preg_replace()仅字符串的特定部分 [英] preg_replace() Only Specific Part Of String

问题描述

我总是在使用正则表达式时遇到麻烦，例如，我基本上有一个网址:

I always have trouble with regex, I basically have a url, for example:

http://somedomain.com/something_here/bla/bla/bla/bla.jpg

我需要的是preg_replace()，用空字符串替换something_here，并保持其他所有内容不变.

What I need is a preg_replace() to replace the something_here with an empty string, and leave everything else in tact.

我尝试了以下方法，它替换了错误的部分:

I have tried the following and it replaces the wrong parts:

$image[0] = preg_replace('/http:\/\/(.*)\/(.*)\/wp-content\/uploads\/(.*)/','$2' . '',$image[0]);

这最终只剩下我要替换的部分，而不是实际替换它！

This ends up leaving only the part I want to replace, rather than actually replacing it!

推荐答案

以下代码基于您提供的描述:

The following code is based on the description you provided:

$url = 'http://somedomain.com/something_here/bla/bla/bla/bla.jpg';
$output = preg_replace('#^(https?://[^/]+/)[^/]+/(.*)$#', '$1$2', $url);
echo $output; // http://somedomain.com/bla/bla/bla/bla.jpg

说明:

• ^:匹配行的开头
• (:开始匹配组1
  • https?://:匹配http或https协议
  • [^/]+:匹配/以外的任何内容一次或多次
  • /:匹配/
  • ^ : match begin of line
  • ( : start matching group 1
    • https?:// : match http or https protocol
    • [^/]+ : match anything except / one or more times
    • / : match /
    • .*:匹配任何零次或多次(贪婪)
    • .* : match anything zero or more times (greedy)

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