Question

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‘

Answer 1

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 $() since you can nest them (e.g., ‘a=$(expr 1 + $(expr 2 + 3))‘).