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s.ouchene
Asked: 2018-01-07 14:25:16 +0800 CST

What does ${0%/*} in shell scripts do?

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Sorry if this is a stupid question, but I searched about it without success.

What does exactly the second line do ?: 

#!/bin/sh
cd ${0%/*} || exit 1

I know the first is the shebang, the second tries to change directory but the confusing part is ${0%/*}.

Can you please explain me that second line?

command-line

1 Answers

  1. Best Answer

    ${0} is the first argument of the script, i.e. the script name or path. If you launch your script as path/to/script.sh, then ${0} will be exactly that string: path/to/script.sh.

    The %/* part modifies the value of ${0}. It means: take all characters until / followed by a file name. In the example above, ${0%/*} will be path/to.

    You can see it in action on your shell:

    $ x=path/to/script.sh
$ echo "${x%/*}"
path/to

    Sh supports many other kinds of "parameter substitution". Here is, for example, how to take the file name instead of the path:

    $ echo "${x##*/}"
script.sh

    In general, % and %% strip suffixes, while # and ## strip prefixes. You can read more about parameter substitution.

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