vBulletin的复杂(?)名称匹配正则表达式 [英] Complex(?) Name Matching Regex for vBulletin

问题描述

我正在为论坛创建一些自定义BBcode.我正在尝试使正则表达式正确，但是已经使我迷恋了好几个星期.任何专家的建议都欢迎.

I'm creating some custom BBcode for a forum. I'm trying to get the regular expression right, but it has been eluding me for weeks. Any expert advice is welcome.

样本输入(一个非常基本的示例):

Sample input (a very basic example):

[quote=Bob]I like Candace. She is nice.[/quote]

Ashley Ryan Thomas

本质上，我想将[user] [/user] BBcode中的任何名称(从指定列表中)括起来...，当然，除了那些被引用的名称，因为这样做会导致一些严重的解析错误.

Essentially, I want to encase any names (from a specified list) in [user][/user] BBcode... except, of course, those being quoted, because doing that causes some terrible parsing errors.

所需的输出:

[quote=Bob]I like [user]Candace[/user]. She is nice.[/quote]

[user]Ashley[/user] [user]Ryan[/user] [user]Thomas[/user]

我当前的代码:

 $searchArray = array(
 '/(?i)([^=]|\b|\s|\/|\r|\n|\t|^)(Ashley|Bob|Candace|Ryan|Thomas)(\s|\r|\n|\t|,|\.(\b|\s|\.|$)|;|:|\'|"|-|!|\?|\)|\/|\[|$)/'
 );

 $replaceArray = array(
   "\\1[user]\\2[/user]\\3"
 );

 $text = preg_replace($searchArray, $replaceArray, $input);

当前产生的内容:

 [quote=Bob]I like [user]Candace[/user]. She is nice.[/quote]

 [user]Ashley[/user] Ryan [user]Thomas[/user]

请注意，[用户]标签未封装Ryan.另请注意，许多额外的正则表达式匹配字符是在论坛上出现时按需添加的，因此删除它们只会使它在其他情况下不匹配(即，不可以).当然，除非您发现正则表达式本身存在明显错误，在这种情况下，请务必指出.

Notice that Ryan isn't encapsulated by [user] tags. Also note that much of the additional regex matching characters were added on an as-needed basis as they cropped up on the forums, so removing them will simply make it fail to match in other situations (i.e. a no-no). Unless, of course, you spot a glaring error in the regex itself, in which case please do point it out.

尽管如此，任何帮助将不胜感激！谢谢.

Really, though, any assistance would be greatly appreciated! Thank you.

推荐答案

很简单，就是在搜索名称的两端匹配定界符(\s|\r|...).较差的Ashley和Ryan在测试字符串中共享一个空格字符.但是正则表达式只能匹配一次-左边界或右边界.

It's quite simply that you are matching delimiters (\s|\r|...) at both ends of the searched names. The poor Ashley and Ryan share a single space character in your test string. But the regex can only match it once - as left or right border.

这里的解决方案是使用断言.用(?<= )括住左列表，用(?= )括住右列表，使它们成为:

The solution here is to use assertions. Enclose the left list in (?<= ) and the right in (?= ) so they become:

 (?<=[^=]|\b|\s|\/|^)
 (?=\s|,|\.(\b|\s|\.|$)|;|:|\'|"|-|!|\?|\)|\/|\[|$)

顺便说一句，\s已经包含\r|\n|\t，因此您可以删除它.

Btw, \s already contains \r|\n|\t so you can probably remove that.

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