bsh脚本获取脚本目录的zsh等效项是什么? [英] What is the zsh equivalent of a bash script getting the script's directory?

问题描述

我想将此bash脚本翻译为zsh脚本.因此，我对此没有经验，希望可以在这里获得帮助:

I want to translate this bash-script intro a zsh-script. Hence I have no experience with this I hope I may get help here:

bash脚本:

SCRIPT_PATH="${BASH_SOURCE[0]}";
if([ -h "${SCRIPT_PATH}" ]) then
    while([ -h "${SCRIPT_PATH}" ]) do SCRIPT_PATH=`readlink "${SCRIPT_PATH}"`; done
fi
pushd . > /dev/null
cd `dirname ${SCRIPT_PATH}` > /dev/null
SCRIPT_PATH=`pwd`;
popd  > /dev/null

我已经知道我可以使用 SCRIPT_PATH="$0";获取脚本所在的路径.但是随后我的"readlink"语句出现了错误.

What I already know is that I can use SCRIPT_PATH="$0"; to get the path were the script is located at. But then I get errors with the "readlink" statement.

感谢您的帮助

推荐答案

除了BASH_SOURCE，我看不到需要进行任何更改.但是脚本的目的是什么?如果要获取目录，脚本位于${0:A:h}(:A将解析所有符号链接，:h将截断最后一个路径组件，并为您提供目录名称):

Except for BASH_SOURCE I see no changes that you need to make. But what is the purpose of the script? If you want to get directory your script is located at there is ${0:A:h} (:A will resolve all symlinks, :h will truncate last path component leaving you with a directory name):

SCRIPT_PATH="${0:A:h}"

仅此而已.请注意，原始脚本会发生一些奇怪的事情:

and that’s all. Note that original script has something strange going on:

1. if(…)和while(…)在子Shell中启动….您无需在这里使用subshel​​l，只需使用if …和while …进行这些检查就会更快.
完全不需要

2. pushd ..使用pushd时，通常将cd调用替换为它:

1. if(…) and while(…) launch … in a subshell. You do not need subshell here, it is faster to do these checks using just if … and while ….

2. pushd . is not needed at all. While using pushd you normally replace the cd call with it:

pushd "$(dirname $SCRIPT_PATH)" >/dev/null
SCRIPT_PATH="$(pwd)"
popd >/dev/null

如果…输出带有空格的内容，则

cd `…`将失败.目录可能包含空格.在上面的示例中，我使用"$(…)"，"`…`"也将起作用.

您不需要在变量声明中尾随;.

有readlink -f可以解析所有符号链接，因此您可以考虑将原始脚本缩减为SCRIPT_PATH="$(dirname $(readlink -f "${BASH_SOURCE[0]}"))"(行为可能会发生变化，因为您的脚本似乎仅在最后一个组件中解析符号链接):这相当于${0:A:h}

if [ -h "$SCRIPT_PATH" ]是多余的，因为除非脚本路径是符号链接，否则不会执行具有相同条件的while正文.

readlink $SCRIPT_PATH将相对于包含$SCRIPT_PATH 的目录返回符号链接.因此，原始脚本可能无法用于解析最后一个组件中的符号链接.

if(…)和then之间没有;.我很惊讶bash接受了这个.

cd `…` will fail if … outputs something with spaces. It is possible for a directory to contain a space. In the above example I use "$(…)", "`…`" will also work.

You do not need trailing ; in variable declarations.

There is readlink -f that will resolve all symlinks thus you may consider reducing original script to SCRIPT_PATH="$(dirname $(readlink -f "${BASH_SOURCE[0]}"))" (the behavior may change as your script seems to resolve symlinks only in last component): this is bash equivalent to ${0:A:h}.

if [ -h "$SCRIPT_PATH" ] is redundant since while body with the same condition will not be executed unless script path is a symlink.

readlink $SCRIPT_PATH will return symlink relative to the directory containing $SCRIPT_PATH. Thus original script cannot possibly used to resolve symlinks in last component.

There is no ; between if(…) and then. I am surprised bash accepts this.

以上所有语句均适用于bash和zsh.

All of the above statements apply both to bash and zsh.

如果仅在最后一个组件中仅解析符号链接是必不可少的，则应这样编写:

If resolving only symlinks only in last component is essential you should write it like this:

SCRIPT_PATH="$0:a"
function ResolveLastComponent()
{
    pushd "$1:h" >/dev/null
    local R="$(readlink "$1")"
    R="$R:a"
    popd >/dev/null
    echo $R
}
while test -h "$SCRIPT_PATH" ; do
    SCRIPT_PATH="$(ResolveLastComponent "$SCRIPT_PATH")"
done

.

为说明第7条语句，有以下示例:

To illustrate 7th statement there is the following example:

1. 创建目录$R/bash($R是任何目录，例如/tmp).
2. 将脚本原样放置在此处，例如在名称$R/bash/script_path.bash下.在测试的末尾添加echo "$SCRIPT_PATH"行，并在测试的开始处添加#!/bin/bash行.
3. 使其可执行:chmod +x $R/bash/script_path.bash.
4. 为其创建符号链接:cd $R/bash && ln -s script_path.bash link.
5. cd $R
6. 启动$R/bash/1.现在，您将看到脚本输出$R，而脚本应该输出$R/bash，就像启动$R/bash/script_path.bash时一样.

1. Create directory $R/bash ($R is any directory, e.g. /tmp).
2. Put your script there without modifications, e.g. under name $R/bash/script_path.bash. Add line echo "$SCRIPT_PATH" at the end of it and line #!/bin/bash at the start for testing.
3. Make it executable: chmod +x $R/bash/script_path.bash.
4. Create a symlink to it: cd $R/bash && ln -s script_path.bash link.
5. cd $R
6. Launch $R/bash/1. Now you will see that your script outputs $R while it should output $R/bash like it does when you launch $R/bash/script_path.bash.

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