从shell脚本(bash)的参数列表中删除最后一个参数

问题描述：

此问题涉及在automator osx中运行的bash脚本.我正在使用自动操作来从查找程序中获取和过滤一堆文件引用.然后，通过自动操作将父文件夹的名称附加到该列表中.然后，Automator将这些参数提供给一个名为运行shell脚本"的操作.我不确定确切如何自动程序调用脚本，但是在用echo "$@"

This question concerns a bash script that is run in automator osx. I am using automator actions to get and filter a bunch of file references from the finder. Then I append to that list the name of the parent folder, also via an automator action. Automator then feeds these arguments to an action called "run shell script". I am not sure exactly how automator invokes the script but the argument list looks like this when echoed with: echo "$@"

/卷/G-Raid/在线/WAV_TEST/Testbok 50/01/01000 43-001.wav/卷/卷/G-Raid/在线/WAV_TEST/Testbok 50/02/02000 43-002.wav/卷/G-Raid/Online/WAV_TEST/Testbok 50/03/03000 43-003.wav/Volumes/G-Raid/Online/WAV_TEST/Testbok 50

/Volumes/G-Raid/Online/WAV_TEST/Testbok 50/01/01000 43-001.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/02/02000 43-002.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/03/03000 43-003.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50

在这种情况下，指向3个文件和一个文件夹的路径.

In this case path to 3 files and a folder.

在shell脚本中，我启动了一个名为ripcheckc *的应用程序，该程序使用从automator传递的args减去列表中的最后一个参数(文件夹).

In the shell script I launch an application called ripcheckc* with the args passed from automator minus the last argument(the folder) in the list.

我用它删除最后一个参数:

I use this to remove the last argument:

_args=( "$@" )
unset _args[${#_args[@]}-1]

这是echo $_args:

/卷/G-Raid/在线/WAV_TEST/Testbok 50/01/01000 43-001.wav/卷/卷/G-Raid/在线/WAV_TEST/Testbok 50/02/02000 43-002.wav/卷/G-Raid/在线/WAV_TEST/Testbok 50/03/03000 43-003.wav

/Volumes/G-Raid/Online/WAV_TEST/Testbok 50/01/01000 43-001.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/02/02000 43-002.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/03/03000 43-003.wav

与以前相同，但没有文件夹.

Same as before but without the folder.

现在，如果我以"$@"作为参数运行ripcheckc，它可以工作(但稍后会失败，因为参数列表中的最后一条路径).如果我使用${_args[@]}，应用程序将无提示地中止.当我回显$@和_args时，除了最后一个参数外，输出看起来相同.

Now, if I run ripcheckc with "$@" as argument it works (but fails later on because of that last path in the argument list) If I use ${_args[@]} the application will just abort silently. When I echo $@ and _args the output looks identical except for the last argument.

我的问题是-$ @和$ _args产生第一个有效输入而第二个无效输入有什么区别?

My question is - what is the difference between $@ and $_args that make the first valid input and the second not?

*该应用程序是 ripcheckc

我希望我的问题有意义.

I hope my question makes sense.

已解决.

答

假设您已经有array，您可以说:

Assuming that you already have an array, you can say:

unset "array[${#array[@]}-1]"

例如，如果您的脚本包含:

For example, if your script contains:

array=( "$@" )
unset "array[${#array[@]}-1]"    # Removes last element -- also see: help unset
for i in "${array[@]}"; do
  echo "$i"
done

调用它:bash scriptname foo bar baz产生:

foo
bar

从shell脚本(bash)的参数列表中删除最后一个参数

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