preg_match 第二个斜杠后向后看

Question

这是我的字符串：

stringa/stringb/123456789,abc,cde

和 preg_match 之后：

preg_match('/(?<=\/).*?(?=,)/',$array,$matches);

输出为：

stringb/123456789

如何更改 preg_match 以在第二个斜线（或最后一个斜线之后）后提取字符串？

期望的输出：

123456789

Answer 1

您可以将 / 以外的任何内容匹配为

/(?<=\/)[^\/,]*(?=,)/

[^\/,]* 否定字符 class 匹配 , 或 \

Regex Demo

例子

preg_match('/(?<=\/)[^\/,]*(?=,)/',$array,$matches);
// $matches[0]
// => 123456789

Answer 2

应该这样做。

<?php
$array = 'stringa/stringb/123456789,abc,cde';
preg_match('~.*/(.*?),~',$array,$matches);
echo $matches[1];
?>

忽略最后一个正斜杠 (.*/) 之前的所有内容。找到最后一个正斜杠后，保留所有数据，直到第一个逗号 ((.*?),).

Answer 3

您不需要使用 lookbehind，即：

$string = "stringa/stringb/123456789,abc,cde";
$string = preg_replace('%.*/(.*?),.*%', '', $string );
echo $string;
//123456789

演示：

http://ideone.com/IxdNbZ

正则表达式解释：

.*/(.*?),.*

Match any single character that is NOT a line break character «.*»
   Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Match the character “/” literally «/»
Match the regex below and capture its match into backreference number 1 «(.*?)»
   Match any single character that is NOT a line break character «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “,” literally «,»
Match any single character that is NOT a line break character «.*»
   Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»



Insert the text that was last matched by capturing group number 1 «»

preg_match 第二个斜杠后向后看

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