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下面的查询似乎有一个问题来计算“ratingValue”,因为SUM内有SUM(h.liked)。不能做SUM的SUM吗?
SELECT d.itemID1 as item,
sum(d.sum + d.count*SUM(h.liked))/sum(d.count) as ratingValue
FROM history h, dev d
WHERE h.userID=:id_user
AND d.itemID1<>h.itemID
AND d.itemID2=h.itemID
GROUP BY d.itemID1,h.itemID
为了更好地理解这是当初和工作查询(从趿拉一种算法):
我刚刚替补的“评级”表“历史”,因为在我的情况r.ratingValue是所有的 “喜欢” 一个用户给出一个ITEMID的(=> r.ratingValue = SELECT SUM(喜欢)从历史中GROUP BY h.itemID)的总和:
SELECT d.itemID1 as item,
sum(d.sum + d.count*r.ratingValue)/sum(d.count) as ratingValue
FROM rating r, dev d
WHERE r.userID=$userID
AND d.itemID1<>r.itemID
AND d.itemID2=r.itemID
GROUP BY d.itemID1
你得到了什么错误? – 2012-08-01 02:16:38
我有“组功能无效” – Anon 2012-08-01 02:32:00