尽管设置了“ -e”，但如何“捕获”非零退出代码，然后回显错误代码

Question

I have script

#!/bin/bash

set -e

if [[ ! $(asd) ]]; then
   echo "caught command failure with exit code ${?}"
fi

echo "end of script" 

purpose of script is to terminate execution on any non zero command exit code with set -e except when command is "caught" (comming from Java) as in case of bad command asd

if [[ ! $(asd) ]]; then
   echo "caught command failure with exit code ${?}"
fi

however, though I "catch" the error and end of script prints to terminal, the error code is 0

echo "caught command failure with exit code ${?}"

so my question is how can I "catch" a bad command, and also print the exit code of that command?

edit

I have refactored script with same result, exit code is still 0

#!/bin/bash

set -e

if ! asd ; then
   echo "caught command failure with exit code ${?}"
fi

echo "end of script"

Answer 1

只需使用短路：

asd || echo "asd exited with $?" >&2

Answer 2

how can I "catch" a bad command, and also print the exit code of that command?

Often enough I do this:

asd && ret=$? || ret=$? 
echo asd exited with $ret

The exit status of the whole expression is 0 , so set -e doesn't exit. If asd succedess, then the first ret=$? executes with $? set to 0 , if it fails, then the first ret=$? is omitted, and the second executes.

Sometimes I do this:

ret=0
asd || ret=$?
echo asd exited with $ret

which works just the same and I can forget if the && or || should go first. And you can also do this:

if asd; then
   ret=0
else
   ret=$?
fi
echo asd exited with $ret