MySQL - 自己选择查询顺序

Question

请帮帮我。我有桌子：

id|count
1 |1
1 |3
2 |1
2 |2
3 |1
3 |2

我需要为同一个count选择id（列）1和2的ID：

id
2
3

Answer 1

您可以使用group by和having：

执行此操作

select id
from t
where count in (1, 2)
group by id
having count(*) = 2;

Answer 2

你可以试试这个

select  t1.id
from    myTable t1
join    myTable t2
on      t1.id = t2.id
where   t1.count = 1 and t2.count = 2

Answer 3

使用INTERSECT查找包含1和2的ID：

SELECT id FROM table WHERE count = 1
INTERSECT
SELECT id FROM table WHERE count = 2

编辑：哎呀，MySQL是否支持INTERSECT？

Answer 4

使用GROUP BY子句：

 SELECT DISTINCT id FROM Your_Table 
 WHERE [count] IN (1, 2)
 GROUP BY id 
 HAVING COUNT(*) > 2

Answer 5

 Select id 
 FROM TABLE_NAME  
 WHERE count = 1 
 INTERSECT 
 Select id
 FROM TABLE_NAME 
 WHERE count = 2;

那应该有用。以下是使用交叉https://www.tutorialspoint.com/sql/sql-intersect-clause.htm

的链接