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.
Related
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:
Difference between sh and Bash
(11 answers)
String comparison in bash. [[: not found
(9 answers)
Closed 4 years ago.
I have a shell script where I want to check if it's being run by root. If it's not then I set the variable SUDO='sudo' to prepend to subsequent commands. That way the user gets one nice permission denied error message from the shell and if they're running with sudo, it properly does its business. However, for the following code I get:
myscript.sh: [: -ne: unexpected operator
Code is:
#!/bin/sh
SUDO=""
if [ $EUID -ne 0 ]; then
SUDO='sudo'
fi
I am running it like: sh myscript.sh as this is not a bash script. This is not an issue of sh vs bash afaik. The shebang says /bin/sh and I'm running it with sh.
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)
This question already has answers here:
Why does shell ignore quoting characters in arguments passed to it through variables? [duplicate]
(3 answers)
Closed 6 years ago.
I'm currently having problems to execute a command from a shell variable.
In general the following works as expected:
COMMAND="echo A"
echo $COMMAND
$COMMAND
produces:
echo A
A
But if I do
COMMAND="su aUser -s /bin/bash -c 'echo A'"
echo $COMMAND
$COMMAND
I get
su aUser -s /bin/bash -c 'echo A'
Password:
A': -c: line 0: unexpected EOF while looking for matching `''
A': -c: line 1: syntax error: unexpected end of file
If I enter the line
su aUser -s /bin/bash -c 'echo A'
directly it works as expected.
It seems my assumption that $COMMAND is equal to entering the content as command directly is wrong.
Questions
1) Does anyone know how I can run the command from a variable?
2) What exactly is the difference between
COMMAND="command"
$COMMAND
and
command
?
Arrays are useful to keep your parameters whole:
command=(su aUser -s /bin/bash -c 'echo A')
and invoke it exactly like this:
"${command[#]}"
You need eval.
$ eval $VARIABLE