如何检查两个选择是否返回相同的ID

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*/)

如果两个查询都为您提供相同的id,则结果应为空。

如果集合相等,则不会返回任何结果:

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

您可以使用 EXCEPT .

EXCEPT返回左侧输入查询中的不同行 通过右输入查询输出。

EXCEPT 在您的案例中示例:

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

作为使用的替代方法 Full Join 在tsql中 :

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

我认为这是一种愚蠢的方式。