Question

我需要一些MySQL查询的帮助。

我有这张桌子：

Name  | Date       | City
------+------------+------
Peter | 2013-04-01 | Berlin
Paul  | 2013-03-01 | London
Peter | 2013-03-01 | New York 
Peter | 2013-01-28 | Tokio

对于每个名字，我需要表格中的第一个日期和相应的城市。

我试过了：

SELECT Name, MIN(Date) as MD, MIN(City) as CN FROM table GROUP BY Name

但是日期与城市名称不对应，结果是：

Peter, 2013-01-028, Berlin

有人可以给我一个提示吗？

由于

Answer 1

您需要使用子查询来获取min(date) name，然后将name和date上的结果加入到您的表中：

SELECT t1.Name, 
  t1.Date as MD, 
  t1.City as CN 
FROM yourtable t1
inner join
(
  select min(date) MinDate, name
  from yourtable
  group by name
) t2
  on t1.name = t2.name
  and t1.date = t2.mindate

请参阅SQL Fiddle with Demo

结果是：

|  NAME |                             MD |     CN |
---------------------------------------------------
|  Paul |   March, 01 2013 00:00:00+0000 | London |
| Peter | January, 28 2013 00:00:00+0000 |  Tokio |

Answer 2

试试这个::

SELECT Name, MIN(Date) as MD, City as CN FROM table GROUP BY Name

Answer 3

LEFT JOIN应该提供您想要的结果，而不会使查询过于复杂;

SELECT m1.* FROM myTable m1
LEFT JOIN myTable m2
  ON m1.Name = m2.Name
 AND m1.`date` > m2.`date`
WHERE m2.Name is NULL

An SQLfiddle for testing

它基本上会选择所有没有旧行同名的行。

NAME    DATE          CITY
Paul    2013-03-01    London
Peter   2013-01-28    Tokio

Answer 4

以下是如何做到这一点。

SELECT
  m.Name,
  m.Date,
  m.City
FROM mytable as m
INNER JOIN (select min(date) as LDate, name from mytable group by name) as l
    on l.LDate = m.date
      and l.name = m.name
group by m.name

Demo

输出

Name  | Date        | City
-----------------------------
Paul  | 2013-03-01  | London
Peter | 2013-01-28  | Tokio

Answer 5

您可以使用IN子句：

SELECT yourtable.*
FROM   yourtable
WHERE  (Name, Date) IN (SELECT Name, Min(Date)
                        FROM yourtable
                        GROUP BY Name)

子查询将为每个名称返回名称及其最短日期。外部查询返回名称和日期对应于子查询返回的行的所有行。

mysql查询组的名称

5 个答案:

Demo