in a shell script, i'm trying to find out if another file is a shell script. i'm doing that by grepping the shebang line. but my grep statement doesn't work:
if [[ $($(cat $file) | grep '^#! /bin' | wc -l) -gt 0 ]]
then
echo 'file is a script!'
else
echo "no script"
fi
i always get the error "bash: #!: command not found". i tried several things to escape the shebang but that didn't work.
maybe you can help me with that? :)
cheers, narf
I would suggest that you change your condition to this:
if grep -q '^#! */bin' "$file"
The -q option to grep is useful in this case as it tells grep to produce no output, exiting successfully if the pattern is matched. This can be used with if directly; there's no need to wrap everything in a test [[ (and especially no need for a useless use of cat).
I also modified your pattern slightly so that the space between #! and /bin is optional.
It's worth noting that this will produce false positives in cases where the match is on a different line of the file, or when another shebang is used. You could work around the first issue by piping head -n 1 to grep, so that only the first line would be checked:
if head -n 1 "$file" | grep -q '^#! */bin'
If you are searching for a known list of shebangs, e.g. /bin/sh and /bin/bash, you could change the pattern to something like ^#! */bin/\(sh\|bash\).