正则表达式删除[标题]

Question

我试图从我的Wordpress生成的数据库的文本字符串中删除一些HTML.

我要这个:

Marnie Stanton led us through the process first and then everyone went crazy. 
[caption id="attachment_76" align="alignnone" width="191"] One of the work stations[/caption]
[caption id="attachment_78" align="alignnone" width="300"] The group is getting some great results[/caption]
[caption id="attachment_83" align="alignnone" width="224"] You can see the prints multiplying[/caption]  

变成这个:

Marnie Stanton led us through the process first and then everyone went crazy. 

所以我想要的是从第一个[caption]到最后一个[/caption]被删除的所有内容.

我从这开始:

(\[caption\s+?[^]]+\])

这只删除了第一个标签.

Answer 1

你可能想要使用这样的东西

$string = 'Marnie Stanton led us through the process first and then everyone went crazy. 
[caption id="attachment_76" align="alignnone" width="191"] One of the work stations[/caption]
[caption id="attachment_78" align="alignnone" width="300"] The group is getting some great results[/caption]
I want to keep this !
[caption id="attachment_83" align="alignnone" width="224"] You can see the prints multiplying[/caption]';

$new_string = preg_replace('#\s*\[caption[^]]*\].*?\[/caption\]\s*#is', '', $string);
echo $new_string;

输出:

Marnie Stanton首先引导我们完成整个过程然后每个人都疯了.我想保留这个!

说明:

• 修饰符is:i表示匹配不区分大小写,s表示将新行与点匹配.
• \s* :匹配空格0次或更多次
• \[caption : 比赛 [caption
• [^]]*:匹配除]0或更多次以外的任何内容
• \] : 比赛 ]
• .*?\[/caption\]:匹配任何东西,直到[/caption]找到(并匹配[/caption])
• \s* :匹配空格0次或更多次

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