Question

我从中获取所有图像。

preg_match_all( !http://.+.(?:jpe?g|png|gif)!Ui  , $content , $matches);

如果它有“坏字”的话,那么它如何阻止所有形象。

 /(list|of|bad|words)/i

How can I combine them and save the result instead in variable instead of print_r? My purpose is to delete from content all those images and produce the "clean" content.

谢谢!

Answer 1

视<代码>连续<>/代码”的格式而定,视具体情况而定,需要一些 t:

<?php
    $content =  http://www.example.com/images/image001.jpg 
                http://www.example.com/images/image002.jpeg 
                http://www.example.com/images/list001.jpg 
                http://www.example.com/images/image003.png 
                http://www.example.com/images/bad002.png 
                http://www.example.com/images/image004.gif 
                http://www.example.com/images/words003.jpg ;

    preg_match_all( !http://.+.(?:jpe?g|png|gif)!Ui , $content, $matches);

    preg_match_all( /S+(list|of|bad|words)S+/i , $content, $bads);

    $filtered = array_values(array_diff($matches[0], $bads[0]));

    print_r($filtered);
?>

<<>Output>:

Array
(
    [0] => http://www.example.com/images/image001.jpg
    [1] => http://www.example.com/images/image002.jpeg
    [2] => http://www.example.com/images/image003.png
    [3] => http://www.example.com/images/image004.gif
)

Answer 2

或许最快的办法是管理第一个ex,然后在以后过滤坏.。也许,你会以非常容易的反馈来这样做。不过,我不想把reg子结合起来。

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