mysql - SQL "GROUP BY"和 "INNER JOIN"语句一起

Question

我有两个 ANIMAL 和 NEED 表:

        ANIMAL                                      NEED        
Name    Species         Birthday                A_Species      Type 
Koala1  Phascolarctidae 02-10-2014           Phascolarctidae    Veg.    
Bear1   Ursinae         03-10-2016           Ursinae            Veg.    
Koala2  Phascolarctidae 04-09-2015           Ursinae            Meet    
Cattle1 Bovidae         20.03.2017           Ursinae            Fish    
Whale1  Cetacea         08.05.2010           Bovidae            Veg.    
                                             Cetacea            Fish    

我要选择下表

    Name    Type                
    Koala1  Veg.                
    Koala2  Veg.                
    Cattle1 Veg.                
    Whale1  Fish                

这是只需要一种食物的动物名称及其食物的列表!

我必须合并两个声明

SELECT A_Species 
FROM NEED GROUP BY A_Species
HAVING COUNT(A_Species)=1;

和

SELECT ANIMAL.NAME, NEED.Type 
FROM ANIMAL 
INNER JOIN NEED ON ANIMAL.Species = NEED.A_Species;                 

我试过了

SELECT ANIMAL.NAME, NEED.Type 
FROM ANIMAL 
INNER JOIN NEED ON ANIMAL.Species = NEED.A_Species 
WHERE EXISTS(SELECT A_Species 
FROM NEED GROUP BY A_Species
HAVING COUNT(A_Species)=1); 

这是行不通的! 你能帮我把它们放在一起吗？

Answer 1

一种方法使用聚合:

SELECT a.NAME, MAX(n.Type)
FROM ANIMAL a INNER JOIN
     NEED n
     ON a.SPECIES = n.A_SPECIES 
GROUP BY a.Name
HAVING COUNT(*) = 1;

或者另一种方法是使用 NOT EXISTS:

SELECT a.NAME, n.Type
FROM ANIMAL a INNER JOIN
     NEED n
     ON a.SPECIES = n.A_SPECIES 
WHERE NOT EXISTS (SELECT 1
                  FROM need n2
                  WHERE n2.A_SPECIES = n.A_SPECIES AND
                        n2.Type <> n.Type
                 );

在实践中，这可能有更好的性能，特别是在 need(a_species, type) 上的索引。