I have a script like this, named judge:
#!/bin/bash
echo "last exit status is $?"
It always outputs "last exit status is 0". Eg:
ls -l; judge # correctly reports 0
ls -z; judge # incorrectly reports 0
beedogs; judge # incorrectly reports 0
Why?
There are different bash processes executing each line of code and $? isn't shared between the processes. You can work around this by making judge a bash function:
[root@xxx httpd]# type judge
judge is a function
judge ()
{
echo "last exit status is $?"
}
[root@xxx httpd]# ls -l / >/dev/null 2>&1; judge
last exit status is 0
[root@xxx httpd]# ls -l /doesntExist >/dev/null 2>&1; judge
last exit status is 2
[root@xxx httpd]#
As discussed in the comments, the $? variable holds the value from the last process that returned a value to the shell.
If judge needs to do something based on a previous command state, you could have it accept a parameter, and pass in the state.
#!/bin/bash
echo "last exit status is $1"
# Or even
return $1
So:
[cmd args...]; judge $?
bashprocesses executing each line of code and$?isn't shared between the processes.