我在一个允许用户查看照片的网站上工作..我希望他们能够按下照片并将照片直接发送到他们的 facebook 帐户...到目前为止,我找到了允许用户从本地上传照片的文档mashine操作方法:使用 Graph API 将照片上传到用户的个人资料。有没有办法修改此代码,以便从另一个网站的 http 帖子中接收 url?
$app_id = "YOUR_APP_ID";
$app_secret = "YOUR_APP_SECRET";
$post_login_url = "YOUR_POST-LOGIN_URL";
$album_name = 'YOUR_ALBUM_NAME';
$album_description = 'YOUR_ALBUM_DESCRIPTION';
$code = $_REQUEST["code"];
//Obtain the access_token with publish_stream permission
if(empty($code))
{
$dialog_url= "http://www.facebook.com/dialog/oauth?"
. "client_id=" . $app_id
. "&redirect_uri=" . urlencode($post_login_url)
. "&scope=publish_stream";
echo("<script>top.location.href='" . $dialog_url .
"'</script>");
}
else {
$token_url= "https://graph.facebook.com/oauth/"
. "access_token?"
. "client_id=" . $app_id
. "&redirect_uri=" . urlencode( $post_login_url)
. "&client_secret=" . $app_secret
. "&code=" . $code;
$response = file_get_contents($token_url);
$params = null;
parse_str($response, $params);
$access_token = $params['access_token'];
// Create a new album
$graph_url = "https://graph.facebook.com/me/albums?"
. "access_token=". $access_token;
$postdata = http_build_query(
array(
'name' => $album_name,
'message' => $album_description
)
);
$opts = array('http' =>
array(
'method'=> 'POST',
'header'=>
'Content-type: application/x-www-form-urlencoded',
'content' => $postdata
)
);
$context = stream_context_create($opts);
$result = json_decode(file_get_contents($graph_url, false,
$context));
// Get the new album ID
$album_id = $result->id;
//Show photo upload form and post to the Graph URL
$graph_url = "https://graph.facebook.com/". $album_id
. "/photos?access_token=" . $access_token;
echo '<html><body>';
echo '<form enctype="multipart/form-data" action="'
.$graph_url. ' "method="POST">';
echo 'Adding photo to album: ' . $album_name .'<br/><br/>';
echo 'Please choose a photo: ';
echo '<input name="source" type="file"><br/><br/>';
echo 'Say something about this photo: ';
echo '<input name="message" type="text"
value=""><br/><br/>';
echo '<input type="submit" value="Upload" /><br/>';
echo '</form>';
echo '</body></html>';
}
?>