如何在 Bash 脚本中迭代参数

本文介绍了如何在 Bash 脚本中迭代参数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个复杂的命令，我想为其制作一个 shell/bash 脚本.我可以很容易地用 $1 来写:

I have a complex command that I'd like to make a shell/bash script of. I can write it in terms of $1 easily:

foo $1 args -o $1.ext

我希望能够将多个输入名称传递给脚本.正确的做法是什么?

I want to be able to pass multiple input names to the script. What's the right way to do it?

当然，我想处理带有空格的文件名.

And, of course, I want to handle filenames with spaces in them.

推荐答案

使用 "$@" 表示所有参数:

Use "$@" to represent all the arguments:

for var in "$@" do echo "$var" done

这将遍历每个参数并将其打印在单独的行上.$@ 的行为与 $* 类似，只是在引用时，如果参数中有空格，则会正确分解:

This will iterate over each argument and print it out on a separate line. $@ behaves like $* except that when quoted the arguments are broken up properly if there are spaces in them:

sh test.sh 1 2 '3 4' 1 2 3 4