Question

好的，所以我有以下问题，但是我试图围绕如何在给定的日期范围内给我每日结果。截至目前，这基本上是我的查询：

SELECT Source = CASE WHEN Project = 'CODE' OR Code = '1' THEN 'SOURCE1'
                       WHEN Project <> 'CODE' AND Code = '2' THEN 'SOURCE2'
                       WHEN Project <> 'CODE' AND Code = '3' THEN 'SOURCE3'
                       WHEN Project <> 'CODE' AND Code = '4' THEN 'SOURCE4'
                       ELSE Code END, Count(Document) as TotalCount, SUM(Amount) as TotalAmount                       

-- Joins here to gather and refine data collected

WHERE Date >= '03/20/2017'
       AND Date < '03/23/2017'
GROUP BY CASE WHEN Project = 'CODE' OR Code = '1' THEN 'SOURCE1'
                           WHEN Project <> 'CODE' AND Code = '2' THEN 'SOURCE2'
                           WHEN Project <> 'CODE' AND Code = '3' THEN 'SOURCE3'
                           WHEN Project <> 'CODE' AND Code = '4' THEN 'SOURCE4'
                           ELSE Code END
ORDER BY TotalCount DESC

给出了：

SOURCE1 123 1225
SOURCE2 23  126
SOURCE3 17  78
SOURCE4 12  56
MISCSOURCE1 6   78

理想情况下，它会给我：

Date         Source   TotalCount  TotalAmount
2017-03-20   SOURCE1  60          625
2017-03-20   SOURCE2  13          60
-- etc
2017-03-21   SOURCE1  63          600
2017-03-21   SOURCE2  10          66
-- etc

它还必须提供单独的MISCSOURCE（即除了重新标记的主要4之外的1 / +额外来源）总计数和金额。我该怎么做呢？在此先感谢帮助新手！

Answer 1

听起来你正想在日期字段上聚合。所以将你的逻辑调整为这样的东西可能对你有用：

SELECT [Date]
, 'SOURCE' + Code AS Source
, COUNT(Document) as TotalCount
, SUM(Amount) as TotalAmount
--FROM...                       
GROUP BY [Date], 'SOURCE' + Code
ORDER BY COUNT(Document) DESC

这应该会为您提供与您请求的输出更相似的输出。如果确实存在“Code”有时不等于Source1的实例，则可能需要添加一些额外的逻辑。进行以下调整可能有效：

SELECT [Date]
, 'SOURCE' + CASE WHEN Project = 'CODE' OR Code = '1' THEN '1' ELSE Code END AS Source
, COUNT(Document) as TotalCount
, SUM(Amount) as TotalAmount     
--FROM...                  
GROUP BY [Date]
, 'SOURCE' + CASE WHEN Project = 'CODE' OR Code = '1' THEN '1' ELSE Code END
ORDER BY COUNT(Document) DESC

应该这样做。

Answer 2

如果您只是将日期添加到选择和组中，您应该得到您想要的内容。像这样：

SELECT Date,Source = CASE WHEN Project = 'CODE' OR Code = '1' THEN 'SOURCE1'
                       WHEN Project <> 'CODE' AND Code = '2' THEN 'SOURCE2'
                       WHEN Project <> 'CODE' AND Code = '3' THEN 'SOURCE3'
                       WHEN Project <> 'CODE' AND Code = '4' THEN 'SOURCE4'
                       ELSE Code END, Count(Document) as TotalCount, SUM(Amount) as TotalAmount                       

-- Joins here to gather and refine data collected

WHERE Date >= '03/20/2017'
       AND Date < '03/23/2017'
GROUP BY CASE WHEN Project = 'CODE' OR Code = '1' THEN 'SOURCE1'
                           WHEN Project <> 'CODE' AND Code = '2' THEN 'SOURCE2'
                           WHEN Project <> 'CODE' AND Code = '3' THEN 'SOURCE3'
                           WHEN Project <> 'CODE' AND Code = '4' THEN 'SOURCE4'
                           ELSE Code END
ORDER BY date,TotalCount DESC

Answer 3

@ChadPortman很接近，只是在Group By部分中缺少Date。最终查询如下所示：

SELECT Date, Source = CASE WHEN Project = 'CODE' OR Code = '1' THEN 'SOURCE1'
                   WHEN Project <> 'CODE' AND Code = '2' THEN 'SOURCE2'
                   WHEN Project <> 'CODE' AND Code = '3' THEN 'SOURCE3'
                   WHEN Project <> 'CODE' AND Code = '4' THEN 'SOURCE4'
                   ELSE Code END, Count(Document) as TotalCount, SUM(Amount) as TotalAmount                       

-- Joins here to gather and refine data collected

WHERE Date >= '03/20/2017'
       AND Date < '03/23/2017'
GROUP BY Date, CASE WHEN Project = 'CODE' OR Code = '1' THEN 'SOURCE1'
                           WHEN Project <> 'CODE' AND Code = '2' THEN 'SOURCE2'
                           WHEN Project <> 'CODE' AND Code = '3' THEN 'SOURCE3'
                           WHEN Project <> 'CODE' AND Code = '4' THEN 'SOURCE4'
                           ELSE Code END
ORDER BY Date, TotalCount DESC

这正是我所需要的。谢谢大家的回复！

SQL新手。如何让以下查询给出每日结果？

3 个答案: