bash脚本中${0%/*}的含义是什么?

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

我正在尝试理解一个测试脚本，它包括以下部分:

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

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

${0%/*}代表什么?谢谢

1 个解决方案

#1

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

它被称为参数展开。看看这个页面和网站的其他部分。

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.

${0%/*}做的是，它将在参数0(即调用脚本的路径)中删除字符串/*后缀后的值展开。

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.

所以，0美元等于{0}，这和其他的论点一样。$1，你可以写成${1}。正如我所说的$0是特殊的，因为它不是一个真正的参数，它总是在那里并且表示脚本的名称。参数扩展在{}-花括号内工作，%是参数扩展的一种类型。

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

%/*匹配/并删除该字符后的任何(*表示任何)内容。看看这个简单的例子:

$ var="foo/bar/baz"
$ echo "$var"
foo/bar/baz
$ echo "${var}"
foo/bar/baz
$ echo "${var%/*}"
foo/bar

#1