我有以下查询：
SELECT DISTINCT
x.CountryID [code],
c.Description [name],
COUNT(x.username) OVER(PARTITION BY x.countryID) [count],
COUNT(x.username) OVER () [tcount],
COALESCE(c.Map,'world') [map]
FROM
(
SELECT DISTINCT
CountryID,
username,
Score
FROM global.dbo.UserCategories
WHERE TLCID LIKE @TLCID OR TLCID IN (@TLCID)
) x
JOIN global.dbo.countrydetails c on x.CountryID=c.country
AND x.Score BETWEEN @ScoreMin AND @ScoreMax
GROUP BY x.CountryID, c.Description, c.Map, x.username --WITH ROLLUP
--HAVING COUNT(x.username) OVER(PARTITION BY x.countryID)>@Min
ORDER BY [name] ASC
为每个国家/地区显示具有特定分数且在某些类别中有效的用户数 我想从列表中排除用户很少的国家/地区 我想添加HAVING子句 但是sql告诉我
＆＃34;窗口函数只能出现在SELECT或ORDER BY中 条款＆＃34;
最重要的表是CountryDetails，如下所示：
CountryID username score TLCID
DE 12-bu 19598 131090
DE 12-bu 19598 220
DE 12-bu 19598 11700
DE 12-bu 19598 131090
DE 1buy3 19648 11700
DE 1buy3 19648 11450
DE 1buy3 19648 11700
DE 1buy3 19648 11700
CN 2012f 22780 11450
CN 2012f 22780 11450
CN 2012f 22780 11450
CN 2012f 22780 11450
CN 2012f 22780 11450
GB agood 74539 1
GB agood 74539 11450
GB agood 74539 12576
GB agood 74539 131090
GB agood 74539 1
GB agood 74539 11233
GB agood 74539 1
GB agood 74539 1
GB agood 74539 1
GB agood 74539 220
BG best0 3173 2984
IT bestb 13777 15032
IT bestb 13777 293
IT bestb 13777 293
IT bestb 13777 15032
IT bestb 13777 15032
IT bestb 13777 15032
IT bestb 13777 58058
IT bestb 13777 293
DE bestb 55165 131090
DE bestb 55165 293
DE bestb 55165 293
DE bestb 55165 293
DE bestb 55165 293
CN besto 24537 11450
GB bidan 4584 58058
GB bidan 4584 58058
IT bimbu 4145 11700
GB buy_d 97093 12576
GB buy_d 97093 888
GB buy_d 97093 888
GB buy_d 97093 888
GB buy_d 97093 159912
GB buy_d 97093 888
GB buy_d 97093 159912
GB buy_d 97093 11700
GB buy_d 97093 11700
GB buy_d 97093 159912
GB buy_d 97093 20710
GB buy_d 97093 159912
GB buy_d 97093 888
GB buy_s 988538 11232
GB buy_s 988538 1
GB buy_s 988538 11700
GB buy_s 988538 26395
关于结果..现在我得到了这个
code name count tcount map
AD Andorra 3 83141 andorra
AU Australia 14 83141 australia
AT Austria 345 83141 austria
.....
BE Belgium 28 83141 belgium
CN China 1070 83141 china
FR France 4732 83141 france
DE Germany 33782 83141 germany
HK Hong Kong 364 83141 hongKong
HU Hungary 9 83141 hungary
.....
IN India 10 83141 india
IR Iran 1 83141 iran
IQ Iraq 17 83141 iraq
IE Ireland 36 83141 ireland
IL Israel 16 83141 israel
IT Italy 5427 83141 italy
JE Jersey 25 83141 world
....
GB United Kingdom 35136 83141 unitedKingdom
...
假设将@Min设置为20我想得到这个：
code name count tcount map
AT Austria 345 80945 austria
BE Belgium 28 80945 belgium
CN China 1070 80945 china
FR France 4732 80945 france
DE Germany 33782 80945 germany
HK Hong Kong 364 80945 hongKong
IE Ireland 36 80945 ireland
IT Italy 5427 80945 italy
JE Jersey 25 80945 world
GB United Kingdom 35136 80945 unitedKingdom
其中tcount是计数总和
Pls可以建议解决方法的路径吗？
谢谢！
答案 0 :(得分：1)
you might be able to do something as simple as this..
SELECT c.countryid,
c.[description],
x.[count],
SUM([count]) OVER () [tcount],
COALESCE(c.Map,'world') [map]
FROM global.dbo.countrydetails c
OUTER APPLY (
SELECT COUNT(DISTINCT uc.username) [count]
FROM global.dbo.UserCategories uc
WHERE uc.countryid = c.countryid
AND TLCID LIKE @TLCID OR TLCID IN (@TLCID)
AND uc.Score BETWEEN @ScoreMin AND @ScoreMax
) x
WHERE x.[count] > @Min
it will select the individual username counts and apply them to the countries.. then you can sum the counts to get tcount and filter on the individual counts
答案 1 :(得分：0)
你似乎离这里很近。我认为你让它变得比它需要的更复杂。此外，subselect中的where子句不太正确。它可以简化为等于。
这应该非常接近。
SELECT DISTINCT
x.CountryID [code],
c.Description [name],
COUNT(x.username) as [tcount],
COALESCE(c.Map,'world') [map]
FROM
(
SELECT DISTINCT
CountryID,
username,
Score
FROM global.dbo.UserCategories
WHERE TLCID = @TLCID
) x
JOIN global.dbo.countrydetails c on x.CountryID=c.country
AND x.Score BETWEEN @ScoreMin AND @ScoreMax
GROUP BY x.CountryID, c.Description, c.Map
HAVING COUNT(x.username) > @Min
ORDER BY [name] ASC
- 编辑 -
要弄清楚你想要什么是很有挑战性的，因为我们没有表格，没有样本数据，没有所需的输出。简而言之，任何人现在都可以做的最好的事情就是猜你想要什么。这是一篇很好的文章，解释了你应该在网上论坛上发布t-sql帮助的内容。 http://spaghettidba.com/2015/04/24/how-to-post-a-t-sql-question-on-a-public-forum/
这是你可能想要的另一个猜测。我注意到你有不同的分组。这是多余的。
with CompiledValues as
(
SELECT x.CountryID as [code],
c.Description as [name],
COUNT(x.username) OVER(PARTITION BY x.countryID) as [count],
COUNT(x.username) OVER () as [tcount],
COALESCE(c.Map,'world') [map]
FROM
(
SELECT DISTINCT
CountryID,
username
FROM global.dbo.UserCategories
WHERE TLCID = @TLCID
) x
JOIN global.dbo.countrydetails c on x.CountryID = c.country
AND x.Score BETWEEN @ScoreMin AND @ScoreMax
GROUP BY x.CountryID
, c.Description
, c.Map
)
select *
from CompiledValues
where tcount > @Min
ORDER BY [name] ASC
答案 2 :(得分：0)
您似乎希望在选择国家/用户名时做出此决定。问题是用户是否可以拥有＆gt; 1分，如果你想要这部分确定你的门槛数。
-- Filtering by Count of CountryID, username, Score
SELECT DISTINCT
x.CountryID [code],
c.Description [name],
COUNT(x.username) OVER(PARTITION BY x.countryID) [count],
COUNT(x.username) OVER () [tcount],
COALESCE(c.Map,'world') [map]
FROM
(
SELECT COUNT(*),
CountryID,
username,
Score
FROM global.dbo.UserCategories
WHERE TLCID LIKE @TLCID OR TLCID IN (@TLCID)
GROUP BY CountryID,
username,
Score
HAVING COUNT(*) > [Your_Threshold_Number]
) x
JOIN global.dbo.countrydetails c on x.CountryID=c.country
AND x.Score BETWEEN @ScoreMin AND @ScoreMax
GROUP BY x.CountryID, c.Description, c.Map, x.username
ORDER BY [name] ASC
-- Filtering by Count of CountryID, username
SELECT DISTINCT
x.CountryID [code],
c.Description [name],
COUNT(x.username) OVER(PARTITION BY x.countryID) [count],
COUNT(x.username) OVER () [tcount],
COALESCE(c.Map,'world') [map]
FROM
(
SELECT COUNT(*),
CountryID,
username,
Score
FROM global.dbo.UserCategories AS g1
JOIN ( SELECT COUNT(*) AS Cnt,
CountryID,
username
FROM global.dbo.UserCategories
GROUP BY CountryID,
username
HAVING COUNT(*) > [Your_Threshold_Number]
) g2 ON g1.CountryID = g2.CountryID
WHERE TLCID LIKE @TLCID OR TLCID IN (@TLCID)
GROUP BY CountryID,
username,
Score
) x
JOIN global.dbo.countrydetails c on x.CountryID=c.country
AND x.Score BETWEEN @ScoreMin AND @ScoreMax
GROUP BY x.CountryID, c.Description, c.Map, x.username
ORDER BY [name] ASC