Question

我正在加入3个表并得到错误的结果。目标是列出餐馆餐桌中的所有餐厅，如果评级存在，则显示任何餐厅的评级，否则显示为空，仅适用于汉堡的评级。

这是SQL：

    SELECT r.RestaurantID, RestaurantName, cr.Rating FROM Restaurant r
      LEFT JOIN CustRating cr ON cr.RestaurantID = r.RestaurantID
      LEFT JOIN FoodType ft ON ft.FoodTypeID = cr.FoodTypeID AND 
        ft.FoodTypeName = 'Burger'

结果如下：

enter image description here

然而'Cafe C'应该有Rating = null，因为我只想显示Burgers的评级。什么是正确的SQL？

用于创建表并填充数据的SQL语句：

CREATE TABLE [dbo].[Restaurant](
    [RestaurantID] [int]  NOT NULL,
    [RestaurantName] [varchar](250) NOT NULL

)

CREATE TABLE [dbo].[FoodType](
    [FoodTypeID] [int]  NOT NULL,
    [FoodTypeName] [varchar](50) NOT NULL

)
CREATE TABLE [dbo].[CustRating](
    [RestaurantID] [int] NOT NULL,
    [FoodTypeID] [int] NOT NULL,
    [Rating] [smallint] NOT NULL
 ) 

 BEGIN TRANSACTION;
INSERT INTO [dbo].[CustRating]([RestaurantID], [FoodTypeID], [Rating])
SELECT 2, 1, 3 UNION ALL
SELECT 3, 2, 2
COMMIT;
GO

BEGIN TRANSACTION;
INSERT INTO [dbo].[FoodType]([FoodTypeID], [FoodTypeName])
SELECT 1, N'Burger' UNION ALL
SELECT 2, N'Taco'
COMMIT;
GO

BEGIN TRANSACTION;
INSERT INTO [dbo].[Restaurant]([RestaurantID], [RestaurantName])
SELECT 1, N'Cafe A' UNION ALL
SELECT 2, N'Cafe B' UNION ALL
SELECT 3, N'Cafe C'
COMMIT;
GO

Answer 1

SELECT              
                    Restaurant.*
                    ,CASE  WHEN FoodType.FoodTypeID IS NULL THEN NULL ELSE CustRating.Rating END AS Rating

 FROM               Restaurant Restaurant
 LEFT OUTER JOIN    CustRating CustRating 
 ON                 Restaurant.RestaurantID =   CustRating.RestaurantID
 LEFT OUTER JOIN    FoodType FoodType 
 ON                 FoodType.FoodTypeID = CustRating.FoodTypeID
 AND                FoodType.FoodTypeName = 'Burger'

Answer 2

尝试：

SELECT r.RestaurantID, RestaurantName, T.Rating 
FROM Restaurant r
  LEFT JOIN (
    select cr.RestaurantID, cr.Rating 
    from FoodType f 
    inner join CustRating cr on f.FoodTypeID = cr.FoodTypeID
    where f.FoodTypeName = 'Burger'
  ) T on r.RestaurantID = t.RestaurantID

Answer 3

使用此查询：

SELECT DISTINCT
  r.RestaurantID, 
  RestaurantName, 
  CASE WHEN ft.FoodTypeID IS NULL THEN NULL ELSE cr.Rating END Rating
FROM Restaurant r
LEFT JOIN CustRating cr ON cr.RestaurantID = r.RestaurantID
LEFT JOIN FoodType ft ON ft.FoodTypeID = cr.FoodTypeID AND 
ft.FoodTypeName = 'Burger'

Answer 4

这样的事情应该有效;在实际加入之前你需要删除“TACO” -

SELECT r.RestaurantID, RestaurantName, crft.Rating 
   FROM Restaurant r    
        LEFT JOIN (SELECT cr.RestaurantID, cr.Rating
                     FROM CustRating cr JOIN FoodType ft  
                       ON ft.FoodTypeID = cr.FoodTypeID 
                    WHERE ft.FoodTypeName = 'Burger' ) as crft 
        ON crft.RestaurantID = r.RestaurantID

也许语法不完全正确，因为我没有SQL-Server来测试它，但是这个想法应该是这样的，并且像这样它在PostgreSQL上完全可行

SELECT "r"."RestaurantID", "RestaurantName", "crft"."Rating" 
   FROM "Restaurant" r    
        LEFT JOIN (SELECT * 
                     FROM "CustRating" cr JOIN "FoodType" ft  
                       ON "ft"."FoodTypeID" = "cr"."FoodTypeID"
                    WHERE "ft"."FoodTypeName" = 'Burger' ) as crft 
        ON "crft"."RestaurantID" = "r"."RestaurantID"

Answer 5

你在写一个外连接时遇到了问题所以这里有一个避免它们的替代方法（以及null值，外连接明确地设计为返回：）

SELECT DISTINCT r.RestaurantID, r.RestaurantName, 
       N'Burger' AS FoodTypeName, cr.Rating
  FROM dbo.Restaurant r
       JOIN dbo.CustRating cr 
          ON r.RestaurantID = cr.RestaurantID
       JOIN dbo.FoodType ft
          ON ft.FoodTypeID = cr.FoodTypeID
             AND ft.FoodTypeName = N'Burger'
UNION
SELECT DISTINCT r.RestaurantID, r.RestaurantName, 
       N'Burger' AS FoodTypeName, -1 AS Rating
  FROM dbo.Restaurant r
 WHERE NOT EXISTS (
                   SELECT * 
                     FROM dbo.CustRating cr 
                          JOIN dbo.FoodType ft
                             ON ft.FoodTypeID = cr.FoodTypeID
                                AND ft.FoodTypeName = N'Burger'
                    WHERE r.RestaurantID = cr.RestaurantID
                   );

Answer 6

尝试：

  LEFT JOIN FoodType ft ON ft.FoodTypeID = cr.FoodTypeID WHERE 
    ft.FoodTypeID=1

思考可以帮助。

在SQL连接中需要帮助

6 个答案: