Question

我有table1：

+------+---------+
| id   | name    |
+------+---------+
| 1    | name1   |
| 2    | name2   |
| 3    | name3   |
| 4    | name4   |
+------+---------+

我有table2：

+------+---------+
| id   | object  |
+------+---------+
| 1    |    4    |
| 1    |    8    |
| 2    |    23   |
| 2    |    8    |
| 2    |    9    |
| 3    |    2    |
| 3    |    8    |
| 4    |    9    |
| 4    |    23   |
+------+---------+

我想选择name = name2，它有对象8,23，但是忽略只有objet 8或者只是对象23的其他id，结果将是：

+------+---------+
| id   | name    |
+------+---------+
| 2    | name2   |
+------+---------+

感谢您的帮助，掌握。

Answer 1

当您想要找到一个SELF JOIN有两个不同的值时，您可以使用id。

你可以尝试一下吗？你可以在这里测试http://www.sqlfiddle.com/#!2/a10879/1/0

SELECT *
FROM
(
    SELECT a.id
    FROM table2 AS a INNER JOIN table2 AS b
      ON a.id = b.id
    WHERE a.object = 8 AND b.object = 23
) x INNER JOIN table1 ON x.id = table1.id;

Answer 2

试试这个：

select a.id, a.name
from table1 a
where exists (select 1 from table2 c
              where object = 8 and c.id = a.id)
and exists (select 1 from table2 c
            where object = 23  and c.id = a.id)

Answer 3

怎么样：

SELECT DISTINCT table1.id, table1.name FROM table1, table2 WHERE table1.name = "name2" AND
table1.id = table2.id and (table2.object = 8 OR table2.object = 23);

Mysql：选择具有多个值的id

3 个答案: