Question

我有两张桌子：

表a：
--- ID --- Name ---
表b：
--- ID --- ID_of_a --- Date ---

现在，我想按table a Date的最新日期table b（desc）排序Table a行。例： Table b包含ID为“1”和“2”的行 SELECT DISTINCT a.ID, a.Name FROM a, b WHERE a.ID=b.ID_of_a ORDER BY b.Date desc包含以下行：{ID，ID_of_a，Date} {1,1，“2013-06-30”}，{2,1，“2013-07-01”}，{3,2， “2013-07-02”} 表a中正确的ID顺序为：1 --- 2
查询：

{{1}}

但这有时不起作用。

Answer 1

假设表之间存在1-1关系：

select a.*
from a left outer join
     b
     on a.id = b.id
order by b.date desc

如果每个b的{{1}}中有多行，那么您需要a：

group by

Answer 2

怎么样

select a.name,a.id from a join (select id_of_a as id_a,max(dt) as max_dt from b group by id_of_a order by max_dt desc) e on e.id_a=a.id

实施例

mysql> select * from a;
+----+------+
| id | name |
+----+------+
|  1 | a    |
|  2 | b    |
+----+------+
2 rows in set (0.00 sec)

mysql> select * from b;
+----+---------+------------+
| id | id_of_a | dt         |
+----+---------+------------+
|  1 |       1 | 2013-06-20 |
|  2 |       2 | 2013-07-01 |
|  3 |       1 | 2013-07-02 |
+----+---------+------------+
3 rows in set (0.00 sec)

mysql> select a.name,a.id from a join (select id_of_a as id_a,max(dt) as max_dt from b group by id_of_a order by max_dt desc) e on e.id_a=a.id;
+------+----+
| name | id |
+------+----+
| a    |  1 |
| b    |  2 |
+------+----+
2 rows in set (0.00 sec)

基本上，您首先选择每个id_of_a的最长日期，然后将其加入a

我认为可以重写它以更有效地运行

MySQL按其他表排序

2 个答案: