Mysql的具体计算行属性

Question

所以，这里是我的表：

Id | name | Store 
001 | John | A 
001 | John | A  
001 | John | A  
001 | John | B 
001 | John | B 
001 | John | D 
002 | Bob | B 
002 | Bob | C 
003 | Dave | C 
004 | Pamela | A 
004 | Pamela | B 
004 | Pamela | C 
005 | Nick | D 
005 | Nick | D 
005 | Nick | D 

1. 我如何选择再算上在A和B已经逛过所有的人，然后
2. A和B ONLY？
3. A或B但别无其他？
4. D ONLY？

在我的例子预期的结果是：（1）约翰+帕梅拉，（2）约翰，（3）约翰（4）尼克

Answer 1

的人在A处和乙购物名称（和可能在别处）：

的具有甲 或乙人

SELECT name 
FROM yourTable 
GROUP BY name 
HAVING SUM(CASE WHEN Store = 'A' THEN 1 END) > 0 AND -- A is present 
     SUM(CASE WHEN Store = 'B' THEN 1 END) > 0  -- B is present 

名称（但不包括其他）：

SELECT name 
FROM yourTable 
GROUP BY name 
HAVING (SUM(CASE WHEN Store = 'A' THEN 1 END) > 0 OR   -- A is present 
     SUM(CASE WHEN Store = 'B' THEN 1 END) > 0) AND   -- B is present 
     SUM(CASE WHEN Store NOT IN ('A', 'B') THEN 1 END) = 0 -- only A or B 

人谁在一个逛过 仅和B个

名称：

SELECT name 
FROM yourTable 
GROUP BY name 
HAVING SUM(CASE WHEN Store <> 'D' THEN 1 END) = 0  -- only store D 

演示了第二个查询这里：

的谁只在商店d逛过人

SELECT name 
FROM yourTable 
GROUP BY name 
HAVING SUM(CASE WHEN Store = 'A' THEN 1 END) > 0 AND -- A is present 
     SUM(CASE WHEN Store = 'B' THEN 1 END) > 0 AND -- B is present 
     COUNT(DISTINCT Store) = 2      -- only A and B are present 

名称

SQLFiddle

Answer 2

我喜欢接近他们使用group by和havingË查询：

select name 
from t 
group by name 
having sum(store = 'A') > 0 and 
     sum(store = 'B') > 0; 

这得到谁已经在两家超市购物的人。如果你只想要那两个商店，没有其他的：

select name 
from t 
group by name 
having sum(store = 'A') > 0 and 
     sum(store = 'B') > 0 and 
     sum(store not in ('A', 'B')) = 0;