Question

我正在尝试在SQL Server 2008中选择具有第二个最新日期的行。我想为每个ID提取第二个最新日期。

+----+-----------+
|ID  |Date       |
+----+-----------+
|01  |2017-01-01 |
|01  |2016-04-25 |
|01  |2015-01-25 |
|02  |2017-02-01 |
|02  |2016-03-15 |
|02  |2015-09-24 |
|03  |2017-02-12 |
|03  |2016-03-05 |
|03  |2015-01-02 |
+----+-----------+

所以我想最终：

+---+-----------+
|01 |2016-04-25 |
|02 |2016-03-15 |
|03 |2016-03-05 |
+---+-----------+

我已经想出如何拉出maxdate而不是第二个最新日期。任何帮助是极大的赞赏。这个论坛对像我这样的新手很有帮助！

Answer 1

你可以用这个：

SELECT id, date 
FROM
(
    SELECT id, date, 
        ROW_NUMBER() OVER (PARTITION BY id ORDER BY date DESC) rn
    FROM table_name
) t
WHERE rn = 2;

如果您有多行包含第二个最新日期，则应考虑将ROW_NUMBER()更改为RANK()

Answer 2

SELECT MAX（日期）来自员工WHERE日期＆lt; （SELECT Max（date）FROM table）;

或

SELECT MAX（date）FROM Employee WHERE date NOT IN（SELECT Max（date）FROM table）;

了解详情：http://www.java67.com/2015/01/second-highest-salary-in-mysql-and-sql-server.html#ixzz4ZUWYpZ1p

Answer 3

试试这个：

     Create table SampleTable(ID INT ,Dates Date)

     insert into SampleTable values(01 ,'2017-01-01')
     insert into SampleTable values(01 ,'2016-04-25')
     insert into SampleTable values (01 ,'2015-01-25')
     insert into SampleTable values(02 ,'2017-02-01') 
     insert into SampleTable values(02 ,'2016-03-15')
     insert into SampleTable values (02 ,'2015-09-24')
     insert into SampleTable values (03 ,'2017-02-12') 
     insert into SampleTable values (03 ,'2016-03-05' )
     insert into SampleTable values (03 ,'2015-01-02')


select id, dates from (select id,dates,row_number() over (partition by id order by id desc )rowno from SampleTable)x where x.rowno=2

选择SQL Server 2008中具有第二大日期的行

3 个答案: