preg_match查找第二个斜杠后面的内容

Question

这是我的字符串：

stringa/stringb/123456789,abc,cde

并且在 preg_match 之后：

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

输出结果为：

stringb/123456789

我该如何修改preg_match函数来提取第二个斜杠（或最后一个斜杠）后面的字符串？

期望的结果是：

123456789

- Driver

Answer 1

您可以匹配除/以外的任何内容。

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

• [^\/,]* 表示匹配除了 , 或者 \ 以外的任何字符

正则表达式演示

示例

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

- nu11p01n73R

---

谢谢，它有效！难道不应该是[...]而不是,或/吗？ - Driver

@Driver 在正则表达式中，[^ ] 表示否定字符集。因此，如果要匹配除了逗号和斜杠之外的任何字符，应该使用 [^,\/]。 - nu11p01n73R

Answer 2

这应该可以解决问题。

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

忽略掉最后一个反斜杠（.*/）之前的所有内容。一旦找到最后一个反斜杠，保留所有数据直到第一个逗号出现（(.*?),）。

- chris85

Answer 3

你不需要使用回顾后发表达式，即：

$string = "stringa/stringb/123456789,abc,cde";
$string = preg_replace('%.*/(.*?),.*%', '$1', $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) «*»

$1

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

- Pedro Lobito