遍历数组，防止通配符扩展(*) [英] Loop over array, preventing wildcard expansion (*)

问题描述

我试图弄清我认为BASH中的一个小问题，但是我很难找到正确的语法.我想遍历一组值，其中一个是星号(*)，我不希望在此过程中发生任何通配符扩展.

I'm trying to figure out what I thought would be a trivial issue in BASH, but I'm having difficulty finding the correct syntax. I want to loop over an array of values, one of them being an asterisk (*), I do not wish to have any wildcard expansion happening during the process.

 WHITELIST_DOMAINS="* *.foo.com *.bar.com"
 for domain in $WHITELIST_DOMAINS
 do
    echo "$domain"
 done

我已经有了上面的内容，并且正在尝试获得以下输出:

I have the above, and I'm trying to get the following output:

 *
 *.foo.com
 *.bar.com

代替上面的内容，我在当前目录中得到一个目录列表，然后是* .foo.com和* .bar.com

Instead of the above, I get a directory listing on the current directory, followed by *.foo.com and *.bar.com

我知道我需要在某处逃避或引用..清晨的雾霾仍然笼罩着我的脑袋.

I know I need some escaping or quoting somewhere.. the early morning haze is still thick on my brain.

我已经审查了以下问题:

I've reviewed these questions:

如何在bash?

停止shell通配符扩展?

推荐答案

您的问题是您需要一个数组，但是您编写了一个包含元素的单个字符串，这些元素之间有空格.请改用数组.

Your problem is that you want an array, but you wrote a single string that contains the elements with spaces between them. Use an array instead.

WHITELIST_DOMAINS=('*' '*.foo.com' '*.bar.com')

始终在变量替换处使用双引号(即"$foo")，否则shell将变量的值拆分为单独的单词并将每个单词视为文件名通配符模式.命令替换也是如此:"$(somecommand)".对于数组变量，使用"${array[@]}"展开到数组元素的列表.

Always use double quotes around variable substitutions (i.e. "$foo"), otherwise the shell splits the the value of the variable into separate words and treats each word as a filename wildcard pattern. The same goes for command substitution: "$(somecommand)". For an array variable, use "${array[@]}" to expand to the list of the elements of the array.

for domain in "${WHITELIST_DOMAINS[@]}"
 do
    echo "$domain"
 done

有关更多信息，请参见关于数组的bash常见问题解答.

For more information, see the bash FAQ about arrays.

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