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一个网页的图片名的截取,该怎么处理

WBOY
WBOYOriginal
2016-06-13 13:29:26928browse

一个网页的图片名的截取
网页的代码里很多:

HTML code
<!--

Code highlighting produced by Actipro CodeHighlighter (freeware)
http://www.CodeHighlighter.com/

--><a href="/index.php/Image:EISL_1.1.6.gif" class="image" title="Image:EISL 1.1.6.gif"><img src="/images/0/04/EISL_1.1.6.gif" alt="Image:EISL 1.1.6.gif"    style="max-width:90%"  style="max-width:90%" longdesc="/index.php/Image:EISL_1.1.6.gif"></a>
<br>
铺砌搅拌的均质池
<br>
<a href="/index.php/Image:EISL_1.1.7.gif" class="image" title="Image:EISL 1.1.7.gif"><img src="/images/d/da/EISL_1.1.7.gif" alt="Image:EISL 1.1.7.gif"    style="max-width:90%"  style="max-width:90%" longdesc="/index.php/Image:EISL_1.1.7.gif"></a>
<br>
穿孔导流槽式均质池
<br>
<a href="/index.php/Image:EISL_1.1.8.gif" class="image" title="Image:EISL 1.1.8.gif"><img src="/images/e/e9/EISL_1.1.8.gif" alt="Image:EISL 1.1.8.gif"    style="max-width:90%"  style="max-width:90%" longdesc="/index.php/Image:EISL_1.1.8.gif"></a>

我现在只需要得到图片名(EISL 1.1.6.gif),该怎么做呢?
用正则截取还是别的什么办法呢?求指导


------解决方案--------------------
$contents = preg_replace('/poweredby_mediawiki_88x31\.png/s','',$contents); //加一步,过滤掉它。
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