compton 2>&1)
if [[ -n "$status" ]]
then
killall compton
else
exec compton -b
fi
3
if pgrep compton >/dev/null; then
is the better practice, so one doesn't directly inspect$?
at all. In addition to avoiding bugs where$?
refers to a different command than you think it does when editing code to add logging or such, this also means thatset -e
will no longer treatpgrep
returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").
– Charles Duffy
Mar 19 at 23:41
add a comment |
You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"
so
#!/bin/bash
pgrep compton >/dev/null
if [[ $? -eq 0 ]]
then
killall compton
else
exec compton -b
fi
or
#!/bin/bash
status=$(pgrep compton 2>&1)
if [[ -n "$status" ]]
then
killall compton
else
exec compton -b
fi
You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"
so
#!/bin/bash
pgrep compton >/dev/null
if [[ $? -eq 0 ]]
then
killall compton
else
exec compton -b
fi
or
#!/bin/bash
status=$(pgrep compton 2>&1)
if [[ -n "$status" ]]
then
killall compton
else
exec compton -b
fi
answered Mar 19 at 21:23
Jakub JindraJakub Jindra
399310
399310
3
if pgrep compton >/dev/null; then
is the better practice, so one doesn't directly inspect$?
at all. In addition to avoiding bugs where$?
refers to a different command than you think it does when editing code to add logging or such, this also means thatset -e
will no longer treatpgrep
returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").
– Charles Duffy
Mar 19 at 23:41
add a comment |
3
if pgrep compton >/dev/null; then
is the better practice, so one doesn't directly inspect$?
at all. In addition to avoiding bugs where$?
refers to a different command than you think it does when editing code to add logging or such, this also means thatset -e
will no longer treatpgrep
returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").
– Charles Duffy
Mar 19 at 23:41
3
3
if pgrep compton >/dev/null; then
is the better practice, so one doesn't directly inspect $?
at all. In addition to avoiding bugs where $?
refers to a different command than you think it does when editing code to add logging or such, this also means that set -e
will no longer treat pgrep
returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").– Charles Duffy
Mar 19 at 23:41
if pgrep compton >/dev/null; then
is the better practice, so one doesn't directly inspect $?
at all. In addition to avoiding bugs where $?
refers to a different command than you think it does when editing code to add logging or such, this also means that set -e
will no longer treat pgrep
returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").– Charles Duffy
Mar 19 at 23:41
add a comment |
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-bash, exit-status, pgrep, stderr, stdout
1
0
or1
is exit status, but$(...)
captures output.– Charles Duffy
Mar 19 at 23:41
grab
$?
. doman bash
.– user2497
Mar 20 at 19:09