Question

我有这个表：

表1：

id   Name
1   Example1
2   Example2

表2：

id  Date..............
1   5.2.2014........
1   6.2.2014.........
1   6.2.2014........
2   16.1.2014.......
2   17.1.2014.......

我需要从table1获取id和name并加入table1.id = table2.id并从table2只取前1行... 例如：

id    Name         Date
1     Example1   5.2.2014
2     Example2   16.1.2014

有可能吗？

Answer 1

您可以row_number()使用id过滤除最新行以外的所有行：

select  *
from    (
        select  row_number() over (partition by id order by Date desc) as rn
        ,       *
        from    Table2
        ) as t2
join    Table1 as t1
on      t1.id = t2.id
where   t2.rn = 1 -- Only latest row

Answer 2

嗯，一个简单的尝试就是

SELECT  t1.*,
        (SELECT TOP 1 t2.Date FROM Table2 t2 WHERE t2.ID = t1.ID t2.Date) t2Date
FROM    Table1 t1

如果您使用的是SQL Server，则可以使用ROW_NUMBER

像

这样的东西

;WITH Vals AS (
            SELECT  t1.ID,
                    t1.Name,
                    t2.Date,
                    ROW_NUMBER() OVER(PARTITION BY t1.ID ORDER BY t2.Date) RowID
            FROm    Table1 t1 LEFT JOIN
                    Table2 t2   ON  t1.ID
    )
    SELECT  *
    FROM    Vals
    WHERE   RowID = 1

Answer 3

Select t1.id, t1.name , MIN(t2.date)
From table1 t1
Inner Join table2 t2
 On t1.id=t2.id
Group By t1.id, t1.name

从表2中加入两个表和顶部的数据

3 个答案: