找到没有订购所有3种比萨饼的顾客

Question

customers: 
+-----------+-----------+ 
| cid  | name  | 
+-----------+-----------+ 
| 1   | a   | 
| 2   | b   | 
| 3   | c   | 
+-----------+-----------+ 

pizza: 
+-----------+-----------+ 
| pid  | type  | 
+-----------+-----------+ 
| 1   | sausage | 
| 2   | cheese | 
| 3   | veggies | 
| 4   | sausage | 
| 5   | veggies | 
| 6   | sausage | 
| 7   | sausage | 
+-----------+-----------+ 

orders: 
+-----------+-----------+-----------+ 
| oid  | cid  | pid  | 
+-----------+-----------+-----------+ 
| 1   | 1   | 1   | 
| 2   | 1   | 2   | 
| 3   | 2   | 3   | 
| 4   | 3   | 4   | 
| 5   | 1   | 5   | 
| 6   | 3   | 6   | 
| 7   | 3   | 7   | 
+-----------+-----------+-----------+ 

我在围绕sql逻辑时遇到了一些麻烦。我如何找到没有订购所有3种比萨饼的顾客？这三种类型是香肠，奶酪和蔬菜。我需要使用NOT EXIST吗？

Answer 1

你可以使用一个having子句来寻找那些已经订购了不到3种比萨饼类型的客户：

select c.name 
from customers c 
join orders c 
on  c.cid = c.id 
join pizza p 
on  p.id = c.pid 
group by 
     c.name 
having 
     count(distinct p.type) < 3 

Answer 2

您可以使用这样的事情：

select c.cid, 
    c.name 
from customers c 
left join orders o 
    on c.cid = o.cid 
left join pizza p 
    on o.pid = p.pid 
where p."type" in ('sausage', 'cheese', 'veggies') -- if you have more pizza types list them here 
group by c.cid, c.name 
having count(distinct "type") <> (select count(distinct "type") 
            from pizza) 

见SQL Fiddle with Demo

或者

select c.cid, 
    c.name 
from customers c 
left join orders o 
    on c.cid = o.cid 
left join pizza p 
    on o.pid = p.pid 
where p."type" in ('sausage', 'cheese', 'veggies') 
group by c.cid, c.name 
having count(distinct "type") <> 3 -- this is equal to the number of pizza types from the IN clause above 

Answer 3

select cid, count(distinct type) as c from orders join pizza on orders.pid = pizza.pid group by cid having c < 3;