Question

我试图在多个条件下从连接中获取一些id-s。例加入等后

A B
x 1
x 2
x 3
y 1
y 2
y 3
y 5
z 1
z 5

所以我想要那些id-s（A），例如B是1，也是5

所以结果是y，z

由于

Answer 1

使用两个EXISTS可能看起来更干净：

SELECT DISTINCT A
FROM Table1 t1
WHERE EXISTS(SELECT A FROM Table1 WHERE A = t1.A and B = 1)
  AND EXISTS(SELECT A FROM Table1 WHERE A = t1.A and B = 5)

Answer 2

您可以通过WHERE和GROUP + HAVING实现此目的，HAVING将确保满足2个条件。

SELECT T1.A
FROM Table1 AS T1
JOIN Table2 AS T2
  ON T2.ID = T1.ID
WHERE T2.B IN(1,5)
GROUP BY T1.A
HAVING COUNT(*) = 2

或者，对于相同的A，如果B可能有1行或5行，则可以这样做：

SELECT DISTINCT T1.A
FROM Table1 AS T1
JOIN Table2 AS T2
  ON T2.ID = T1.ID
 AND T2.B = 1  
JOIN Table2 AS T3
  ON T3.ID = T1.ID
 AND T3.B = 5

Answer 3

select A from Table group by A
having max(decode(B,1,0)) + max(decode(B,5,0)) = 0

fiddle

Answer 4

如果条件是＆＃34;不同＆＃34;在某个意义上，给定的行只能满足条件的一个，那么可以使用类似的东西：

select a
  from table1
 group by a
having count(
   case
      when b = 1 then 1
      when b = 5 then 1
   end
) = 2
 order by a

在案例结构中设置每个条件，并测试计数是否等于条件数。

或者类似于Egor Skriptunoff的方法：

select a
  from table1
 group by a
having max(case when b=1 then 1 end)=1
   and max(case when b=5 then 1 end)=1
 order by a

每种条件下的最大值（）。

两种方法都不需要自联接。

如果陈述为真，则获得1

4 个答案: