从PHP URL保存图像

9 回答

• 22

  如果 allow_url_fopen 设置为 true ：

  $url = 'http://example.com/image.php';
  $img = '/my/folder/flower.gif';
  file_put_contents($img, file_get_contents($url));
  

  否则使用cURL：

  $ch = curl_init('http://example.com/image.php');
  $fp = fopen('/my/folder/flower.gif', 'wb');
  curl_setopt($ch, CURLOPT_FILE, $fp);
  curl_setopt($ch, CURLOPT_HEADER, 0);
  curl_exec($ch);
  curl_close($ch);
  fclose($fp);
  

  回复于 2024-04-25T23:50:22+08:00
• 3

  copy('http://example.com/image.php', 'local/folder/flower.jpg');
  

  回复于 2024-04-25T23:50:22+08:00
• 3

  $content = file_get_contents('http://example.com/image.php');
  file_put_contents('/my/folder/flower.jpg', $content);
  

  回复于 2024-04-25T23:50:22+08:00
• 224

  在这里，示例将远程图像保存到image.jpg .

  function save_image($inPath,$outPath)
  { //Download images from remote server
      $in=    fopen($inPath, "rb");
      $out=   fopen($outPath, "wb");
      while ($chunk = fread($in,8192))
      {
          fwrite($out, $chunk, 8192);
      }
      fclose($in);
      fclose($out);
  }
  
  save_image('http://www.someimagesite.com/img.jpg','image.jpg');
  

  回复于 2024-04-25T23:50:22+08:00
• 59

  Vartec's answer与cURL对我不起作用 . 由于我的具体问题，它确实略有改善 .

  e.g.,

  当服务器上有重定向时（例如当您尝试保存Facebook Profiles 图像时），您将需要以下选项集：

  curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
  

  ---

  The full solution becomes:

  $ch = curl_init('http://example.com/image.php');
  $fp = fopen('/my/folder/flower.gif', 'wb');
  curl_setopt($ch, CURLOPT_FILE, $fp);
  curl_setopt($ch, CURLOPT_HEADER, 0);
  curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
  curl_exec($ch);
  curl_close($ch);
  fclose($fp);
  

  回复于 2024-04-25T23:50:22+08:00
• 646

  我无法使任何其他解决方案工作，但我能够使用wget：

  $tempDir = '/download/file/here';
  $finalDir = '/keep/file/here';
  $imageUrl = 'http://www.example.com/image.jpg';
  
  exec("cd $tempDir && wget --quiet $imageUrl");
  
  if (!file_exists("$tempDir/image.jpg")) {
      throw new Exception('Failed while trying to download image');
  }
  
  if (rename("$tempDir/image.jpg", "$finalDir/new-image-name.jpg") === false) {
      throw new Exception('Failed while trying to move image file from temp dir to final dir');
  }
  

  回复于 2024-04-25T23:50:22+08:00
• 9

  $img_file='http://www.somedomain.com/someimage.jpg'
  
  $img_file=file_get_contents($img_file);
  
  $file_loc=$_SERVER['DOCUMENT_ROOT'].'/some_dir/test.jpg';
  
  $file_handler=fopen($file_loc,'w');
  
  if(fwrite($file_handler,$img_file)==false){
      echo 'error';
  }
  
  fclose($file_handler);
  

  回复于 2024-04-25T23:50:22+08:00
• 1

  见file()PHP Manual：

  $url    = 'http://mixednews.ru/wp-content/uploads/2011/10/0ed9320413f3ba172471860e77b15587.jpg';
  $img    = 'miki.png';
  $file   = file($url);
  $result = file_put_contents($img, $file)
  

  回复于 2024-04-25T23:50:22+08:00
• 26

  创建一个名为images的文件夹，该文件夹位于您计划放置要创建的php脚本的路径中 . 确保它具有每个人的写权限，否则脚本将无法工作（它将无法将文件上载到目录中） .

  回复于 2024-04-25T23:50:22+08:00