查询以收集前一行的数据

Question

我有一个记录的表格，在CO.Nr下面的示例数据是TH-123，Th-456等......我需要收集数据..

Nr.  CO.Nr   Employee  Resp   Description  Date 

1   TH-123   ABC   NULL    HELLO   10.05.2010 
2   TH-123   NULL   S14    NULL   18.05.2010 
3   TH-123   DEF        NULL   13.05.2010 
4   TH-456   XYZ   NULL    NULL    1.07.2010 
5   TH-456   NULL   S19    SOME    NULL 
6   TH-456           TEXT   08.05.2010 
7   TH-456      NULL        28.05.2010 

对于TH-123, 如果Nr。是最大的，这是我需要以CO.Nr组开始的记录，所以如果其他列中的值为NULL或空格，则记录的Nr为3， ，转到上面的记录是记录Nr为2，即使它具有空值，也会使用Nr转到记录上方的记录。在这种情况下为1。 在我需要采取最大日期的3条记录。 对于上述数据，我需要有作为输出，

 CO.Nr   Employee  Resp   Description  Date 

     TH-123   DEF   S14    HELLO   18.05.2010 
     TH-456   XYZ   S19    TEXT    01.07.2010 

提前感谢！

Answer 1

你可以做到这一点很多方面

select [co.nr], 
(select top(1) employee from mytable b where b.[co.nr]=a.[co.nr] and 
         employee is not null order by nr desc) as employee, 
(select top(1) resp from mytable b where b.[co.nr]=a.[co.nr] and 
         resp is not null order by nr desc) as resp, 
(select top(1) description from mytable b where b.[co.nr]=a.[co.nr] and 
         description is not null order by nr desc) as description, 
(select max([date]) from mytable b where b.[co.nr]=a.[co.nr]) as Date 
from (
select distinct [co.nr] 
from mytable) as a 

Answer 2

您可以使用子选择选择您想要的记录，然后加入。像对于员工下列操作之一（我将离开列的其余部分的练习）：

SELECT MyTable.[CO.Nr], Employees.Employee 
FROM MyTable 
LEFT OUTER JOIN (SELECT FIRST(Employee) as Employee, [CO.Nr] 
       FROM MyTable 
       WHERE Employee IS NOT NULL AND Employee <> '' 
       GROUP BY [CO.Nr] 
       ORDER BY [Nr.] DESC) Employees 
      ON MyTable.[CO.Nr] = Employees.[CO.Nr] 
GROUP BY MyTable.[CO.Nr] 

或者，如果FIRST()不是有效的聚合函数，如您的评论中提到，你可以试试子选择在SELECT条款，如：

SELECT t.MyTable.[CO.Nr], 
     (SELECT TOP(1) x.Employee 
     FROM MyTable x 
     WHERE x.[CO.Nr] = t.[CO.Nr] 
     AND x.Employee IS NOT NULL AND x.Employee <> '' 
     ORDER BY [Nr.] DESC) as Employee 
FROM MyTable t 
GROUP BY t.[CO.Nr]