bash - 如何在Bash脚本中迭代参数(How to iterate over arguments in a Bash script)

Question

I have a complex command that I'd like to make a shell/bash script of.

(我有一个复杂的命令，我想制作一个shell / bash脚本。)

I can write it in terms of $1 easily:

(我可以轻松地用$1来写它：)

foo $1 args -o $1.ext

I want to be able to pass multiple input names to the script.

(我希望能够将多个输入名称传递给脚本。)

What's the right way to do it?

(什么是正确的方法呢？)

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

(当然，我想处理其中包含空格的文件名。)

  ask by Thelema translate from so

Answer 1

Use "$@" to represent all the arguments:

(使用"$@"表示所有参数：)

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

This will iterate over each argument and print it out on a separate line.

(这将迭代每个参数并在单独的行上打印出来。)

$@ behaves like $* except that when quoted the arguments are broken up properly if there are spaces in them:

($ @的行为类似于$ *，但是当引用时，如果参数中有空格，则它们会被正确分解：)

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