How do I run multiple scripts within a script, even if one script fails
0
I have a simple bash script that executes a number of other scripts …
#/bin/bash
…
./update_artifact1.sh $ARTIFACT_VERSION
./update_artifact2.sh $ARTIFACT_VERSION
./update_artifact3.sh $ARTIFACT_VERSION
The problem is, if the first command (or second) dies with an error, none of the subsequent commands are run. Is there a way I can run all three commands, save the exit status of each, and then return a successful exit status if all three execute successfully? The scripts don’t have to run concurrently.
bash bash-scripting exit-code
asked Feb 21 '14 at 21:15
DaveDave
2325
add a comment |
0
I have a simple bash script that executes a number of other scripts …
#/bin/bash
…
./update_artifact1.sh $ARTIFACT_VERSION
./update_artifact2.sh $ARTIFACT_VERSION
./update_artifact3.sh $ARTIFACT_VERSION
The problem is, if the first command (or second) dies with an error, none of the subsequent commands are run. Is there a way I can run all three commands, save the exit status of each, and then return a successful exit status if all three execute successfully? The scripts don’t have to run concurrently.
bash bash-scripting exit-code
asked Feb 21 '14 at 21:15
DaveDave
2325
2
What you show does not indicate that the main script should die. What is behind "…"?
– glenn jackman
Feb 21 '14 at 21:26
If the command "./update_artifact1.sh $ARTIFACT_VERSION" fails, the other two commands ("./update_artifact2.sh $ARTIFACT_VERSION" and "./update_artifact3.sh $ARTIFACT_VERSION") don't run. Do you understand?
– Dave
Feb 21 '14 at 21:45
3
yes I do. But if you trycd /unknown_dir; echo ok, the echo command executed even though cd failed. Why do the 2nd and 3rd commands fail in your script? Do you haveset -eturned on? Do the 2nd and 3rd commands rely on something the 1st provides? Is there any error output? You need to provide more information.
– glenn jackman
Feb 21 '14 at 22:17
add a comment |
0
0
0
I have a simple bash script that executes a number of other scripts …
#/bin/bash
…
./update_artifact1.sh $ARTIFACT_VERSION
./update_artifact2.sh $ARTIFACT_VERSION
./update_artifact3.sh $ARTIFACT_VERSION
The problem is, if the first command (or second) dies with an error, none of the subsequent commands are run. Is there a way I can run all three commands, save the exit status of each, and then return a successful exit status if all three execute successfully? The scripts don’t have to run concurrently.
bash bash-scripting exit-code
asked Feb 21 '14 at 21:15
DaveDave
2325
I have a simple bash script that executes a number of other scripts …
#/bin/bash
…
./update_artifact1.sh $ARTIFACT_VERSION
./update_artifact2.sh $ARTIFACT_VERSION
./update_artifact3.sh $ARTIFACT_VERSION
The problem is, if the first command (or second) dies with an error, none of the subsequent commands are run. Is there a way I can run all three commands, save the exit status of each, and then return a successful exit status if all three execute successfully? The scripts don’t have to run concurrently.
bash bash-scripting exit-code
bash bash-scripting exit-code
asked Feb 21 '14 at 21:15
DaveDave
2325
asked Feb 21 '14 at 21:15
DaveDave
2325
asked Feb 21 '14 at 21:15
DaveDave
2325
asked Feb 21 '14 at 21:15
DaveDave
2325
asked Feb 21 '14 at 21:15
DaveDave
2325
2325
2
What you show does not indicate that the main script should die. What is behind "…"?
– glenn jackman
Feb 21 '14 at 21:26
If the command "./update_artifact1.sh $ARTIFACT_VERSION" fails, the other two commands ("./update_artifact2.sh $ARTIFACT_VERSION" and "./update_artifact3.sh $ARTIFACT_VERSION") don't run. Do you understand?
– Dave
Feb 21 '14 at 21:45
3
yes I do. But if you trycd /unknown_dir; echo ok, the echo command executed even though cd failed. Why do the 2nd and 3rd commands fail in your script? Do you haveset -eturned on? Do the 2nd and 3rd commands rely on something the 1st provides? Is there any error output? You need to provide more information.
– glenn jackman
Feb 21 '14 at 22:17
add a comment |
2
What you show does not indicate that the main script should die. What is behind "…"?
– glenn jackman
Feb 21 '14 at 21:26
If the command "./update_artifact1.sh $ARTIFACT_VERSION" fails, the other two commands ("./update_artifact2.sh $ARTIFACT_VERSION" and "./update_artifact3.sh $ARTIFACT_VERSION") don't run. Do you understand?
– Dave
Feb 21 '14 at 21:45
3
yes I do. But if you trycd /unknown_dir; echo ok, the echo command executed even though cd failed. Why do the 2nd and 3rd commands fail in your script? Do you haveset -eturned on? Do the 2nd and 3rd commands rely on something the 1st provides? Is there any error output? You need to provide more information.
– glenn jackman
Feb 21 '14 at 22:17
2
2
What you show does not indicate that the main script should die. What is behind "…"?
– glenn jackman
Feb 21 '14 at 21:26
What you show does not indicate that the main script should die. What is behind "…"?
– glenn jackman
Feb 21 '14 at 21:26
If the command "./update_artifact1.sh $ARTIFACT_VERSION" fails, the other two commands ("./update_artifact2.sh $ARTIFACT_VERSION" and "./update_artifact3.sh $ARTIFACT_VERSION") don't run. Do you understand?
– Dave
Feb 21 '14 at 21:45
If the command "./update_artifact1.sh $ARTIFACT_VERSION" fails, the other two commands ("./update_artifact2.sh $ARTIFACT_VERSION" and "./update_artifact3.sh $ARTIFACT_VERSION") don't run. Do you understand?
– Dave
Feb 21 '14 at 21:45
3
3
yes I do. But if you try
cd /unknown_dir; echo ok, the echo command executed even though cd failed. Why do the 2nd and 3rd commands fail in your script? Do you have set -e turned on? Do the 2nd and 3rd commands rely on something the 1st provides? Is there any error output? You need to provide more information.– glenn jackman
Feb 21 '14 at 22:17
yes I do. But if you try
cd /unknown_dir; echo ok, the echo command executed even though cd failed. Why do the 2nd and 3rd commands fail in your script? Do you have set -e turned on? Do the 2nd and 3rd commands rely on something the 1st provides? Is there any error output? You need to provide more information.– glenn jackman
Feb 21 '14 at 22:17
add a comment |
1 Answer
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You can slightly modify your script in this way
#!/bin/bash
…
./update_artifact1.sh $ARTIFACT_VERSION
exit_state_1=$?
./update_artifact2.sh $ARTIFACT_VERSION
exit_state_2=$?
./update_artifact3.sh $ARTIFACT_VERSION
exit_state_3=$?
exit_state_all=$[ $exit_state_1 && $exit_state_2 && $exit_state_3 ]
exit $exit_state_all
Note:
- Each command return an exit code that
bashstores in$?. It is0if there is no error, or a different number if there is an error.
After every line the value of$?is updated so you need to store in a variable (in this case we usedexit_state_1...exit_state_n).
At the end you want that your script will return you only an error code with 0 if it is all ok and we realize it with$[ $a || $b || $c ].
Note that$[ 0|| 0 || 0 ]is 0 meanwhile$[ $a || $b || $c ]with at least one on$a,$b,$cnot equal to0returns 1. - It is important the ! in the first line when you have to write
#!/bin/bashto say to the system that is a script that have to be executed in abashshell.
You may find interesting to read more about the shebang here.
- As noted in a comment too if you put
set -esomewhere in your script before those line, the script will be interrupted at the first error without process the other line, and it will return an error code (the one of the last command executed).
answered Jun 25 '14 at 14:26
HasturHastur
13.2k53268
add a comment |
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0
You can slightly modify your script in this way
#!/bin/bash
…
./update_artifact1.sh $ARTIFACT_VERSION
exit_state_1=$?
./update_artifact2.sh $ARTIFACT_VERSION
exit_state_2=$?
./update_artifact3.sh $ARTIFACT_VERSION
exit_state_3=$?
exit_state_all=$[ $exit_state_1 && $exit_state_2 && $exit_state_3 ]
exit $exit_state_all
Note:
- Each command return an exit code that
bashstores in$?. It is0if there is no error, or a different number if there is an error.
After every line the value of$?is updated so you need to store in a variable (in this case we usedexit_state_1...exit_state_n).
At the end you want that your script will return you only an error code with 0 if it is all ok and we realize it with$[ $a || $b || $c ].
Note that$[ 0|| 0 || 0 ]is 0 meanwhile$[ $a || $b || $c ]with at least one on$a,$b,$cnot equal to0returns 1. - It is important the ! in the first line when you have to write
#!/bin/bashto say to the system that is a script that have to be executed in abashshell.
You may find interesting to read more about the shebang here.
- As noted in a comment too if you put
set -esomewhere in your script before those line, the script will be interrupted at the first error without process the other line, and it will return an error code (the one of the last command executed).
answered Jun 25 '14 at 14:26
HasturHastur
13.2k53268
add a comment |
0
You can slightly modify your script in this way
#!/bin/bash
…
./update_artifact1.sh $ARTIFACT_VERSION
exit_state_1=$?
./update_artifact2.sh $ARTIFACT_VERSION
exit_state_2=$?
./update_artifact3.sh $ARTIFACT_VERSION
exit_state_3=$?
exit_state_all=$[ $exit_state_1 && $exit_state_2 && $exit_state_3 ]
exit $exit_state_all
Note:
- Each command return an exit code that
bashstores in$?. It is0if there is no error, or a different number if there is an error.
After every line the value of$?is updated so you need to store in a variable (in this case we usedexit_state_1...exit_state_n).
At the end you want that your script will return you only an error code with 0 if it is all ok and we realize it with$[ $a || $b || $c ].
Note that$[ 0|| 0 || 0 ]is 0 meanwhile$[ $a || $b || $c ]with at least one on$a,$b,$cnot equal to0returns 1. - It is important the ! in the first line when you have to write
#!/bin/bashto say to the system that is a script that have to be executed in abashshell.
You may find interesting to read more about the shebang here.
- As noted in a comment too if you put
set -esomewhere in your script before those line, the script will be interrupted at the first error without process the other line, and it will return an error code (the one of the last command executed).
answered Jun 25 '14 at 14:26
HasturHastur
13.2k53268
add a comment |
0
0
0
You can slightly modify your script in this way
#!/bin/bash
…
./update_artifact1.sh $ARTIFACT_VERSION
exit_state_1=$?
./update_artifact2.sh $ARTIFACT_VERSION
exit_state_2=$?
./update_artifact3.sh $ARTIFACT_VERSION
exit_state_3=$?
exit_state_all=$[ $exit_state_1 && $exit_state_2 && $exit_state_3 ]
exit $exit_state_all
Note:
- Each command return an exit code that
bashstores in$?. It is0if there is no error, or a different number if there is an error.
After every line the value of$?is updated so you need to store in a variable (in this case we usedexit_state_1...exit_state_n).
At the end you want that your script will return you only an error code with 0 if it is all ok and we realize it with$[ $a || $b || $c ].
Note that$[ 0|| 0 || 0 ]is 0 meanwhile$[ $a || $b || $c ]with at least one on$a,$b,$cnot equal to0returns 1. - It is important the ! in the first line when you have to write
#!/bin/bashto say to the system that is a script that have to be executed in abashshell.
You may find interesting to read more about the shebang here.
- As noted in a comment too if you put
set -esomewhere in your script before those line, the script will be interrupted at the first error without process the other line, and it will return an error code (the one of the last command executed).
answered Jun 25 '14 at 14:26
HasturHastur
13.2k53268
You can slightly modify your script in this way
#!/bin/bash
…
./update_artifact1.sh $ARTIFACT_VERSION
exit_state_1=$?
./update_artifact2.sh $ARTIFACT_VERSION
exit_state_2=$?
./update_artifact3.sh $ARTIFACT_VERSION
exit_state_3=$?
exit_state_all=$[ $exit_state_1 && $exit_state_2 && $exit_state_3 ]
exit $exit_state_all
Note:
- Each command return an exit code that
bashstores in$?. It is0if there is no error, or a different number if there is an error.
After every line the value of$?is updated so you need to store in a variable (in this case we usedexit_state_1...exit_state_n).
At the end you want that your script will return you only an error code with 0 if it is all ok and we realize it with$[ $a || $b || $c ].
Note that$[ 0|| 0 || 0 ]is 0 meanwhile$[ $a || $b || $c ]with at least one on$a,$b,$cnot equal to0returns 1. - It is important the ! in the first line when you have to write
#!/bin/bashto say to the system that is a script that have to be executed in abashshell.
You may find interesting to read more about the shebang here.
- As noted in a comment too if you put
set -esomewhere in your script before those line, the script will be interrupted at the first error without process the other line, and it will return an error code (the one of the last command executed).
answered Jun 25 '14 at 14:26
HasturHastur
13.2k53268
edited Jun 25 '14 at 15:21
answered Jun 25 '14 at 14:26
HasturHastur
13.2k53268
answered Jun 25 '14 at 14:26
HasturHastur
13.2k53268
answered Jun 25 '14 at 14:26
HasturHastur
13.2k53268
13.2k53268
add a comment |
add a comment |
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2
What you show does not indicate that the main script should die. What is behind "…"?
– glenn jackman
Feb 21 '14 at 21:26
If the command "./update_artifact1.sh $ARTIFACT_VERSION" fails, the other two commands ("./update_artifact2.sh $ARTIFACT_VERSION" and "./update_artifact3.sh $ARTIFACT_VERSION") don't run. Do you understand?
– Dave
Feb 21 '14 at 21:45
3
yes I do. But if you try
cd /unknown_dir; echo ok, the echo command executed even though cd failed. Why do the 2nd and 3rd commands fail in your script? Do you haveset -eturned on? Do the 2nd and 3rd commands rely on something the 1st provides? Is there any error output? You need to provide more information.– glenn jackman
Feb 21 '14 at 22:17