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FQL query SlectTuser_id from such... Return food sign, no wrong?
原标题:FQL query SELECT user_id FROM like... returns empty array, no errors?

我的文字是使用<代码>api方法使用PHP SDK的harm。 我可以成功制作《手册》、上载照片,在用户是管理者的地方打上名人。

我愿获得<代码>用户_id。 我发现,这并非是使用图象A普朗西语,而是转至FQL的档案,最后是:

// Use current page id to build query and get likes
$query  = sprintf( SELECT user_id FROM like WHERE object_id="%s" , $this->id);
$args   = array( method  =>  fql.query ,  query  => $query);
$result = $this->sdk->api($args); // $this->sdk is Facebook configured SDK

var_dump($args, $this->sdk, $args); 

产出是空洞的。 <>? 没有错误,没有警告例外。 详细情况(可用数据sh)如下:

array
   method  => string  fql.query  (length=9)
   query  => string  SELECT user_id FROM like WHERE object_id="273552079362810"  (length=58)

object(Facebook)[3]
  protected  appId  => string  292315032146649  (length=15)
  protected  appSecret  => string  360abceefda40484db536cda4010813a  (length=32)
  protected  user  => null
  protected  signedRequest  => null
  protected  state  => null
  protected  accessToken  => string  AAAEJ2ZBX4gBkBAOQvllIySNhlu6trtuZB5uCeWuxvBcnGvq3JBuy5FULZAObdgsHZBRiPgBCcqkFDjwzdaWi4B7MIKp5AZCzPenwxSIfAGgZDZD  (length=112)
  protected  fileUploadSupport  => boolean true

array
  empty
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