2017-02-14 96 views
0

USER_KEY有多个结果存在问题。 我必须按部门总结用户的小时数。所以,每个用户在某个部门都会有一定的时间。按数据透视表分组

一切都很酷,除了用户行正在重复。我需要在查询上执行group by,但没有成功。

下面是该查询:

DECLARE @DATEFROM DATETIME = DATEADD(DAY, -14, GETDATE()) 
DECLARE @DATETO DATETIME = DATEADD(DAY, -12, GETDATE()) 

DECLARE @COLDEPARTMENTS NVARCHAR(MAX) 
SELECT @COLDEPARTMENTS = STUFF((SELECT DISTINCT ',' + QUOTENAME(DEPA_KEY, '[') FROM CADEPA FOR XML PATH('')), 1, 1, '') 
--SELECT @COLDEPARTMENTS 

DECLARE @QUERY AS NVARCHAR(MAX) 

DECLARE @USERS TABLE 
(
    USER_KEY INT, 
    USDE_HSU DECIMAL(8,2) 
) 

DECLARE @USERS_STR NVARCHAR(MAX) 

INSERT INTO @USERS (USER_KEY, USDE_HSU) 
SELECT USERS_.USER_KEY, SUMMARY FROM (
SELECT DISTINCT USER_KEY, SUM(USDE_HSU) SUMMARY 
FROM CAUSDE_TAS 
WHERE USDE_DAT >= @DATEFROM AND USDE_DAT <= @DATETO 
GROUP BY USER_KEY 
HAVING SUM(USDE_HSU) IS NOT NULL AND SUM(USDE_HSU) > 0) USERS_ 

SELECT @USERS_STR = STUFF((SELECT DISTINCT ',' + CAST(USER_KEY AS NVARCHAR(9)) FROM @USERS FOR XML PATH('')), 1, 1, '') 

SELECT @QUERY = 'SELECT DISTINCT USER_KEY, ' + @COLDEPARTMENTS + ' 
FROM CAUSDE_TAS 
PIVOT 
(
SUM(USDE_HSU) 
FOR DEPA_KEY IN (' + @COLDEPARTMENTS + ') 
) PIVOT_LOCATIONS 
WHERE USDE_DAT >= ''' + format(@DATEFROM, 'MM.dd.yyyy') + ''' AND USDE_DAT <= ''' + format(@DATETO, 'MM.dd.yyyy') + ''' 
AND USER_KEY IN (' + @USERS_STR + ')' 

EXECUTE (@QUERY) 

问题是与查询的最后一部分:

SELECT @QUERY = 'SELECT DISTINCT USER_KEY, ' + @COLDEPARTMENTS + ' 
FROM CAUSDE_TAS 
PIVOT 
(
SUM(USDE_HSU) 
FOR DEPA_KEY IN (' + @COLDEPARTMENTS + ') 
) PIVOT_LOCATIONS 
WHERE USDE_DAT >= ''' + format(@DATEFROM, 'MM.dd.yyyy') + ''' AND USDE_DAT <= ''' + format(@DATETO, 'MM.dd.yyyy') + ''' 
AND USER_KEY IN (' + @USERS_STR + ')' 

我不知道如何使group by这里?

的结果是这样的:

enter image description here

我已经试过是添加GROUP BY这里:

SELECT @QUERY = 'SELECT DISTINCT USER_KEY, ' + @COLDEPARTMENTS + ' 
FROM CAUSDE_TAS 
PIVOT 
(
SUM(USDE_HSU) 
FOR DEPA_KEY IN (' + @COLDEPARTMENTS + ') 
) PIVOT_LOCATIONS 
WHERE USDE_DAT >= ''' + format(@DATEFROM, 'MM.dd.yyyy') + ''' AND USDE_DAT <= ''' + format(@DATETO, 'MM.dd.yyyy') + ''' 
AND USER_KEY IN (' + @USERS_STR + ') 
GROUP BY USER_KEY' 

但错误是:

列“PIVOT_LOCATIONS .110000003'在选择列表 中无效,因为它不是包含在聚合函数或 GROUP BY子句中。

更新: 我已经添加group by user_key, ' + @COLDEPARTMENTS末,但USER_KEY仍然重复。 (没有错误,但结果并不好)

这里是在年底加入group by后查询的外观:由那些你不通过,仅组

SELECT DISTINCT USER_KEY, [110000003],[120000002],[120000003],[120000004],[120000005],[120000006],[120000007],[120000008],[120000009],[120000010],[120000011],[120000012],[120000013],[120000015],[120000016],[120000017],[120000021],[120000022],[120000023],[120000025],[120000026],[120000027],[120000028],[120000029],[120000030],[120000039],[120000040],[120000042],[120000043],[120000044],[120000045],[120000046],[120000047],[120000048],[120000049],[120000050],[120000051],[130000001],[130000002],[130000003],[130000004],[130000005],[130000006],[130000007],[140000001],[140000002],[140000003],[140000004],[140000005],[140000006],[140000007],[140000008],[140000009],[140000010],[140000011],[140000012],[140000013],[140000014],[140000015],[140000016],[140000017],[140000018],[150000001],[150000002],[150000003],[150000004],[150000005],[150000006],[150000007],[150000008],[150000009],[150000010],[150000011],[150000012],[160000001],[160000002],[160000003],[160000004],[160000005] 
FROM CAUSDE_TAS 
PIVOT 
(
SUM(USDE_HSU) 
FOR DEPA_KEY IN ([110000003],[120000002],[120000003],[120000004],[120000005],[120000006],[120000007],[120000008],[120000009],[120000010],[120000011],[120000012],[120000013],[120000015],[120000016],[120000017],[120000021],[120000022],[120000023],[120000025],[120000026],[120000027],[120000028],[120000029],[120000030],[120000039],[120000040],[120000042],[120000043],[120000044],[120000045],[120000046],[120000047],[120000048],[120000049],[120000050],[120000051],[130000001],[130000002],[130000003],[130000004],[130000005],[130000006],[130000007],[140000001],[140000002],[140000003],[140000004],[140000005],[140000006],[140000007],[140000008],[140000009],[140000010],[140000011],[140000012],[140000013],[140000014],[140000015],[140000016],[140000017],[140000018],[150000001],[150000002],[150000003],[150000004],[150000005],[150000006],[150000007],[150000008],[150000009],[150000010],[150000011],[150000012],[160000001],[160000002],[160000003],[160000004],[160000005]) 
) PIVOT_LOCATIONS 
WHERE USDE_DAT >= '01.31.2017' AND USDE_DAT <= '02.02.2017' 
AND USER_KEY IN (100000002,100000008,100000013,100000014,100000042,100000043,100000052,100000066,110000025,110000030,110000057,120000030,120000033,120000037,120000039,120000052,120000064,130000007,130000017,130000021,130000033,130000041,130000069,130000073,130000096,130000109,130000115,140000031,140000054,140000066,140000073,140000074,150000018,150000019,150000023,150000024,150000045,150000067,150000072,150000095,150000101,150000102,150000115,150000205,150000215,150000281,160000012,160000057,160000058,160000071,160000078,160000107,160000109,160000145,160000146,160000151,160000181,160000182,160000192,160000204,160000220,170000001,170000006,170000008) 
group by user_key, [110000003],[120000002],[120000003],[120000004],[120000005],[120000006],[120000007],[120000008],[120000009],[120000010],[120000011],[120000012],[120000013],[120000015],[120000016],[120000017],[120000021],[120000022],[120000023],[120000025],[120000026],[120000027],[120000028],[120000029],[120000030],[120000039],[120000040],[120000042],[120000043],[120000044],[120000045],[120000046],[120000047],[120000048],[120000049],[120000050],[120000051],[130000001],[130000002],[130000003],[130000004],[130000005],[130000006],[130000007],[140000001],[140000002],[140000003],[140000004],[140000005],[140000006],[140000007],[140000008],[140000009],[140000010],[140000011],[140000012],[140000013],[140000014],[140000015],[140000016],[140000017],[140000018],[150000001],[150000002],[150000003],[150000004],[150000005],[150000006],[150000007],[150000008],[150000009],[150000010],[150000011],[150000012],[160000001],[160000002],[160000003],[160000004],[160000005] 
+0

你试过了什么,得到了什么错误? –

+0

我已更新问题。 – FrenkyB

+0

好的,错误消息告诉你,当你正在构建'@ COLDEPARTMENTS'时,你需要将列名放在聚合函数中,以便'GROUP BY'工作。 –

回答

1

取而代之的将每列该组不要重复。剩下的其他内容。类似这样的:

SELECT USER_KEY, SUM(col1), SUM(col2), ... SUM(colN) 
FROM {all the stuff in the middle} 
GROUP BY USER_KEY 
+0

这个工程...虽然我不明白如何:) – FrenkyB

+0

Google,并通过一些关于'SQL GROUP BY'的教程。那里有很多详细的解释。 –