What is the meaning of ${0%/*} in a bash script?
I am trying to understand a test script, which includes the following segment:
SCRIPT_PATH=${0%/*}
if [ "$0" != "$SCRIPT_PATH" ] && [ "$SCRIPT_PATH" != "" ]; then
cd $SCRIPT_PATH
fi
What does the ${0%/*}
stand for? Thanks
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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