Question

我有一个脚本，该脚本从数组中的URL下载图像并将图像复制到服务器上的文件夹中。 / folder /具有写权限（CHMOD 777）。我收到错误消息（dowwload失败...），此脚本出了什么问题？

<?php
$IMAGES = 'https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/10s%20Or%20Better.png,https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/Alice%20and%20the%20Red%20Queen.png,https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/All%20American.png,https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/Astro%20Roulette.png';

    $images = explode(', ',$IMAGES); 

    foreach($images as $image) {

       $name = basename($image);
       $newfile = 'https://myserver.com/folder/'.$name;

       if(copy($image, $newfile)){
          echo 'Successfully downloaded '. $image;
       }else{
          echo 'Download failed for '. $image;
       }
    }

    ?>

Answer 1

有些事情在这里有些不合时宜

您的网址字符串由一个逗号（,）分隔。

$IMAGES = 'https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/10s%20Or%20Better.png,https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/Alice%20and%20the%20Red%20Queen.png,https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/All%20American.png,https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/Astro%20Roulette.png';

但是，您explode()是一个逗号，后跟一个空格（,）。

$images = explode(', ',$IMAGES);

您的数组无法正确生成：

var_dump($images);
/* Output:
array(1) {
  [0]=>
  string(429) "https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/10s%20Or%20Better.png,https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/Alice%20and%20the%20Red%20Queen.png,https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/All%20American.png,https://s3-eu-west-1.amazonaws.com/groove-marketing-assets/games-prod/1X2%20Gaming/Astro%20Roulette.png"
}
*/

更改您的explode()调用以使用正确的定界符（单个逗号）将项目分开：

$images = explode(',',$IMAGES);

您正在尝试复制到HTTP路径：
```
$newfile = 'https://myserver.com/folder/'.$name;
```
From the PHP docts on copy()：

如果目标是URL，则如果包装程序不支持覆盖现有文件，则复制操作可能会失败。

相反，请参考本地系统文件路径。（正确的路径字符串将取决于服务器的操作系统。）

用于在Array PHP中下载图像的脚本

1 个答案: