Question

我有一个包含这样数据的表

Road  Item Response  added_on
1     82   Yes       7/11/16
1     83   Yes       7/11/16
1     84   Yes       7/11/16
2     82   Yes       8/11/16 
2     83   No        8/11/16
2     85   Yes       8/11/16

这反映了对道路的评估，其中＆＃39;项目＆＃39;正在评估的事情。有些项目将始终在评估期间完成（82,83），其他项目是可选的（84,85）。我想返回一个结合了道路/日期的所有评估结果的东西，如果没有评估该项目，则返回null。并且只返回上个月的结果。例如

Road  82  83  84  85   added_on
1     Yes Yes Yes      7/11/16
2     Yes No      Yes  8/11/16

我尝试了多次这样的自我加入，但它没有返回任何内容。

FROM assess AS A
JOIN assess AS B
ON A.road = B.road AND a.added_on = B.added on
JOIN assess AS C
ON A.road = C.road AND a.added_on = C.added on
JOIN assess AS D
ON A.road = D.road AND a.added_on = D.added on

WHERE A.item = '81'
AND B.item = '82'
AND (C.item = '83' OR C.item IS NULL)
AND (D.item = '84' OR D.item IS NULL)
AND datepart(month,A.added_on) = datepart(month,getdate()) -1

澄清，

- 道路每天评估不止一次 - 每个项目仅评估一次，有时为NULL，即不适用 - 每天评估多条道路 - 这张表有其他评估，但我们并不担心这些。

有什么想法吗？使用SQL Server 2008.谢谢。

Answer 1

假设你需要动态

Declare @SQL varchar(max) 
Select  @SQL = Stuff((Select Distinct ',' + QuoteName(Item) From YourTable Order By 1 For XML Path('')),1,1,'') 
Select  @SQL = 'Select [Road],' + @SQL + ',[added_on] 
                From YourTable
                Pivot (max(Response) For Item in (' + @SQL + ') ) p'
Exec(@SQL);

返回

编辑 - SQL Generated如下。（以防万一你不能去动态）

Select [Road],[82],[83],[84],[85],[added_on]
 From  YourTable
 Pivot (max(Response) For Item in ([82],[83],[84],[85]) ) p

Answer 2

实现此目标的另一种方法不太优雅，但如果您不想使用pivot，则使用基本操作。

加载测试数据

create table #assess ( road int, item varchar(10), response varchar(3), added_on date )
insert #assess( road, item, response, added_on )
values 
  (1, '82', 'Yes', '2016-07-11' )
, (1, '83', 'Yes', '2016-07-11' )
, (1, '84', 'Yes', '2016-07-11' )
, (2, '82', 'Yes', '2016-08-11' )
, (2, '83', 'No', '2016-08-11' )
, (2, '85', 'Yes', '2016-08-11' )

处理数据

-- Get every possible `item`
select distinct item into #items from #assess

-- Ensure every road/added_on combination has all possible values of `item`
-- If the combination does not exist in original data, leave `response` as blank
select road, added_on, i.item, cast('' as varchar(3)) as response into #assess2
from #items as i cross join #assess AS A
group by road, added_on, i.item 

update a set response = b.response
from #assess2 a inner join #assess b on A.road = B.road AND a.added_on = B.added_on AND a.item = b.item

-- Join table to itself 4 times - inner join if `item` must exist or left join if `item` is optional
select a.road, a.added_on, a.response as '82', b.response as '83', c.response as '84', d.response as '85'
FROM #assess2 AS A
INNER JOIN #assess2 AS B    ON A.road = B.road AND a.added_on = B.added_on
LEFT JOIN #assess2 AS C     ON A.road = C.road AND a.added_on = C.added_on
LEFT JOIN #assess2 AS D     ON A.road = D.road AND a.added_on = D.added_on

WHERE A.item = '82'
AND B.item = '83'
AND (C.item = '84' OR C.item IS NULL)
AND (D.item = '85' OR D.item IS NULL)
--AND datepart(month,A.added_on) = datepart(month,getdate()) -1

结果集是：

road    added_on    82  83  84  85
1       2016-07-11  Yes Yes Yes 
2       2016-08-11  Yes No      Yes

Answer 3

我会使用条件聚合来做到这一点：

select road,
       max(case when item = 82 then response end) as response_82,
       max(case when item = 83 then response end) as response_83,
       max(case when item = 84 then response end) as response_84,
       max(case when item = 85 then response end) as response_85,
       added_on
from t
group by road, added_on
order by road;

对于月份组件，您可以添加where子句。一种方法是：

where year(date_added) * 12 + month(date_added) = year(getdate())*12 + month(getdate()) - 1

或者，你可以使用这样的逻辑：

where date_added < dateadd(day, 1 - day(getdate()), cast(getdate() as date)) and
      date_added >= dateadd(month, -1, dateadd(day, 1 - day(getdate()), cast(getdate() as date)))

第二个看起来更复杂，但它是 sargable ，这意味着可以使用date_added上的索引（如果有的话）。

在TSQL查询中组合多行

3 个答案: