在正则表达式模式中包含 OR (PHP)

发布于 2024-12-08 16:56:44 字数 402 浏览 0 评论 0原文

尝试创建包含或条件的模式。

$pattern["body"]='/(<\/a><\/li><\/ul><\/div><p>|<h2>)(.*)<div class="like">/s';

该模式应识别一系列“<\/a><\/li><\/ul><\/div>

”之间的任何文本或“

`”作为开始分隔符，但显然 or 条件在这种情况下不起作用。任何人都可以提供正确语法的提示！？我努力了1个小时，我已经失去了对这条规则的专注和耐心。先感谢您。`

原文

Trying to create a pattern that includes an or condition.

$pattern["body"]='/(<\/a><\/li><\/ul><\/div><p>|<h2>)(.*)<div class="like">/s';

The pattern should recognize any text between a series of "<\/a><\/li><\/ul><\/div><p>" or a "<h2>" as the beginning delimiter, but apparently the or condition does not work in this case.
Anyone can offer a hint for the correct syntax!? I am striving for 1 hour and I have lost the focus and patience in this rule.
THank you in advance.

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ヅ她的身影、若隐若现 2024-12-15 16:56:44

"(?<=</a></li></ul></div>|<h2>).*"

请参阅下面的 grep 测试：

kent$  echo "</a></li></ul></div>something"|grep -Po "(?<=</a></li></ul></div>|<h2>).*"                                                  
something

kent$  echo "<h2>something"|grep -Po "(?<=</a></li></ul></div>|<h2>).*"                                                                  
something

"(?<=</a></li></ul></div>|<h2>).*"

see below grep test:

kent$  echo "</a></li></ul></div>something"|grep -Po "(?<=</a></li></ul></div>|<h2>).*"                                                  
something

kent$  echo "<h2>something"|grep -Po "(?<=</a></li></ul></div>|<h2>).*"                                                                  
something

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杯别 2024-12-15 16:56:44

'%(</a></li></ul></div><p>|<h2>)(.*)<div class="like">%s'

我没有任何样本数据可供测试，但看起来您就在那里。我只是改变了分隔符，这样就不必转义了，我发现它更容易理解。

'%(</a></li></ul></div><p>|<h2>)(.*)<div class="like">%s'

I don't have any sample data to test against but it looks like you were there. I only changed the delimiter so it wouldn't have to be escaped, I find it's easier to understand.

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谁人与我共长歌 2024-12-15 16:56:44

这可能有效 http://www.ideone.com/zYB1n
但是，我不是 php 专家。

$regex = '/(?:<\/a>\s*<\/li>\s*<\/ul>\s*<\/div>\s*<p>|<h2>)(?<Text>[\S\s]*)(?=<div class="like">)/';
if (preg_match( $regex, $str1, $matches ))
  print_r( $matches['Text'] );

This might work http://www.ideone.com/zYB1n
But, I'm no php expert.

$regex = '/(?:<\/a>\s*<\/li>\s*<\/ul>\s*<\/div>\s*<p>|<h2>)(?<Text>[\S\s]*)(?=<div class="like">)/';
if (preg_match( $regex, $str1, $matches ))
  print_r( $matches['Text'] );

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~没有更多了~