This question already has answers here:
Can a shell script set environment variables of the calling shell? [duplicate]
(20 answers)
Closed 14 days ago.
I wrote a bash script to set the environment variable VAR if it is currently not set:
example.sh
#!/bin/bash
if [ -z $VAR ]; then
export VAR=abc
fi
Now I type this in the command line: ./example.sh && echo $VAR. I expect abc, but the result is blank. Why?
Use source example.sh or . example.sh
Related
This question already has answers here:
Exit code of variable assignment to command substitution in Bash
(5 answers)
Closed 1 year ago.
#!/bin/bash
set -e
set -o pipefail
shopt -s inherit_errexit
declare _x=`command-with-error`
echo "_x=${_x}"
Run the script shows:
bash xx.sh
xx.sh: line 6: command-with-error: command not found
_x=
Apparently line 6 did not exit the shell. What option should I use to make the script exit when the subshell command on declare line fails?
The successful exit status of declare is overriding the unsuccessful exit status of command-with-error.
Break it into two separate commands:
declare _x
_x=$(command-with-error)
...as you can see running correctly (which is to say, without writing anything to stdout) at https://ideone.com/TGyFCZ
This question already has answers here:
How to detect if a script is being sourced
(22 answers)
Closed 3 years ago.
I have a bash script that has inside it:
exit 1
When I "source" this script instead of running it, it causes the caller to exit.
Is there a way that the script can determine that it's being run with "source" and not as its script?
You can use this check inside your script:
[[ $0 = $BASH_SOURCE ]] && echo "normal run" || echo "sourced run"
Or using if/else/fi wherever you're calling exit:
if [[ $0 = $BASH_SOURCE ]]; then
exit 1
else
# don't call exit
echo "some error..."
fi
This question already has answers here:
Automatic exit from Bash shell script on error [duplicate]
(8 answers)
Error handling in Bash [closed]
(15 answers)
Closed 4 years ago.
In bash how do we make the script to automatically exit if a command line return code is not zero. For example:
#!/bin/bash
cd /something_something
mv file_a /somedir/file_a # this produce an error
echo $? # This produce a non-zero output
echo "We should not continue to this line"
I know we can debug bash script with #!/bin/bash -x but sometime the script is too long, it run so fast, and we missed important error.
And I don't want to keep writing
[[ $? -ne 0 ]] && run next_command
There are lots of problems with using set -e. Just join the commands with &&, and test the result with an if statement.
if cd /something_something && mv file_a /somedir/file_a; then
echo $?
exit
fi
echo "Both cd and mv worked"
This question already has answers here:
Why should there be spaces around '[' and ']' in Bash?
(5 answers)
Closed 5 years ago.
This is the code of the my shell script:
#! /bin/bash
if ["$SHELL" = "/bin/bash"];then
echo "this is bash"
elif ["$SHELL" = "aa"];then
echo "this is aa"
else
echo "this is not /bin/bash, but $SHELL"
fi
why I execute the test_bash_03 script file gets the else result? shouldn't it be:this is bash ？
aircraftdeMacBook-Pro:bash_demo ldl$ ./test_bash_03
./test_bash_03: line 3: [/bin/bash: No such file or directory
./test_bash_03: line 5: [/bin/bash: No such file or directory
this is not /bin/bash, but /bin/bash
And I echo the $SHELL I also get the /bin/bash
aircraftdeMacBook-Pro:bash_demo ldl$ echo $SHELL
/bin/bash
You are missing a space after [ and before ].
The bash tries to execute a command named [/bin/bash instead of [ (which is test), then doesn't find that and has an exit code of 1 (false). So you end up in the else case.
This question already has an answer here:
Blank first line of shell script: explain behavior of UID variable
(1 answer)
Closed 6 years ago.
Hi I have a question about bash.
and I'm new to it.
I made a file named "test.sh" and its contents is
#!/bin/bash
set -x
echo $UID
echo "$UID"
echo "$(id -u)"
and the result is blank!!
nothing shows up
However, when i just type "echo $UID" on terminal
it shows "1011"
is there anything i missed for bash?
Please help
UPDATED
bash version is 4.3.11 and I typed "sh test.sh" to execute.
and the result is
+ echo
+ echo
+ id -u
+ echo 1011
1011
thanks!
$UID is a Bash variable that is not set under sh, that may be why it outputs blank lines.
Try bash test.sh or make your script executable with chmod u+x test.sh, the program defined in shebang will then be used (/bin/bash)