我正在测试这个 fql 多查询
"status":"SELECT status_id,time,message,place_id,source,uid FROM status WHERE uid = \''.$pageID.'\' AND status_id IN (SELECT object_id FROM #query1))",
这必须返回所选“用户”（$pageID）在所选 status_id 上的状态列表
结果是错误的，我得到相同的结果
"status":"SELECT status_id,time,message,place_id,source,uid FROM status WHERE status_id IN (SELECT object_id FROM #query1))",
第二个查询打印具有许多不同 UID 的 27 状态，第一个查询打印相同的 27 状态但都具有相同的 UID？
任何想法/建议？
更新：我在这里粘贴完整的多查询使用
$multiQuery = '{
"query1":"SELECT object_id,post_id,object_type FROM like WHERE user_id = me() limit 200",
"status":"SELECT status_id,time,message,place_id,source,uid FROM status WHERE (uid = \''.$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)"
}';