bash 脚本中的 ${0%/*} 是什么意思? [英] What is the meaning of ${0%/*} in a bash script?

问题描述

我正在尝试理解一个测试脚本，其中包括以下部分:

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

${0%/*} 代表什么?谢谢

推荐答案

它叫做参数扩展.查看此页面和网站的其余部分.

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

${0%/*} 的作用是，在删除参数 0(即调用脚本的路径)中包含的值string /* 从它的末尾开始.

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.

因此，$0 与 ${0} 相同，就像任何其他参数一样，例如.$1 可以写成 ${1}.正如我所说的 $0 是特殊的，因为它不是 真正的 参数，它总是存在并代表脚本的名称.参数扩展在 { } -- 大括号内起作用，而 % 是参数扩展的一种类型.

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.

%/* 匹配最后一次出现的 / 并删除该字符之后的任何内容(* 表示任何内容).看看这个简单的例子:

%/* 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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