Question

需要查询帮助才能从表中获取不同的记录。这是为了在收件箱区域显示。

SELECT DISTINCT(from_id),to_id,message,sent_date,status,message_status from messages where to_id='$mid'

有了这个，我得到了两个from_id

的记录

SSM5130     testing         7 December
SSM4228     testing         9 December
SSM4228     testing         7 December
SSM12       Hai             7 December
SSM5130     aaaaaaaaaaaaaaaaaaaaaa      6 December

但我需要得到这样的东西。

SSM5130     testing         7 December
SSM4228     testing         9 December
SSM12       Hai             7 December

我的数据库是这样的........

mes_id  to_id   from_id message                 date                read_status message_status
----------------------------------------------------------------------------------------------
1       SSM11   SSM13   dfsdfsfsfsdfsdfsd       2013-12-06 10:49:44 1           0
2       SSM11   SSM5130 aaaaaaaaaaaaaaaaaaaaaa  2013-12-06 10:52:01 1           0
3       SSM11   SSM5130 aaaaaaaaaaaaaaaaaaaaaa  2013-12-06 10:52:14 1           0
4       SSM11   SSM4228 aaaaaaaaaaaaaaaaaaaaaa  2013-11-01 09:30:30 1           0
41      SSM11   SSM12    Hai                    2013-12-07 12:50:13 1           0
43      SSM11   SSM5130 aaaaaaaaaaaaaaaaaaaaaa  2013-11-07 09:30:30 1           0
44      SSM11   SSM100  aaaaaaaaaaaaaaaaaaaaaa  2013-11-07 09:30:30 1           0
46      SSM11   SSM5130 testing                 2013-12-09 15:16:25 1           1
47          SSM5130   SSM11 testing                 2013-12-09 15:16:25     1           1

Answer 1

您需要按to_id对每一行进行排序，并为每个to_id组选择第一行。

SELECT
    t.mes_id,
    t.to_id,
    t.from_id,
    t.message,
    t.date,
    t.read_status,
    t.message_status
FROM
    (SELECT
        @row_num := IF(@prev_value=o.to_id, @row_num + 1, 1) AS RowNumber,
        @prev_value := o.to_id,
        o.*
    FROM
        my_table o,
        (SELECT @row_num := 1) X, (SELECT @prev_value := '') Y
    ORDER BY
        o.to_id) t
 WHERE RowNumber = 1

Answer 2

我的猜测是你现有的from_id包含尾随空格。

使用此查询：

select distinct (ltrim(rtrim(from_id))), to_id,message,sent_date,status,message_status
  from messages
 where to_id='$mid'

Answer 3

更新：如果理解正确，请尝试这种方式

SELECT m.* 
  FROM 
(
  SELECT to_id, from_id, MAX(date) date
    FROM messages  
   WHERE to_id='SSM11'   -- if you need it for a specific to_id 
   GROUP BY to_id, from_id
) q JOIN messages m
    ON q.to_id = m.to_id
   AND q.froM_id = m.from_id
   AND q.date = m.date

输出：

+--------+-------+---------+------------------------+---------------------+-------------+----------------+
| mes_id | to_id | from_id | message                | date                | read_status | message_status |
+--------+-------+---------+------------------------+---------------------+-------------+----------------+
|      1 | SSM11 | SSM13   | dfsdfsfsfsdfsdfsd      | 2013-12-06 10:49:44 |           1 |              0 |
|     41 | SSM11 | SSM12   | Hai                    | 2013-12-07 12:50:13 |           1 |              0 |
|     44 | SSM11 | SSM100  | aaaaaaaaaaaaaaaaaaaaaa | 2013-11-07 09:30:30 |           1 |              0 |
|     46 | SSM11 | SSM5130 | testing                | 2013-12-09 15:16:25 |           1 |              1 |
|     51 | SSM11 | SSM4228 | testing                | 2013-12-09 15:03:51 |           0 |              1 |
+--------+-------+---------+------------------------+---------------------+-------------+----------------+

这是 SQLFiddle 演示

sql获取不同的记录

3 个答案: