如何从已编辑的变量加载图像？

问题描述：

Hello and sorry for the noobie question
Let's say i have a variable with value url

<input type="text" name="link">

Which submits the following link :

domain.com/profile/deCKfkiriCJFff

Since the link above is not a valid img src i would like to display an image with the same source but with one slight change :

domain.com/profile/images/deCKfkiriCJFff

^ valid image link

对于noobie问题，您好抱歉
我说我有一个值为url的变量 p>

 ＆lt; input type =“text”name =“link”＆gt; 
  code>  pre> 
 
 提交以下链接：  p> 
 
 
  domain.com/profile/deCKfkiriCJFff 
  code>  pre> 
 
 由于上面的链接不是有效的img src，我想 显示具有相同来源但略有改变的图像： p> 
 
 
  domain.com/profile/images/deCKfkiriCJFff 
  code>  pre> 
 
   ^有效图片链接 p> 
  div>

答

Use explode

$domain = 'domain.com/profile/deCKfkiriCJFff';
$exploded = explode('profile', $domain);
$validUrl =  $exploded[0].'profile/images'.$exploded[1];
echo $validUrl;

如何从已编辑的变量加载图像？

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