Question

我目前有以下表格：

小屋

LodgeID (PK)
Lodge
etc

分数

ScoreID (PK)
Score
CategoryID (FK)
LodgeID (FK)

（也用于计算ScoreCount和AverageScore）

categoriesFull

CategoryID (PK)
Category

国家

CountryID (PK)
Country

提名

NominationID (PK)
LodgeID (FK)
CategoriesFullID (FK) (--to categoriesFull.CategoryID)
SW_finalist
SW_rank

所以，如果我有以下内容：

小屋

LodgeID, Lodge
798, Borana Lodge

分数

ScoreID, LodgeID, CategoryID, Score
1001, 798, 7, 3
1002, 798, 7, 6
1003, 798, 7, 9

categoriesFull

CategoryID 
7

国家

CountryID
3

提名

NominationID, LodgeID, CategoryID, SW_finalist, SW_rank
1234, 798, 7, Yes, 2

我希望能够输出

LodgeID, Lodge, CategoryID, CountryID, ScoreCount, AverageScore, NominationID, SW_finalist, SW_rank
798, Borana Lodge, 7, 3, 3, 6, 1234, Yes, 2

这是我在介绍提名表之前所拥有的：

SELECT 
   ScoreID, COUNT(ScoreID) as scoreCount, AVG(Score) as AverageScore, lodges.LodgeID, 
   Lodge, Country, lodges.CountryID, CategoryID, SW_finalist, SW_rank 
FROM 
   lodges 
INNER JOIN 
   countries ON lodges.CountryID = countries.CountryID 
INNER JOIN 
   nominations ON lodges.LodgeID = nominations.LodgeID 
INNER JOIN 
   scores ON lodges.LodgeID = scores.LodgeID 
WHERE 
   lodges.CountryID = 3  
   AND CategoryID = '7' 
GROUP BY 
   Lodge

这似乎有助于输出：

LodgeID, Lodge, CategoryID, CountryID, ScoreCount, AverageScore
798, Borana Lodge, 7, 3, 3, 6

我只需要在提名表中标记这三个字段。我认为提名的加入也需要在categoriesFull.CategoryID上，但是当我尝试类似

的时候

SELECT 
   ScoreID, COUNT(ScoreID) as scoreCount, AVG(Score) as AverageScore, lodges.LodgeID, 
   Lodge, Country, lodges.CountryID, CategoryID, SW_finalist, SW_rank 
FROM 
   lodges 
INNER JOIN 
   countries ON lodges.CountryID = countries.CountryID 
LEFT OUTER JOIN 
   nominations ON lodges.LodgeID = nominations.LodgeID AND nominations.CategoriesFullID = categoriesFull.CategoryID 
INNER JOIN 
   scores ON lodges.LodgeID = scores.LodgeID 
WHERE 
   lodges.CountryID = 3 
   AND CategoryID = '7' 
GROUP BY 
   Lodge

我收到有关categoriesFull.CategoryID是未知列的错误。

我认为所有的表都有他们需要的ID，但是非常感谢任何有关正确联接以使其工作的帮助。

实际上，它看起来并没有起作用 - 我看起来像是这样的查询：

SELECT ScoreID, COUNT(ScoreID) as scoreCount, AVG(Score) as AverageScore, lodges.LodgeID,       lodges.Lodge, Country, lodges.CountryID, categoriesFull.CategoryID, nominations.NominationID,     SW_finalist, SW_rank, 2015_Awards_Participant FROM lodges 
INNER JOIN countries ON lodges.CountryID = countries.CountryID 
INNER JOIN scores ON lodges.LodgeID = scores.LodgeID 
INNER JOIN nominations ON lodges.LodgeID=nominations.LodgeID 
INNER JOIN categoriesFull ON nominations.CategoriesFullID = categoriesFull.CategoryID AND     scores.categoryID = categoriesFull.CategoryID
WHERE lodges.CountryID = 3 AND scores.CategoryID=7 GROUP BY Lodge

但看起来它正在对ScoreID进行多次计算 - 如果Lodge在提名表中有3项提名，则得分数乘以3。

mySQL查询语法 - 多个表和连接

0 个答案: