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所以現在這個樣子:見最後的評論或流標籤爲每個朋友
$mq = array(
"usr"=>"SELECT uid2 FROM friend WHERE uid1 = me() LIMIT 70",
"basics"=>"SELECT name, uid FROM user WHERE uid IN (SELECT uid2 FROM #usr)",
"q1" =>"SELECT actor_id FROM stream_tag WHERE target_id = me() AND actor_id IN (SELECT uid2 FROM #usr)",
"q2" =>"SELECT target_id FROM stream_tag WHERE target_id IN (SELECT uid2 FROM #usr) AND actor_id = me()"
);
我試圖讓每個活躍用戶朋友之間的相互作用。任何更好的方式來做到這一點?
嗯,我需要的,並因此它只拉標籤,其中兩端相等(原因朋友到用戶/用戶到朋友),基於每個用戶朋友ID(全部壓縮成單個多重查詢) –
>>兩端相等<<這裏什麼是相等的?恐怕不會得到你...... –
基本上我設置的多重查詢(上面)拉你所有的朋友的ID。因此,q1和q2(MFQL數組的)返回每個迭代人的響應;所以如果設置爲「OR」,則會返回每個用戶的XX +而不是X {1}。請欣賞幫助。 –