如何在单引号引起来的字符串中转义单引号 [英] How to escape single quotes within single quoted strings

问题描述

比方说，您有一个Bash alias，例如:

Let's say, you have a Bash alias like:

alias rxvt='urxvt'

效果很好.

但是:

alias rxvt='urxvt -fg '#111111' -bg '#111111''

将不起作用，也将不会:

won't work, and neither will:

alias rxvt='urxvt -fg \'#111111\' -bg \'#111111\''

那么一旦转义了引号，您如何最终匹配字符串中的开始和结束引号?

So how do you end up matching up opening and closing quotes inside a string once you have escaped quotes?

alias rxvt='urxvt -fg'\''#111111'\'' -bg '\''#111111'\''

似乎难以理解，尽管如果允许您将它们串联在一起的话，它将表示相同的字符串.

seems ungainly although it would represent the same string if you're allowed to concatenate them like that.

推荐答案

如果您确实要在最外层使用单引号，请记住可以同时粘贴两种引号.示例:

If you really want to use single quotes in the outermost layer, remember that you can glue both kinds of quotation. Example:

 alias rxvt='urxvt -fg '"'"'#111111'"'"' -bg '"'"'#111111'"'"
 #                     ^^^^^       ^^^^^     ^^^^^       ^^^^
 #                     12345       12345     12345       1234

'"'"'如何解释为'的解释:

1. '使用单引号结束第一个引号.
2. "使用双引号开始第二个报价.
3. '引号字符.
4. "使用双引号结束第二个报价.
5. '使用单引号将第三个引号引起来.

1. ' End first quotation which uses single quotes.
2. " Start second quotation, using double-quotes.
3. ' Quoted character.
4. " End second quotation, using double-quotes.
5. ' Start third quotation, using single quotes.

如果您没有在(1)与(2)之间或(4)与(5)之间放置任何空格，则Shell会将该字符串解释为一个长字.

If you do not place any whitespaces between (1) and (2), or between (4) and (5), the shell will interpret that string as a one long word.

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