问题
How can I check the exit code of a command substitution in bash if the assignment is to a local variable in a function?
Please see the following examples. The second one is where I want to check the exit code.
Does someone have a good work-around or correct solution for this?
$ function testing { test="$(return 1)"; echo $?; }; testing
1
$ function testing { local test="$(return 1)"; echo $?; }; testing
0
回答1:
If you look at the man file for local (which is actually just the BASH builtins man page), it is treated as its own command, which gives an exit code of 0 upon successfully creating the local variable. So local is overwriting the last-executed error code.
Try this:
function testing { local test; test="$(return 1)"; echo $?; }; testing
EDIT: I went ahead and tried it for you, and it works.
来源：https://stackoverflow.com/questions/9524623/exit-code-of-command-substitution-in-bash-local-variable-assignment