获取所有<a>标签的内容与在PHP

Question

类我得到这个代码：

..some code here.. 
<a href="members/username1.12345/" class="username bigs" title="">Username 1</a> 
..some code here.. 
..some code here.. 
<a href="members/username2.55334/" class="username bigs" title="">Username 2</a> 
..some code here.. 
..some code here.. 
<a href="members/username3.23455/" class="username bigs" title="">Username 3</a> 
..some code here.. 
..some code here.. 
<a href="members/username4.45545/" class="username bigs" title="">Username 4</a> 
..some code here.. 

如何获得用户名1，用户名2，用户名3，4名等。与preg_match_all（）？

我不想使用SimpleHTMLdom或其他..我想用preg_match_all（）做到​​这一点...如何做到这一点？我尝试了很多正则表达式的...但他们都没有工作..

Answer 1

正则表达式是一个可怕的想法。但仍然 - 试试这个：@<a href="members/.+/" class="username bigs" title="">(.+)</a>@U

$subject = '..some code here.. 
<a href=\"members/username1.12345/\" class=\"username bigs\" title=\"\">Username 1</a> 
..some code here.. 
..some code here.. 
<a href=\"members/username2.55334/\" class=\"username bigs\" title=\"\">Username 2</a> 
..some code here.. 
..some code here.. 
<a href=\"members/username3.23455/\" class=\"username bigs\" title=\"\">Username 3</a> 
..some code here.. 
..some code here.. 
<a href=\"members/username4.45545/\" class=\"username bigs\" title=\"\">Username 4</a> 
..some code here..'; 

$pattern = '@<a href="members/.+/" class="username bigs" title="">(.+)</a>@U'; 

$result = preg_match_all($pattern, $subject, $subpattern); 

echo '<pre>'.print_r($subpattern, TRUE).'</pre>';