如何检查两个选择是否返回相同的ID

Question

假设我们有这样的查询：

SELECT
1
FROM DUAL WHERE  
(SELECT id FROM table_1 t1 WHERE /*conditions*/)
IN
(SELECT id FROM table_1 t2 WHERE /*conditions*/)

我想检查是否第一次查询 SELECT id FROM table_1 t1 WHERE /*conditions*/ 返回与第二个查询相同的ID。 当然，这个查询（IN语句）不起作用。

Answer 1

尝试：

SELECT id FROM table_1 t1 WHERE /*conditions1*/ and id not in (SELECT id FROM table_1 t2 WHERE /*conditions2*/)
union 
SELECT id FROM table_1 t1 WHERE /*conditions2*/ and id not in (SELECT id FROM table_1 t2 WHERE /*conditions1*/)

如果两个查询都给你相同的id，那么结果应为空。

Answer 2

如果集合相等，则不返回任何内容：

SELECT id FROM table_1 t1 WHERE /*conditions*/
EXCEPT
SELECT id FROM table_1 t2 WHERE /*conditions*/

Answer 3

您可以使用EXCEPT。

  

EXCEPT返回左输入查询中不是的不同行   通过正确的输入查询输出。

---

您案件中的

EXCEPT样本：

SELECT id 
FROM   table_1 AS t1 
WHERE /*conditions*/
EXCEPT
SELECT id 
FROM   table_1 AS t2 
WHERE /*conditions*/

Answer 4

就像在tsql 中使用Full Join 的替代方法一样：

SELECT CASE WHEN isnull(Count(*), 0) > 1 then 1 else 0 end as result
FROM    (SELECT t1.id as t1_id, t2.id as t2_id FROM 
            (SELECT id FROM table1 WHERE /*conditions*/) As t1 
            Full Outer Join 
            (SELECT id FROM table2 WHERE /*conditions*/) As t2 
    On t1.id = t2.id
    ) As ft
WHERE ft.t1_id is null or ft.t2_id is null

我认为这可以说是一种愚蠢的方式。