Question

我有以下两个表：

T1

+----+------+-----+
| id | name | age |
+----+------+-----+
|  1 | bob  |  42 |
|  2 | dave |  12 |
|  3 | mary |  78 |
|  4 | kim  |  41 |
+----+------+-----+

T2

+----+-------+------------------+
| id | t1_id |       date       |
+----+-------+------------------+
|  1 |     1 | 12/26/2014 14:32 |
|  2 |     1 | 12/26/2014 16:32 |
|  3 |     1 | 12/27/2014 14:32 |
|  4 |     2 | 12/26/2014 15:32 |
|  5 |     4 | 12/26/2014 14:32 |
|  6 |     4 | 12/27/2014 14:32 |
+----+-------+------------------+

我希望获得以下结果（仅包括t2中最新的JOINed日期）：

+----+------+-----+------------------+
| id | name | age |       date       |
+----+------+-----+------------------+
|  1 | bob  |  42 | 12/27/2014 14:32 |
|  2 | dave |  12 | 12/26/2014 15:32 |
|  3 | mary |  78 | null             |
|  4 | kim  |  41 | 12/27/2014 14:32 |
+----+------+-----+------------------+

以下内容让我接近，但是从t2加入所有比赛。我如何只包括最新的？

SELECT t1.id, t1,name, t1.age, t2.date
FROM t1
LEFT OUTER JOIN t2 ON t2.t1_id=t1.id;

Answer 1

mysql> SELECT t1.id, t1.name, t1.age, t3.max_date 
FROM t1 LEFT JOIN 
(SELECT t1_id, MAX(date) AS max_date FROM t2 GROUP BY t1_id) t3 
ON t1.id = t3.t1_id;

产量

+------+------+------+------------------+
| id   | name | age  | max_date         |
+------+------+------+------------------+
|    1 | bob  |   42 | 12/27/2014 14:32 |
|    2 | dave |   12 | 12/26/2014 15:32 |
|    3 | mary |   78 | NULL             |
|    4 | kim  |   41 | 12/27/2014 14:32 |
+------+------+------+------------------+
4 rows in set (0.00 sec)

Answer 2

只需使用group by和max()：

SELECT t1.id, t1.name, t1.age, MAX(t2.date) as date
FROM t1 LEFT OUTER JOIN
     t2
     ON t2.t1_id = t1.id 
GROUP BY  t1.id, t1.name, t1.age;

编辑：

您可能会发现这更方便，并且可以获得更好的性能：

SELECT t1.*,
       (SELECT MAX(t2.date)
        FROM t2
        WHERE t2.t1_id = t1.id
       ) as date
FROM t1;

为获得最佳性能，您需要t2(t1_id, date)上的索引。

LEFT OUTER JOIN仅限最新记录

2 个答案: