提取“$@"中最后一个参数之前的参数 [英] Extract parameters before last parameter in "$@"

问题描述

我正在尝试创建一个 Bash 脚本，该脚本将从命令行给出的最后一个参数提取到一个变量中，以便在其他地方使用.这是我正在处理的脚本:

I'm trying to create a Bash script that will extract the last parameter given from the command line into a variable to be used elsewhere. Here's the script I'm working on:

#!/bin/bash
# compact - archive and compact file/folder(s)

eval LAST=$$#

FILES="$@"
NAME=$LAST

# Usage - display usage if no parameters are given
if [[ -z $NAME ]]; then
  echo "compact <file> <folder>... <compressed-name>.tar.gz"
  exit
fi

# Check if an archive name has been given
if [[ -f $NAME ]]; then
  echo "File exists or you forgot to enter a filename.  Exiting."
  exit
fi

tar -czvpf "$NAME".tar.gz $FILES

由于第一个参数可以是任意数量，我必须找到一种方法来提取最后一个参数，(例如压缩文件.a文件.b文件.d文件-a-b-d.tar.gz).现在，存档名称将包含在要压缩的文件中.有没有办法做到这一点?

Since the first parameters could be of any number, I have to find a way to extract the last parameter, (e.g. compact file.a file.b file.d files-a-b-d.tar.gz). As it is now the archive name will be included in the files to compact. Is there a way to do this?

推荐答案

要从数组中删除最后一项，您可以使用以下方法:

To remove the last item from the array you could use something like this:

#!/bin/bash

length=$(($#-1))
array=${@:1:$length}
echo $array

更短的方式:

array=${@:1:$#-1}

但是 arays 是一种Bashism，尽量避免使用它们:(.

But arays are a Bashism, try avoid using them :(.

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