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I would like to return exit code "0" from a failed command. Is there any easier way of doing this, rather than:
function a() {
ls aaaaa 2>&1;
}
if ! $(a); then
return 0
else
return 5
fi
776
Exit When Any Command Fails This can actually be done with a single line using the set builtin command with the -e option. Putting this at the top of a bash script will cause the script to exit if any commands return a non-zero exit code.
How to get the exit code of a command. To get the exit code of a command type echo $? at the command prompt. In the following example a file is printed to the terminal using the cat command.
# By convention, an 'exit 0' indicates success, #+ while a non-zero exit value means an error or anomalous condition. # See the "Exit Codes With Special Meanings" appendix. $? is especially useful for testing the result of a command in a script (see Example 16-35 and Example 16-20).
For the bash shell's purposes, a command which exits with a zero (0) exit status has succeeded. A non-zero (1-255) exit status indicates failure. If a command is not found, the child process created to execute it returns a status of 127.
Simply append return 0 to the function to force a function to always exit successful.
function a() {
ls aaaaa 2>&1
return 0
}
a
echo $? # prints 0
If you wish to do it inline for any reason you can append || true to the command:
ls aaaaa 2>&1 || true
echo $? # prints 0
If you wish to invert the exit status simple prepend the command with !
! ls aaaaa 2>&1
echo $? # prints 0
! ls /etc/resolv.conf 2>&1
echo $? # prints 1
Also if you state what you are trying to achieve overall we might be able to guide you to better answers.
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