mysql query为两个给定值选择相似列值的记录

Question

id  | person | cities travelled | distance 

----------+---------+-------------+----------- 

1   | jack | japan    | 8 

2   | Joe | florida   | 11 

3   | Joe | california   | 215 

4   | Jane | ghana    | 3 

5   | Jane | florida   | 8 

6   | Joe | jakarta   | 11 

7   | Joe | california   | 215 

8   | Joe | japan    | 3 

对于两个给定的人“jane”和“joe”，我想找到他们旅行的共同城市。这是什么SQL查询。在这种情况下，我们可以看到，简和乔已经旅行的普通城市是'佛罗里达'。我如何在sql查询的帮助下找到它？

Answer 1

您可以group by和having做到这一点：

select CitiesTravelled 
from t 
group by CitiesTravelled 
having sum(person = 'Joe') > 0 and 
     sum(person = 'Jane') > 0; 

通过将逻辑having子句中，你可以很容易地概括查询。下面是版本简或乔：

select CitiesTravelled 
from t 
group by CitiesTravelled 
having sum(person = 'Joe') > 0 or 
     sum(person = 'Jane') > 0; 

这里是版本为简，乔走过的城市，和杰克：

select CitiesTravelled 
from t 
group by CitiesTravelled 
having sum(person = 'Joe') > 0 and 
     sum(person = 'Jane') > 0 and 
     sum(person = 'Jack') > 0; 

编辑：

如果你想添加行驶距离，您可以将select更改为类似：

select CitiesTravelled, min(DistanceTravelled) 
select CitiesTravelled, avg(DistanceTravelled) 
select CitiesTravelled, max(case when person = 'Jane' then DistanceTravelled end) 

Answer 2

select CitiesTravelled 
from t 
where person = 'Joe' 
and CitiesTravelled in (
    select CitiesTravelled 
    from t 
    where person = 'Jane' 
); 

或

select CitiesTravelled 
from t t1 
where person = 'Joe' 
and exists (
    select t2.CitiesTravelled 
    from t t2 
    where t2.person = 'Jane' 
    and t2.CitiesTravelled = t1.CitiesTravelled 
);