如何在正则表达式中将 $1 与数字连接 [英] How to concatenate $1 with number in a regex

问题描述

我想在两个模式之间插入数字:

I would like to insert the number between two patterns:

$s = 'id="" value="div"';
$s =~ s/(id=")(".*)/$1123$2/;

当然，我得到了错误和 " value="div" 结果.预期结果是 id="123" value="div".

Of course I got the error and " value="div" as a result. The expected result is id="123" value="div".

在替换中，我的意思是 1 美元，然后是数字 123，然后是 2 美元，但不是 1123 美元，然后是 2 美元.正则表达式中的正确替换是什么?我想用一个正则表达式来完成.

In the replacement I meant $1, then number 123 and then $2, but not $1123 and then $2. What is the correct replacement in the regex? I would like to do it in one single regex.

谢谢.

推荐答案

$s =~ s/(id=")(".*)/$1123$2/;  # Use of uninitialized value $1123 in concatenation (.) or string

尽管您希望它替代 $1，但 perl 将其视为变量 $1123.Perl 无法知道您的意思是 $1.因此，您需要将 variablisation 限制为 $1，将其指定为 ${1}:

Though you expected it to substitute for $1, perl sees it as the variable $1123. Perl has no way to know that you meant $1. So you need to limit the variablisation to $1 by specifying it as ${1}:

$s =~ s/(id=")(".*)/${1}123$2/;

在脚本顶部包含以下内容始终是个好主意.它们将为您节省大量时间和精力.

It is always a good idea to include the following at the top of your scripts. They will save you a lot of time and effort.

use strict;
use warnings;

例如，运行包含上述模块的脚本会导致错误消息:

For example, running your script with the above modules included results in the error message:

Use of uninitialized value $1123 in concatenation (.) or string at /tmp/test.pl line 7.

(忽略报告的脚本名称和行号.)它清楚地说明了 perl 的预期.

(Disregard the reported script name and line numbers.) It clearly states what perl expected.

另一种使用后视和前瞻断言的方法:

Another approach using look-behind and look-ahead assertions:

$s =~ s/(?<=id=")(?=")/123/;

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