Question

我需要一个返回ff的查询：

从每个姓名中的最新日期算起
如果最新日期的Count值为-1，那么它将返回最晚日期之前的日期计数
如果最新日期的Count值为-1而另一个日期为-1。然后返回0
如果最新日期的Count值为-1且该名称没有其他日期。然后返回0

示例表：

ID  Name  Date         Count
1   Adj   09/29/2012   2
2   Adj   09/30/2012   4
3   Ped   09/29/2012  -1
4   Ped   09/30/2012   5
5   Mel   09/29/2012   3
6   Mel   09/30/2012  -1
7   Rod   09/30/2012   7
8   Ney   09/30/2012  -1
9   Jin   09/29/2012  -1
10  Jin   09/30/2012  -1

期望的输出：

Name   Count
Adj    4
Ped    5
Mel    3
Rod    7
Ney    0
Jin    0

我对如何在SQL中处理这个问题感到很困惑，因为我只知道简单的查询。

有关如何对此进行查询的任何想法？感谢。

Answer 1

试试这个

SQL FIDDLE EXAMPLE

select A.name, isnull(T.[Count], 0) as [Count]
from (select distinct T.name from table1 as T) as A
    outer apply 
    (
        select top 1 T.[Count]
        from table1 as T
        where T.name = A.name and T.[Count] <> -1 
        order by T.[date] desc
    ) as T
order by A.name asc

更新：对于SQL 2000，您可以使用此类查询

SQL FIDDLE EXAMPLE for SQL 2000

select A.name, isnull(T1.[Count], 0) as [Count]
from 
(
    select T.name, max(case when T.[Count] <> -1 then T.[date] else null end) as [date]
    from table1 as T
    group by T.name
) as A
    left outer join table1 as T1 on T1.name = A.name and T1.[date] = A.[date]

但它依赖于您对name, [date]列

有唯一约束的建议

Answer 2

另一个

Select * from
(
Select Test.name,[Count]
from TEST
Join(
Select name, MAX(Date) as Date from TEST  
where [Count]<>-1
Group by Name) a
on a.Name=test.Name and a.Date=Test.Date
UNION 
Select Distinct name,0 from test o where not Exists(Select * from test where name=o.Name and [count]<>-1)
) res
order by Name

从最新日期开始的SQL查询计数值

2 个答案: