Can someone help explain the following:
If I type:
a=`ls -l`
Then the output of the ls command is saved in the variable a
but if I try:
a=`sh ./somefile`
The result is outputed to the shell ( stdout
) rather than the variable a
What I expected was the result operation of the shell trying to execute a scrip ' somefile
' to be stored in the variable.
Please point out what is wrong with my understanding and a possible way to do this.
Thanks.
EDIT:
Just to clarify, the script ' somefile
' may or may not exist. If it exsists then I want the output of the script to be stored in ' a
'. If not, I want the error message "no such file or dir" stored in ' a
'
I think because the shell probably attaches itself to /dev/tty but I may be wrong. Why wouldn't you just set execute permissions on the script and use:
a=`./somefile`
If you want to capture stderr and stdout to a, just use:
a=`./somefile 2>&1`
To check file is executable first:
if [[ -x ./somefile ]] ; then
a=$(./somefile 2>&1)
else
a="Couldn't find the darned thing."
fi
and you'll notice I'm switching to the $() method instead of backticks. I prefer $() since you can nest them (eg, " a=$(expr 1 + $(expr 2 + 3))
").
You can try the new and improved way of doing command substitution, use $() instead of backticks.
a=$(sh ./somefile)
If it still doesn't work, check if somefile is not actually stderr 'ing.
You are correct, the stdout of ./somefile
is stored in the variable a
. However, I assume somefile outputs to stderr. You can redirect that with 2>&1
directly after ./somefile
.
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