Question

好吧，我在MySQL上有这些专栏：

id
id_conv
associated_statut

associated_statut 是1和7之间的数字。

我想要做的是，如果此id_conv的最后 associated_statut为id_conv，则仅计算2。

示例：

-----------------------------------------------
|  id   |   id_conv   |   associated_statut   |
-----------------------------------------------
|  1    |      15     |          1            |
|  2    |      15     |          2            |
|  3    |      15     |          2            |
|  4    |      15     |          4            |
|  5    |      15     |          2            |
|  6    |      15     |          3            |

如果我想要id_conv = 2，则不会计算associated_statut，因为此id_conv的最后一个associated_statut是3。

我已经尝试过此查询：

SELECT COUNT(DISTINCT id_conv) FROM MyTable WHERE associated_statut = 2

但这并不能归还我想要的东西。

有没有办法在SQL中执行此操作？

感谢。

Answer 1

SELECT COUNT(sub1.id_conv) FROM MyTable
INNER JOIN
(
SELECT DISTINCT id, FIRST(associated_statut ORDER BY id DESC)
group by id_conv
recent FROM MyTable
) sub1 ON sub1.id = MyTable.id
WHERE sub1.recent_associated_statut = 2

Answer 2

也许，这对你有用：

SELECT count(t1.id) FROM mytable t1
  INNER JOIN (SELECT id_conv, MAX(id) id FROM foo GROUP BY id_conv) t2
    ON t1.id = t2.id 
    WHERE t1.associated_statut = 2

Answer 3

我们可以在没有子查询的情况下做同样的事情。拥有更多数据需要更少的时间。

SELECT count(t1.id) FROM 
mytable t1
  LEFT JOIN 
mytable t2
    ON t1.id_conv = t2.id_conv 
    AND t1.id < t2.id
WHERE t2.id IS NULL
AND t1.associated_statut = 2;

MySQL - 仅在最后的associated_statut = 2时计数

3 个答案: