如何使用JOIN从某个列中选择仅具有唯一值的同一个表中的行？

Question

我有三列

Id Color Price 

台苹果和表这个数据

+----+-----------+------+ 
| Id | Color  | Price| 
+----+-----------+------+ 
| 1 | Red  | 5 | 
| 2 | Red  | 1 | 
| 3 | Green  | 3 | 
| 4 | Orange | 4 | 
+----+-----------+------+- 

在结果表我想看看：

+----+-----------+------+ 
| Id | Color  | Price| 
+----+-----------+------+ 
| 3 | Green  | 3 | 
| 4 | Orange | 4 | 
+----+-----------+------+ 

如何做到这一点使用连接没有子查询？

在这一点上我有这样的查询只选择红苹果：

select * 
from apples as a1 
left join apples as a2 
on a1.Id != a2.Id 
where a1.Color = a2.Color; 

Answer 1

的灵感来自India.Rocket：

select A.Id, A.Color, A.Price 
    from Apples A 
    left join Apples A2 on A2.Color= A.Color and A2.Id<>A.Id 
    where A2.Id is null 
    group by A.Id, A.Color, A.Price 

Answer 2

你并不需要一个join ......这样的事情应该工作：

SELECT * 
FROM apples 
GROUP BY Color 
HAVING COUNT(*)=1 

I made a fiddle

Answer 3

内部查询：当您将它们组合在一起时，选择只有一个发生的所有颜色。外部查询：选择所有颜色为内部查询所选颜色之一的苹果。

select * from Apples 
    where Color in (
     select Color from Apples group by Color having count(*)=1 
    ) 

Answer 4

或者只使用联接：

select Apples.* 
    from Apples 
    inner join 
    (
     select Color from Apples group by Color having count(*)=1 
    ) colors on colors.Color=Apples.Color 

Answer 5

修改您的查询了一下。只有1 left join，而不是subqueries或having条款查询，并在大多数的数据库

看看这个工作。

select distinct a1.* 
from apples as a1 
inner join apples as a2 
on a1.Color < a2.Color; 

OR

select distinct a1.* 
from apples as a1 
inner join apples as a2 
on a1.Id > a2.Id 
where a1.Color != a2.Color; 

希望这有助于:-)

Answer 6

该查询在sqlfiddle并在MySQL 5.7的作品。

首先，我连接表以选择具有重复颜色的行。然后我加入结果表并从表中删除重复项。

select a1.* 
from apples as a1 
left join apples as a2 
on (a1.Id <> a2.Id and a1.Color = a2.Color) 
left join apples a3 
on (a2.Color is null and a1.Id = a3.Id) 
where a3.Color is not null; 

http://sqlfiddle.com/#!9/23df971/1