Mixing
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How to output the excit code after the function finished
If in a shell script. There are two function with loop, and I run both function in the script and put both in the background.
For example:
#!/bin/bash
function a {
for 1 in 2; do
if 3.sh; then
echo 'done'
else
exit 1
fi
done
}
function b {
for a in b; do
if c.sh; then
echo 'done'
else
exit 1
fi
done
}
a &
b &
Now since both functions are in background, once I run the script it will be completed right away. What I expected was to capture the exit code of the script so if anything wrong happened during loop a and b I can become acknowledged.
My another concern is that if anything happened during loop a, the script will be terminated right away (since exit code 1 is given), so b got terminated as well even if it's innocent.
function sh exit-code
edited Nov 20 '18 at 22:09
halfer
14.5k758111
asked Nov 20 '18 at 20:48
BeyondTryingToCodeBeyondTryingToCode
134
add a comment |
If in a shell script. There are two function with loop, and I run both function in the script and put both in the background.
For example:
#!/bin/bash
function a {
for 1 in 2; do
if 3.sh; then
echo 'done'
else
exit 1
fi
done
}
function b {
for a in b; do
if c.sh; then
echo 'done'
else
exit 1
fi
done
}
a &
b &
Now since both functions are in background, once I run the script it will be completed right away. What I expected was to capture the exit code of the script so if anything wrong happened during loop a and b I can become acknowledged.
My another concern is that if anything happened during loop a, the script will be terminated right away (since exit code 1 is given), so b got terminated as well even if it's innocent.
function sh exit-code
edited Nov 20 '18 at 22:09
halfer
14.5k758111
asked Nov 20 '18 at 20:48
BeyondTryingToCodeBeyondTryingToCode
134
There's no real reason to useexitinstead ofreturn; since the function is the only thing running in the background job, the background process will exit as soon as the function returns.
– chepner
Nov 20 '18 at 21:03
add a comment |
If in a shell script. There are two function with loop, and I run both function in the script and put both in the background.
For example:
#!/bin/bash
function a {
for 1 in 2; do
if 3.sh; then
echo 'done'
else
exit 1
fi
done
}
function b {
for a in b; do
if c.sh; then
echo 'done'
else
exit 1
fi
done
}
a &
b &
Now since both functions are in background, once I run the script it will be completed right away. What I expected was to capture the exit code of the script so if anything wrong happened during loop a and b I can become acknowledged.
My another concern is that if anything happened during loop a, the script will be terminated right away (since exit code 1 is given), so b got terminated as well even if it's innocent.
function sh exit-code
edited Nov 20 '18 at 22:09
halfer
14.5k758111
asked Nov 20 '18 at 20:48
BeyondTryingToCodeBeyondTryingToCode
134
If in a shell script. There are two function with loop, and I run both function in the script and put both in the background.
For example:
#!/bin/bash
function a {
for 1 in 2; do
if 3.sh; then
echo 'done'
else
exit 1
fi
done
}
function b {
for a in b; do
if c.sh; then
echo 'done'
else
exit 1
fi
done
}
a &
b &
Now since both functions are in background, once I run the script it will be completed right away. What I expected was to capture the exit code of the script so if anything wrong happened during loop a and b I can become acknowledged.
My another concern is that if anything happened during loop a, the script will be terminated right away (since exit code 1 is given), so b got terminated as well even if it's innocent.
function sh exit-code
function sh exit-code
edited Nov 20 '18 at 22:09
halfer
14.5k758111
asked Nov 20 '18 at 20:48
BeyondTryingToCodeBeyondTryingToCode
134
edited Nov 20 '18 at 22:09
halfer
14.5k758111
asked Nov 20 '18 at 20:48
BeyondTryingToCodeBeyondTryingToCode
134
edited Nov 20 '18 at 22:09
halfer
14.5k758111
edited Nov 20 '18 at 22:09
halfer
14.5k758111
edited Nov 20 '18 at 22:09
halfer
14.5k758111
14.5k758111
asked Nov 20 '18 at 20:48
BeyondTryingToCodeBeyondTryingToCode
134
asked Nov 20 '18 at 20:48
BeyondTryingToCodeBeyondTryingToCode
134
asked Nov 20 '18 at 20:48
BeyondTryingToCodeBeyondTryingToCode
134
134
There's no real reason to useexitinstead ofreturn; since the function is the only thing running in the background job, the background process will exit as soon as the function returns.
– chepner
Nov 20 '18 at 21:03
add a comment |
There's no real reason to useexitinstead ofreturn; since the function is the only thing running in the background job, the background process will exit as soon as the function returns.
– chepner
Nov 20 '18 at 21:03
There's no real reason to use
exit instead of return; since the function is the only thing running in the background job, the background process will exit as soon as the function returns.– chepner
Nov 20 '18 at 21:03
There's no real reason to use
exit instead of return; since the function is the only thing running in the background job, the background process will exit as soon as the function returns.– chepner
Nov 20 '18 at 21:03
add a comment |
1 Answer
1
active
oldest
votes
If you wait on a background job, the exit status of wait will be the exit status of the job.
a & a_pid=$!
b & b_pid=$!
wait $a_pid
a_status=$?
wait $b_pid
b_status=$?
You might object that if b finishes before a, then you have to wait to get b's status. That might be a problem, but only if you have anything else to do. Regardless of which job finishes first, the time it takes for both to complete is the same. (If b does finish first, then wait $b_pid will exit immediately once it is run.)
answered Nov 20 '18 at 21:02
chepnerchepner
248k33235330
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you wait on a background job, the exit status of wait will be the exit status of the job.
a & a_pid=$!
b & b_pid=$!
wait $a_pid
a_status=$?
wait $b_pid
b_status=$?
You might object that if b finishes before a, then you have to wait to get b's status. That might be a problem, but only if you have anything else to do. Regardless of which job finishes first, the time it takes for both to complete is the same. (If b does finish first, then wait $b_pid will exit immediately once it is run.)
answered Nov 20 '18 at 21:02
chepnerchepner
248k33235330
add a comment |
If you wait on a background job, the exit status of wait will be the exit status of the job.
a & a_pid=$!
b & b_pid=$!
wait $a_pid
a_status=$?
wait $b_pid
b_status=$?
You might object that if b finishes before a, then you have to wait to get b's status. That might be a problem, but only if you have anything else to do. Regardless of which job finishes first, the time it takes for both to complete is the same. (If b does finish first, then wait $b_pid will exit immediately once it is run.)
answered Nov 20 '18 at 21:02
chepnerchepner
248k33235330
add a comment |
If you wait on a background job, the exit status of wait will be the exit status of the job.
a & a_pid=$!
b & b_pid=$!
wait $a_pid
a_status=$?
wait $b_pid
b_status=$?
You might object that if b finishes before a, then you have to wait to get b's status. That might be a problem, but only if you have anything else to do. Regardless of which job finishes first, the time it takes for both to complete is the same. (If b does finish first, then wait $b_pid will exit immediately once it is run.)
answered Nov 20 '18 at 21:02
chepnerchepner
248k33235330
If you wait on a background job, the exit status of wait will be the exit status of the job.
a & a_pid=$!
b & b_pid=$!
wait $a_pid
a_status=$?
wait $b_pid
b_status=$?
You might object that if b finishes before a, then you have to wait to get b's status. That might be a problem, but only if you have anything else to do. Regardless of which job finishes first, the time it takes for both to complete is the same. (If b does finish first, then wait $b_pid will exit immediately once it is run.)
answered Nov 20 '18 at 21:02
chepnerchepner
248k33235330
answered Nov 20 '18 at 21:02
chepnerchepner
248k33235330
answered Nov 20 '18 at 21:02
chepnerchepner
248k33235330
answered Nov 20 '18 at 21:02
chepnerchepner
248k33235330
248k33235330
add a comment |
add a comment |
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There's no real reason to use
exitinstead ofreturn; since the function is the only thing running in the background job, the background process will exit as soon as the function returns.– chepner
Nov 20 '18 at 21:03