从服务器上的目录中获取图像文件以进行进一步处 [英] get image files from directory on server for further processing

问题描述

我在服务器上有包含图像文件的文件夹。我正在尝试获取这些文件，然后进一步上传，但我认为我使用了错误的功能。

I have folders on the server that contain image files. I'm trying to get those files and then upload them further, but I think I'm using the wrong function.

我的代码：

$dir = "../uploads/".$folderimage."/";
if ($handle = opendir($dir)) {
  while (false !== ($entry = readdir($handle))) {
    if ($entry != "." && $entry != "..") {
      echo "$entry\n";
      $handle =  fopen($entry,"wb");
      $mug->images_upload(array( "AlbumID" => "#####", "File" =>   $handle));
    }
  }
  closedir($handle);
}

不知道我在做什么，我需要做的就是传递文件到类函数 $ mug-> images-> upload 。它适用于 $ _ POST 请求，但我需要将已经上传的文件移动到该文件夹​​。

Not sure what I am doing, all I need to do is pass the file to the class function $mug->images->upload. It works from a $_POST request but I need to move the files already uploaded to the folder.

推荐答案

您发布的代码存在许多明显的问题。

There are a number of apparent issues with the code that you have posted.

$dir = "../uploads/".$folderimage."/";
if ($handle = opendir($dir)) {
  while (false !== ($entry = readdir($handle))) {
    if ($entry != "." && $entry != "..") {
      echo "$entry\n";
      $handle =  fopen($entry,"wb");
      $mug->images_upload(array( "AlbumID" => "#####", "File" =>   $handle));
    }
  }
  closedir($handle);
}

• 冲突变量。

  您已经使用 $ handle vairable来存储目录句柄，但以后才会覆盖它使用循环内的文件资源。一旦你在循环中覆盖它，下一次调用 readdir（$ handle）就没有任何意义，你是在文件资源上调用该函数。这很容易导致无限循环，因为当 readdir（）被给予垃圾时，它可能返回 NULL 。

  You have used the $handle vairable to store the directory handle, only to later overwrite it with a file resource inside the loop. As soon as you overwrite it inside the loop, the next call to readdir($handle) does not make any sense are you are calling the function on a file resource. This could very easily lead to an infinite loop since when readdir() is given rubbish, it might return NULL.

  的路径不正确fopen（）

  Incorrect path for fopen()

  鉴于 $ folderimage =images和 $ entry =photo.jpg ，然后 fopen（）行将尝试在 photo.jpg 而不是打开图像> ../上传/图像/ photo.jpg 。你可能想要使用像 $ dir这样的东西。 $ entry ，但请继续阅读，因为你根本不应该使用 fopen（）。

  Given $folderimage = "images" and $entry = "photo.jpg", then the fopen() line will try to open the image at photo.jpg rather than ../uploads/images/photo.jpg. You likely wanted to use something like $dir . $entry, but read on as you shouldn't be using fopen() at all.

  phpSmug库的使用不正确

  文件参数必须是字符串含的路径，正在上载的本地文件（来源）。您尝试将文件资源从 fopen（）传递给它。

  The File argument must be a string containing "the path to the local file that is being uploaded" (source). You instead try to pass a file resource from fopen() to it.

  opendir（）/ readdir（）目录迭代是古老的

  opendir()/readdir() for directory iteration is ancient

  有更好的方法来遍历PHP中的目录内容。如劳伦斯的答案中所述， glob（）可能有用。

  There are better ways to traverse directory contents in PHP. As mentioned in Lawrence's answer, glob() might be useful.

  我还提倡使用 filesystem iterator SPL 。

  $dir = "../uploads/$folderimage/";
  foreach (new FilesystemIterator($dir) as $fileinfo) {
      $image_pathname = $fileinfo->getPathname();
      $mug->images_upload("AlbumID=#####", "File=$image_pathname");
  }
  




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