我的SQL - 从另一个表行引用数据

Question

我不知道是否可能。 我有两张桌子：

医院表

中的列

• hospitalID
• 名称
• AverageRating

评分表

中的列

• ID
• 额定值1
• rating2
• rating3
• rating4
• rating5
• rating6
• hospitalID

rating1到rating6是评分类型。现在按calculation我可以从average rating表中获取particular hospital来自以下SELECT查询

ratings

以上查询对我来说非常合适，但是SELECT IFNULL((SUM(charges) + SUM(behaviour) + SUM(admission) + SUM(properInformation) + SUM(hygine) + SUM(treatment))/(count(hospitalID) * 6), 0 ) AverageRating,COUNT(ID) RatingCount FROM ratings WHERE hospitalID = '111111' 我还想在我的医院餐桌上计算，因为我想对医院进行排名。

mySQL中是否有任何函数可以通过引用评级表来计算医院表的平均评分。

评级表

查询输出表

Answer 1

根据给定的表结构，您可以尝试这个。

修改

SELECT avgratings.*, @curRow := @curRow + 1 AS hospitalRank
FROM (
    SELECT (SUM(r.`rating1`)+SUM(r.`rating2`)+SUM(r.`rating3`)+SUM(r.`rating4`)+SUM(r.`rating5`)+SUM(r.`rating6`))/(COUNT(r.`hospitalID`)*6) AS AverageRating, h.hospitalID
    FROM hospitals h INNER JOIN ratings r
    ON h.`hospitalID`=r.`hospitalID`
    WHERE 1 GROUP BY r.`hospitalID` 
) avgratings JOIN (SELECT @curRow := 0) rank
ORDER BY avgratings.AverageRating DESC

第二次查询以获得特定医院的排名。

SELECT tablea.*
FROM (
    SELECT avgratings.*, @curRow := @curRow + 1 AS hospitalRank
    FROM (
        SELECT (SUM(r.`rating1`)+SUM(r.`rating2`)+SUM(r.`rating3`)+SUM(r.`rating4`)+SUM(r.`rating5`)+SUM(r.`rating6`))/(COUNT(r.`hospitalID`)*6) AS AverageRating, h.hospitalID
        FROM hospitals h INNER JOIN ratings r
        ON h.`hospitalID`=r.`hospitalID`
        WHERE 1 GROUP BY r.`hospitalID` 
    ) avgratings JOIN (SELECT @curRow := 0) rank

) tablea
WHERE tablea.hospitalID=1 ORDER BY tablea.AverageRating DESC

将1块中的WHERE替换为hospitalID。

Answer 2

我不确定我理解你的问题......

您可以获得单个记录的平均值：

charges + behaviour + admission + properInformation + hygine + treatment / 6

你得到了总的平均值：

avg(charges + behaviour + admission + properInformation + hygine + treatment / 6)

我命令每个医院得到它，你按医院分组：

select
  hospitalid,
  avg(charges + behaviour + admission + properinformation + hygine + treatment / 6) as avr,
  count(*) as rating_count
from ratings
group by hospitalid
order by 2 desc;

您可以通过加入医院表来选择医院数据（例如医院名称）。 E.g。

select
  h.hospitalid,
  h.name,
  avg(r.charges + r.behaviour + r.admission +
      r.properinformation + r.hygine + r.treatment / 6) as average_rating,
  count(*) as rating_count
from hospitals h
left join ratings r on r.hospitalid = h.hospitalid
group by h.hospitalid
order by average_rating desc;