Question

我有这张桌子：

CUSTOMER NAME   JOINED DATE
Daniel          03/01/2013
Daniel          04/01/2013
Daniel          23/05/2013
Zack            27/06/2013
Zack            06/08/2013
Patricia        07/01/2015
John            08/05/2008
John            02/05/2013
John            23/05/2013
John            08/06/2015
Stephanie       16/04/2010
Stephanie       08/05/2012
Stephanie       28/09/2012
Stephanie       23/01/2013
Stephanie       28/03/2014
Elizabeth       18/04/2013
Elizabeth       22/04/2013
Elizabeth       25/04/2013

我看起来并且无法弄清楚如何返回删除[Customer Name]列中重复项的查询，但是将最早的日期保存在属于[customer name]的[Joined date]列中，请参阅下面是所需的输出：

customer name Joined on

Daniel 03/01/2013 Zack 27/06/2013 Patricia 07/01/2015 John 08/05/2008 Stephanie 16/04/2010 Elizabeth 18/04/2013

我是sql和编码的新手一般，我在某处读到你可以使用MIN（DATE）语句？或者在声明中找到这个？但我仍然不确定：

select distinct [customer name], [joined date] from database.dbo.practise where [status ] = 'live' and [joined date] = (select min(joined date) from database.dbo.practise)

Answer 1

试试这个

SELECT [CUSTOMER NAME], MIN([joined date]) AS "Joined on"
FROM database.dbo.practise
GROUP BY [CUSTOMER NAME]

如何检索列的早期日期并将不同列的重复项合并在一起？

1 个答案: