mysql一对多在一行中

mysql一对多在一行中,mysql,sql,select,group-by,sql-order-by,Mysql,Sql,Select,Group By,Sql Order By,我有两张桌子: id_c name surname 1 frank pulio 2 mark adam 3 peter norton id_v id_c data_visited note 1 1 2013-10-12 first visited to frank 2 1 2013-10-16 second visited to frank 3

我有两张桌子:

id_c    name    surname
 1      frank   pulio
 2      mark    adam
 3      peter   norton


id_v   id_c    data_visited    note
  1      1     2013-10-12      first visited to frank
  2      1     2013-10-16      second visited to frank
  3      2     2013-10-15      first visited to mark
  4      1     2013-10-20      third visited to frank
  5      2     2013-10-18      second visited to mark
我会用上次访问的数据获取客户名称并记录

例如:

id_c   name   data_visited   note
  1    frank   2013-10-20    third visited to frank   
  2    mark    2013-10-18    second visited to mark 
试试这个:

SELECT A.id_c, A.name, A.data_visited, A.note 
FROM (SELECT A.id_c, A.name, B.data_visited, B.note 
      FROM tableA A 
      INNER JOIN tableB B ON A.id_c = B.id_c 
      ORDER BY A.id_c, B.data_visited DESC) AS A 
GROUP BY A.id_c
检查

输出

| ID_C |  NAME |                   DATA_VISITED |                   NOTE |
|------|-------|--------------------------------|------------------------|
|    1 | frank | October, 20 2013 00:00:00+0000 | third visited TO frank |
|    2 |  mark | October, 18 2013 00:00:00+0000 | SECOND visited TO mark |

这带来了一切。不仅仅是OP询问的最后一次访问。@mike检查更新的答案。在子查询中不需要使用子查询。你可以直接得到结果。检查查询并用提琴演示你如何展示你的尝试?哇!我没想到这么难…现在工作,非常感谢
 SELECT t1.`id_c`,c.`name`,t1.latest,t1.`note` FROM `tablea`c

 INNER JOIN
 (
 SELECT t2.`id_c`,t2.latest,t3.`note` from `tableb` t3 

 INNER JOIN

 (SELECT `id_c`,MAX(`data_visited`) AS latest from
`tableb` 
GROUP BY(`id_c`))t2
 ON (t3.`id_c`=t2.`id_c`) AND (t3.`data_visited`=t2.latest)

)t1
ON(c.`id_c`=t1.`id_c`)