PHP抓取远程图片

 1 <?php
 2 function getImage($url, $filename=""){
 3     if($url == ""){
 4         return false;
 5     }
 6     $ext = strrchr($url, ".");
 7     if($ext!=".gif" && $ext!=".jpg" && $ext!=".bmp" && $ext!=".png"){
 8         echo "格式不支持";
 9         return false;
10     }
11     if($filename == ""){
12         $filename = time().$ext;
13     }
14     ob_start();
15     readfile($url);
16     $img = ob_get_contents();
17     ob_end_clean();
18     $size = strlen($img);
19     $fp = fopen($filename, "a");
20     fwrite($fp, $img);
21     fclose($fp);
22     return $filename;
23 }
24 $result = getImage("图片路径","图片名称");
25 if($result){
26     echo '<img src="'.$result.'">';
27 }else{
28     echo "抓取失败";
29 }
30 ?>
原文地址:https://www.cnblogs.com/hell0x/p/5485803.html