1

こんにちは私は私のMYSQLクエリの助けが必要です

私はテーブルを持っています

id | tn   | title    | customer_id |create_time | comment              |
1  | 1342 | sample1  | customer1   | 2012-01-01 | hello world          |
2  | 1342 | sample1  | customer1   | 2012-01-02 | hello world          |
3  | 1342 | sample1  | customer1   | 2012-01-03 | hello new world      |
4  | 3362 | sample2  | customer1   | 2012-01-02 | good bye world       |
5  | 3362 | sample2  | customer1   | 2012-01-03 | good bye world       |
6  | 3362 | sample2  | customer1   | 2012-01-04 | good bye world       |
7  | 3362 | sample2  | customer1   | 2012-01-05 | good bye new world   |

私がtnでグループ化するとき私は取った

1  | 1342 | sample1  | customer1   | 2012-01-01 | hello world          |
4  | 3362 | sample2  | customer1   | 2012-01-02 | good bye world       |

しかし、私は取る必要があります

3  | 1342 | sample1  | customer1   | 2012-01-03 | hello new world      |
7  | 3362 | sample2  | customer1   | 2012-01-05 | good bye new world   |

これは、最大IDまたは最大create_timeを使用してtnでグループ化するようなものです。

これどうやってするの？ありがとう！

4

3 に答える 3

2

これを試して： 

mysql> select * from ( select * from tbl2 tn order by id desc ) t group by tn;
+------+------+---------+-------------+-------------+--------------------+
| id   | tn   | title   | customer_id | create_time | comment            |
+------+------+---------+-------------+-------------+--------------------+
|    3 | 1342 | sample1 | customer1   | 2012-01-03  | hello new world    |
|    7 | 3362 | sample2 | customer1   | 2012-01-05  | good bye new world |
+------+------+---------+-------------+-------------+--------------------+
2 rows in set (0.02 sec)
于 2012-06-29T21:10:41.320 に答える
1 
SELECT t2.* FROM
(SELECT MAX(id) AS id,tn FROM my_table GROUP BY tn) AS t1
LEFT JOIN my_table AS t2 USING(id)
于 2012-06-29T21:07:31.723 に答える
0

これを試して

Select t.* from 
table t right join
(Select max(id) as max_id from table group by tn) t1 on (t.id=t1.max_id)
于 2012-06-29T21:09:36.090 に答える