Facebook：使用Graph api获取在线好友列表 [英] Facebook : Get List of Online Friends Using Graph api

问题描述

我想要在线获取用户名单，不需要Facebook SDK.My fql查询如下

  $ FQL =SELECT UID，名称，pic_square，online_presence 
从用户
其中online_presence IN（ '活性'， '空闲'）
和UID IN（
选择UID2 FROM friend WHERE uid1 = $ user 
）; 
  

我的图表URl是

  https://graph.facebook.com/fql?q=SELECT UID，名称，pic_square，online_presence FROM用户WHERE online_presence IN（ '活性'， '空闲'）和UID IN（SELECT UID2 FROM朋友WHERE UID1 = 1234）及的access_token = xxxxxxtokenxxxxx&安培;方法= GET 
  

但是，上面的URL返回null数据。

  {
data：[
 
] 
} 
  

注意：我的令牌有friends_online_presence权限。我有40位用户在线。

'的div类= h2_lin>解决方案

您需要 user_online_presence 或 friends_online_presence 权限查看 online_presence 。 （请参阅 http://developers.facebook.com/docs/reference/fql/user）

如果没有这些权限，您将收到 online_presence 作为 null ，这部分 WHERE 将不被任何内容过滤： online_presence IN（'active'，'idle'）

如果您从查询中删除此限制，您将看到具有状态的朋友列表（如果您没有此信息，则为null许可）

I want to Get list of users online without Facebook SDK.My fql Query is as below

 $fql = "SELECT uid, name, pic_square, online_presence
            FROM user
            WHERE online_presence IN ('active', 'idle')
            AND uid IN (
                SELECT uid2 FROM friend WHERE uid1 = $user
            )";

And my graph URl is

https://graph.facebook.com/fql?q=SELECT uid, name, pic_square, online_presence FROM user WHERE online_presence IN ('active', 'idle') AND uid IN (SELECT uid2 FROM friend WHERE uid1 = 1234)&access_token=xxxxxxtokenxxxxx&method=GET

But Above Url returning null data.

{
   "data": [

   ]
}

Note: my token have friends_online_presence permission .I have 40 users online.

解决方案

You need user_online_presence or friends_online_presence permissions to view online_presence. (see http://developers.facebook.com/docs/reference/fql/user)

Without these permissions you'll receive online_presence as null, and nothing will be filtered by this part of WHERE: online_presence IN ('active', 'idle')

If you'll remove this restriction from your query, you'll see the list of friends with their statuses (or nulls if you don't have this permission)

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