处理除第一个参数以外的所有参数(在bash脚本中) [英] Process all arguments except the first one (in a bash script)

问题描述

我有一个简单的脚本，其中第一个参数保留给文件名，所有其他可选参数应传递给脚本的其他部分.

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.

我使用Google找到了此Wiki ，但它提供了一个字面示例:

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

echo "${@: -1}"

我什么都无法工作，例如:

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

echo "${@:2}"

或

echo "${@:2,1}"

我从终端获得错误替换".

I get "Bad substitution" from the terminal.

问题是什么，除了传递给bash脚本的第一个参数外，我该如何处理?

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

推荐答案

使用此:

echo "${@:2}"

---

以下语法:

---

The following syntax:

echo "${*:2}"

也可以使用，但不建议使用，因为它是 @ Gordon 已经解释过，使用*，它会将所有参数一起作为带有空格的单个参数运行，而@保留它们之间的间隔(即使某些参数本身包含空格).它与echo没什么区别，但是对许多其他命令来说很重要.

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.

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