php 读取目录下图像文件

发布:smiling　来源: PHP粉丝网　添加日期：2014-08-19 15:05:33　浏览: 评论：0　

本款是一款php 读取目录下图像文件代码,利用了opendir来打开目录然后获取文件后缀名,判断是否为指定文件.

php 读取目录下图像文件实例代码如下:

$directory = 'gallery';

$allowed_types=array('jpg','jpeg','gif','png');

$file_parts=array();

$ext='';

$title='';

$i=0;

$dir_handle = @opendir($directory) or die("there is an error with your image directory!");

while ($file = readdir($dir_handle))

{

if($file=='.' || $file == '..') continue;



$file_parts = explode('.',$file);

$ext = strtolower(array_pop($file_parts));

$title = implode('.',$file_parts);

$title = htmlspecialchars($title);



$nomargin='';



if(in_array($ext,$allowed_types))

{

  if(($i+1)%4==0) $nomargin='nomargin';



  echo '

  <div class="pic '.$nomargin.'" style="background:url('.$directory.'/'.$file.') no-repeat 50% 50%;">

  <a href="'.$directory.'/'.$file.'" title="'.$title.'" target="_blank">'.$title.'</a>

  </div>';



  $i++;

}

}

//开源代码phpfensi.com

closedir($dir_handle);