从 shell 脚本 (bash) 的参数列表中删除最后一个参数
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Remove last argument from argument list of shell script (bash)
提问by Nordanfors
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 "$@"
这个问题涉及在 automator osx 中运行的 bash 脚本。我正在使用自动操作从查找器中获取和过滤一堆文件引用。然后我也通过自动操作将父文件夹的名称附加到该列表中。Automator 然后将这些参数提供给名为“运行 shell 脚本”的操作。我不确定 automator 究竟是如何调用脚本的，但是当回显时参数列表看起来像这样：echo "$@"
/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
/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 /在线/WAV_TEST/Testbok 50/03/03000 43-003.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50
In this case path to 3 files and a folder.
在这种情况下，路径为 3 个文件和一个文件夹。
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.
在 shell 脚本中，我启动了一个名为 ripcheckc* 的应用程序，其中包含从 automator 传递的参数减去列表中的最后一个参数（文件夹）。
I use this to remove the last argument:
我用它来删除最后一个参数：
_args=( "$@" )
unset _args[${#_args[@]}-1]
And this is echo $_args:
这是echo $_args：
/Volumes/G-Raid/Online/WAV_TEST/Testbok 50/01/01000 43-001.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/02/02000 43-002.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/03/03000 43-003.wav
/Volumes/G-Raid/Online/WAV_TEST/Testbok 50/01/01000 43-001.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/02/02000 43-002.wav /Volumes/G-Raid /在线/WAV_TEST/Testbok 50/03/03000 43-003.wav
Same as before but without the folder.
和以前一样，但没有文件夹。
Now, if I run ripcheckc with "$@"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 $@and _argsthe output looks identical except for the last argument.
现在，如果我将 ripcheckc"$@"作为参数运行，它可以工作（但由于参数列表中的最后一条路径，稍后会失败）如果我使用${_args[@]}该应用程序，它将静默中止。当我回显时$@，_args除了最后一个参数外，输出看起来相同。
My question is - what is the difference between $@ and $_args that make the first valid input and the second not?
我的问题是 - $@ 和 $_args 使第一个有效输入和第二个无效输入有什么区别？
*The application is ripcheckc
*该应用程序是ripcheckc
I hope my question makes sense.
我希望我的问题是有道理的。
EDIT: Solved.
编辑：解决了。
采纳答案by devnull
Assuming that you already have an array, you can say:
假设你已经有了array，你可以说：
unset "array[${#array[@]}-1]"
For example, if your script contains:
例如，如果您的脚本包含：
array=( "$@" )
unset "array[${#array[@]}-1]" # Removes last element -- also see: help unset
for i in "${array[@]}"; do
echo "$i"
done
invoking it with: bash scriptname foo bar bazproduces:
调用它：bash scriptname foo bar baz产生：
foo
bar
回答by starfry
I have used this bashone-liner before
我以前用过这个bash单线
set -- "${@:1:$(($#-1))}"
It sets the argument list to the current argument list, less the last argument.
它将参数列表设置为当前参数列表，减去最后一个参数。
How it works:
这个怎么运作：
$#is the number of arguments$((...))is an arithmetic expression, so$(($#-1))is one less than the number of arguments.${variable:position:count}is a substring expression: it extractscountcharacters fromvariablestarting at position. In the special casewherevariableis@, which means the argument list, it extractscountargumentsfrom the list beginning atposition. Here,positionis1for the first argument andcountis one less than the number of arguments worked out previously.set -- arg1...argnsets the argument listto the given arguments
$#是参数的数量$((...))是一个算术表达式，所以$(($#-1))比参数的数量少一个。${variable:position:count}是一个子字符串表达式：它count从variable位置开始提取字符。在is的特殊情况下，这意味着参数列表，它从从 开始的列表中提取参数。这里，是第一个参数，比之前计算出的参数数量少一个。variable@countpositionposition1countset -- arg1...argn将参数列表设置为给定的参数
So the end result is that the argument list is replaced with a new list, where the new list is the original list except for the last argument.
所以最终的结果是参数列表被替换为一个新列表，其中新列表是除了最后一个参数之外的原始列表。
回答by chepner
You can also get all but the last argument with
你也可以得到除了最后一个参数之外的所有参数
"${@:0:$#}"
which, honestly, is a little sketchy, since it seems to be (ab)using the fact that arguments are numbered starting with 1, not 0.
老实说，这有点粗略，因为它似乎（ab）使用了这样一个事实，即参数从 1 开始编号，而不是从 0 开始。
Update: This only works due to a bug (fixed in 4.1.2 at the latest) in handling $@. It works in version 3.2.
更新：这仅由于处理$@. 它适用于 3.2 版。
