（PHP）拒绝说明不包括所需要的标签

Question

这是在描述太短怎么我的剧本目前发送错误：

if(strlen($linkres->content) = minStoryLength) { // if description is too short 
     $main_smarty->assign('submit_error', 'incomplete'); 
     $main_smarty->display($the_template . '/submit_errors.tpl'); 
     $error = true; 
    } 

我希望在描述不发送相同的错误信息不包括以下标签/文字中的至少一个：

<img><youtube><googlevideo><xoinks><break><vimeo><revver><myspace> 
<veoh><wmv><dailymotion><ifilm><metacafe><tubeley><guba> 

感谢您的帮助！

Answer 1

$tags = array('img','youtube','googlevideo','xoinks','break','vimeo','revver','myspace','veoh','wmv','dailymotion','ifilm','metacafe','tubeley','guba'); 

$contains = false; 
foreach($tags as $check) 
{ 
    if(preg_match("/<".$check.">/", $linkres->content)) 
    { 
     $contains = true; 
    } 
} 

if(!$contains) 
{ 
    // here goes your error 
} 

这是英迪的解决方案的替代品。

Answer 2

1将阵列中的

2迭代数组的标签和检查值是匹配

foreach() 

http://www.php.net/manual/en/control-structures.foreach.php

3显示错误

Answer 3

我会使用一些正则表达式在这里，做一个负面的比赛

$tags = array("img", "youtube", "googlevideo", "xoinks", "break", "vimeo", "revver", "myspace", "veoh", "wmv", "dailymotion", "ifilm", "metacafe", "tubeley", "guba"); 

if ((strlen($linkres->content) < minStoryLength) // if description is too short 
|| (!preg_match("'[<](".implode("|",$tags).")[^>]*[>]'is",$linkres->content))) 
// or it does not contain any of the above 
{ 
    $main_smarty->assign('submit_error', 'incomplete'); 
    $main_smarty->display($the_template . '/submit_errors.tpl'); 
    $error = true; 
} 

编辑：做一点修正

EDIT2：我有一个缺少参数preg_match