preg_replace（）只有字符串的特定部分

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

我一直有正则表达式的麻烦,我基本上有一个网址,例如:

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

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

我需要的是preg_replace()用一个空字符串替换something_here,并保留其他所有内容。

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

我尝试了以下内容,它取代了错误的部分:

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

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

这最终只留下我要替换的部分,而不是实际替换它!

3 个解决方案

#1

The following code is based on the description you provided:

以下代码基于您提供的说明:

$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

Explanation:

^ : match begin of line

^:匹配行的开头

( : start matching group 1
- https?:// : match http or https protocol
- [^/]+ : match anything except / one or more times
- / : match /

(:开始匹配组1 https?://:匹配http或https协议[^ /] +:除了/一次或多次匹配任何东西/:匹配/

) : end matching group 1

):结束匹配组1

[^/]+ : match anything except / one or more times -/ : match /

[^ /] +:除了/一次或多次匹配任何东西 - /:匹配/

( : start matching group 2
- .* : match anything zero or more times (greedy)

(:开始匹配组2。*:匹配任何零次或多次(贪婪)

) : end matching group 2

):结束匹配组2

$ : match end of line

$:匹配行尾

#2

You could do this:

你可以这样做:

$image[0] = preg_replace('!^(http://[^/]*)/[^/]*!', '$1', $image[0]);

Or you might consider just splitting the string to work on its individual components:

或者您可以考虑将字符串拆分为适用于其各个组件:

$parts = explode('/', $image[0]);
unset($parts[3]);
$image[0] = implode('/', $parts);

#3

You could do this with a simple string replace:

您可以使用简单的字符串替换来执行此操作:

$image[0] = str_replace('/wp-content/uploads/', '/', $image[0]);

Or if you want to use a regular expression:

或者,如果要使用正则表达式:

$image[0] = preg_replace('~(http://.*?)/wp-content/uploads/(.*)~', '$1/$2', $image[0]);

#1