Question

我在MySQL数据库中有两个表：

照片：

PID - PhotoID [PK], 
DateOf - DateTime of uploading, 
UID -UserID (owner of the photo)

评分：

WhoUID - UserID who rated, 
DateOf - DateTime of rate, 
RatingValue - +1 or -1 (positive or negative), 
RatingStrength - coefficient (different for each user who vote)
PID - PhotoID, [FK]

实际评分值= RatingValue * RatingStrength

有什么可能获得“当天的照片”？

规则，例如：

当天的照片必须在现场至少24小时（自上传时间起）
当天的照片必须至少有10票
当天的照片是自上传时间后24小时内Real rating value最多的照片
当天的新照片不得位于photo_of_day表格

UPD1。 10票表示 - 每张照片的表ratings中至少有10条记录

UPD2。是否有可能获得确切日期时间的“当日照片”？例如，如何获取“2011-03-11”或“2011-01-25”的当天照片？

Answer 1

select p.PID from photos p, ratings r where
r.PID = p.PID and                             ; Link ratings to photo
p.DateOf >= '$day' and p.DateOf <= '$day' and ; Get the photos uploaded on the specified date
datediff(p.DateOf, now()) > 1 and             ; photo of the day must be on site at least 24 hours
count(r.*) >= 10 and                          ; photo should have at least 10 ratings
not exists(select p.PID from photo_of_day)    ; photo should not appear in photo_of_day
group by p.PID                                ; group the results by PID
order by sum(r.RatingValue*r.RatingStrength) desc ; order descending by RealRating
limit 1                                       ; limit the results to only one

此查询可能需要一段时间，因此不对每个页面请求执行此操作是有意义的。您可以通过在午夜运行一次的脚本将结果存储在photo_of_day中。

Answer 2

像这样的东西。我不确定'当天的新照片一定不能在photo_of_day表中'，但试试这个 -

SET @exact_datetime = NOW();

SELECT p.*, SUM(r.RatingValue * r.RatingStrength) AS real_rating FROM photos p
  JOIN ratings r
    ON p.PhotoID = r.PhotoID
WHERE
  p.DateOf <= @exact_datetime - INTERVAL 1 DAY
GROUP BY
  p.PhotoID
HAVING
  count(*) >= 10
ORDER BY
  real_rating DESC
LIMIT 1

如何获得当天的照片？

2 个答案: