错误的结果与简单的fql状态

Question

我正在测试这个fql multiquery

"status":"SELECT status_id,time,message,place_id,source,uid FROM status WHERE uid = \''.$pageID.'\' AND status_id IN (SELECT object_id FROM #query1))",

这必须在选定的status_id

上选择“user”（$ pageID）返回状态列表

结果是错误的，我得到的结果与

相同

"status":"SELECT status_id,time,message,place_id,source,uid FROM status WHERE status_id IN (SELECT object_id FROM #query1))",

如果第二个查询打印27状态有许多不同的UID，第一个quety打印相同的27状态，但都具有相同的UID ??

任何想法/建议??

更新：我在这里粘贴了完整的多个查询

$multiQuery = '{ "query1":"SELECT object_id,post_id,object_type FROM like WHERE user_id = me() limit 200", "status":"SELECT status_id,time,message,place_id,source,uid FROM status WHERE (uid = \''.$pageID.'\' AND status_id IN (SELECT object_id FROM #query1))", "photo":"SELECT object_id,created,owner FROM photo WHERE owner = \''.$pageID.'\' AND object_id IN (SELECT object_id FROM #query1)", "link":"SELECT created_time,owner FROM link WHERE owner = \''.$pageID.'\' AND link_id IN (SELECT object_id FROM #query1)" }';