处理除第一个之外的所有参数（在bash脚本中）

I have a simple script where the first argument is reserved for the filename, and all other optional arguments should be passed to other parts of the script.

我有一个简单的脚本，其中第一个参数是为文件名保留的，所有其他可选参数应该传递给脚本的其他部分。

Using Google I found this wiki, but it provided a literal example:

使用Google我发现了这个wiki，但它提供了一个文字示例：

echo "${@: -1}"

I can't get anything else to work, like:

我无法得到任何其他工作，例如：

echo "${@:2}"

要么

echo "${@:2,1}"

I get "Bad substitution" from the terminal.

我从终端得到“坏替换”。

What is the problem, and how can I process all but the first argument passed to a bash script?

问题是什么，除了传递给bash脚本的第一个参数之外，我该如何处理？

3 个解决方案

#1

442

Use this:

用这个：

echo "${@:2}"

The following syntax:

以下语法：

echo "${*:2}"

would work as well, but is not recommended, because as @Gordon already explained, that using *, it runs all of the arguments together as a single argument with spaces, while @ preserves the breaks between them (even if some of the arguments themselves contain spaces). It doesn't make the difference with echo, but it matters for many other commands.

也可以使用，但是不推荐，因为正如@Gordon已经解释的那样，使用*，它将所有参数作为带有空格的单个参数一起运行，而@保留它们之间的中断（即使一些参数本身包含空格）。它与echo没有区别，但对许多其他命令来说很重要。

#2

132

If you want a solution that also works in /bin/sh try

如果你想要一个也适用于/ bin / sh的解决方案

first_arg="$1"
shift
echo First argument: "$first_arg"
echo Remaining arguments: "$@"

shift [n] shifts the positional parameters n times. A shift sets the value of $1 to the value of $2, the value of $2 to the value of $3, and so on, decreasing the value of $# by one.

shift [n]将位置参数移位n次。移位将$ 1的值设置为$ 2的值，$ 2的值设置为$ 3，依此类推，将$＃的值减1。

#3

http://wiki.bash-hackers.org/scripting/posparams

It explains the use of shift (if you want to discard the first N parameters) and then implementing Mass Usage (look for the heading with that title).

它解释了shift的使用（如果你想丢弃前N个参数），然后实现Mass Usage（查找带有该标题的标题）。

#1

442