What is the meaning of ${0%/*} in a bash script?

BashScripting

Bash Problem Overview

I am trying to understand a test script, which includes the following segment:

SCRIPT_PATH=${0%/*}
if [ "$0" != "$SCRIPT_PATH" ] && [ "$SCRIPT_PATH" != "" ]; then 
    cd $SCRIPT_PATH
fi

What does the ${0%/*} stand for? Thanks

Bash Solutions

Solution 1 - Bash

It is called Parameter Expansion. Take a look at this page and the rest of the site.

What ${0%/*} does is, it expands the value contained within the argument 0 (which is the path that called the script) after removing the string /* suffix from the end of it.

So, $0 is the same as ${0} which is like any other argument, eg. $1 which you can write as ${1}. As I said $0 is special, as it's not a real argument, it's always there and represents name of script. Parameter Expansion works within the { } -- curly braces, and % is one type of Parameter Expansion.

%/* matches the last occurrence of / and removes anything (* means anything) after that character. Take a look at this simple example:

$ var="foo/bar/baz"
$ echo "$var"
foo/bar/baz
$ echo "${var}"
foo/bar/baz
$ echo "${var%/*}"
foo/bar
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Questionuser785099View Question on Stackoverflow
Solution 1 - Bashc00kiemon5terView Answer on Stackoverflow