正则表达式 (PCRE):以存在字符串为条件匹配所有数字

问题描述

使用 PCRE，我只想捕获一行中的所有数字，但前提是该行内的任何位置都存在某个字符串(例如STRING99").

Using PCRE, I want to capture only and all digits in a single line, but only if a certain string (say "STRING99") is present anywhere within that line.

例如，考虑以下两种情况:

For example, consider these two cases:

a1 STRING99 2b c3d

a1 2b c3d

在第一种情况下，我希望结果是19923".在第二种情况下，我想要一个空结果.

In the first case, I want the result to be "19923".In the second case, I want an empty result.

我不确定这是否可行.它可能适用于可变长度的lookbehind，但在PCRE 中不受支持.此外，类似于 (?=.*STRING99.*$)(\D|(\d))* 的东西会起作用，但是重复捕获组只会捕获最后一次迭代"，意思是第二个捕获组只捕获最后一位数字.我找不到解决方法.

I am not sure this is possible. It might work with a variable-length lookbehind, but this is not supported in PCRE. Furthermore, something like (?=.*STRING99.*$)(\D|(\d))* WOULD work, but "a repeated capturing group will only capture the last iteration", meaning the second capturing group only captures the last digit. I am unable to find a workaround for this.

(这显然不难通过 2 个连续的 Regex 操作实现，但我希望在一个公式中实现.)

(This is obviously not hard to achieve with 2 consecutive Regex operations, but I want it in one formula.)

推荐答案

您可以在 preg_replace 中使用这个 PCRE 正则表达式:

You may use this PCRE regex in preg_replace:

^(?!.*STRING99).*(*SKIP)|\D+

正则表达式演示

正则表达式详情:

• ^:开始
• (?!.*STRING99):Lokahead 检查输入中是否有 STRING99
• .*(*SKIP):匹配输入的其余部分直到结束并跳过它
• |:或
• \D+:匹配 1+ 个非数字

• ^: Start
• (?!.*STRING99): Lokahead to check if we have STRING99 anywhere in input
• .*(*SKIP): Match rest of the input till end and skip it
• |: OR
• \D+: Match 1+ non-digit

PHP 代码:

$repl = preg_replace('~^(?!.*STRING99).*(*SKIP)|\D+~', '', $str);

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