如何将fb图片网址解析为$_GET？我使用的是javascript和php：
https://fbcdn-profile-a.akamaihd.net/hprofile-ak-xaf1/v/t1.0-1/c0.8.50.50/p50x50/10418963_845746262178119_4978000552642842671_n.jpg?oh=21e838bf609a8f4b0137a786c7434ee3&oe=58144D56&__gda__=1479268869_857fd43604c037bdf53d104a318910c6
$image= $_GET[picture];
答案 0 :(得分：0)
$source="https://fbcdn-profile-a.akamaihd.net/hprofile-ak-xaf1/v/t1.0-1/c0.8.50.50/p50x50/10418963_845746262178119_4978000552642842671_n.jpg?oh=21e838bf609a8f4b0137a786c7434ee3&oe=58144D56&__gda__=1479268869_857fd43604c037bdf53d104a318910c6";
$target="server_file_path/image.jpg";//location and name on local server, may not be a jpg, you need to check the file type.
$Img = file_get_contents($source); //get the image from the remote server
file_put_contents($target, $Img); //save the image locally
答案 1 :(得分：0)
您可以使用
file_get_contents($file)
接受URI