Php读取目录文件

Question

我有一个脚本通过一个有3个图像的目录

$imglist='';
$img_folder = "path to my image";

//use the directory class
$imgs = dir($img_folder);

//read all files from the  directory, checks if are images and ads them to a list 
while ($file = $imgs->read()) {
  if (eregi("gif", $file) || eregi("jpg", $file) || eregi("png", $file))
    $imglist .= "$file ";
} 
closedir($imgs->handle);

//put all images into an array
$imglist = explode(" ", $imglist);

//display image
foreach($imglist as $image) {
  echo '<img src="'.$img_folder.$image.'">';
}

但我遇到的问题是显示没有图像的第4个img ..但我在该文件夹中只有3个图像.

Answer 1

无需构建一串图像,然后将该字符串分解为图像数组,而只需将图像直接添加到数组中,如Radu所述.这是更正后的代码:

$imglist = array();
$img_folder = "path to my image";

//use the directory class
$imgs = dir($img_folder);

//read all files from the  directory, checks if are images and adds them to a list 
while ($file = $imgs->read()) {
   if (eregi("gif", $file) || eregi("jpg", $file) || eregi("png", $file)){
      $imglist[] = $file;
   } 
}
closedir($imgs->handle);

//display image
foreach($imglist as $image) {
    echo '<img src="'.$img_folder.$image.'">';
}