php 读取目录下图像文件
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.'" .$directory.'/'.$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);