我有一个包含两个表的数据库; [客户] 和 [交易] 。我有一个外键分配给 [Transactions] 表中与 [Clients] 表相关的一个或多个记录。
在 [Transactions] 表格中，我有一个名为'URL'的字段，其中填充了网址（就像它在锡上所说的那样）。我想浏览 [Transactions] 表格中的所有记录，找出与'网址'字段中最常见的值，以便与 [客户] 表。
一旦我拥有最常见的值，我想将它们插入 [Clients] 表中，在一个名为'URL'的字段中（就像在< strong> [Transactions] table），针对关联的客户记录。
我确信我能解决大部分问题，我唯一的问题是找到许多不同小组的最常见值。任何帮助表示赞赏！
示例数据：
[Clients]
ID Name URL
-----------------------------------
999999999 Testing Client 1 NULL
999999998 Testing Client 2 NULL
999999997 Testing Client 3 NULL
999999996 Testing Client 4 NULL
999999995 Testing Client 5 NULL
[Transactions]
ID ClientID URL
-----------------------------------------
73611 999999999 http://www.google.com
73612 999999999 http://www.yahoo.com
73613 999999999 http://www.google.com
73626 999999998 http://www.stackoverflow.com
73627 999999998 http://www.stackoverflow.com
73628 999999998 http://www.slack.com
73629 999999997 http://www.dotnetpearls.com
73630 999999997 http://www.c-sharpcorner.com
73631 999999996 http://www.roastmymealdeal.co.uk
73632 999999996 http://www.roastmymealdeal.co.uk
73633 999999996 http://www.roastmymealdeal.co.uk
73634 999999996 NULL
73635 999999995 NULL
73636 999999995 http://www.w3schools.com
73637 999999995 http://www.w3schools.com
答案 0 :(得分：0)
这将使用最常见的URL更新Clients表：
update Clients
set Clients.URL = x2.URL
from(
--This takes the inner query and sorts the rows according to the URLCount (descending), assigning a rank (using the row_number() function).
--The highest URLCount will be given a URLRank of 1. The URLRank resets for each client (partition by clientID).
select ClientID, URL, row_number() over (partition by clientID order by URLCount desc) URLRank
from(
--This groups the clients, giving one row for each client/URL combo, along with how many times that combo occurs.
select t.ClientID, t.URL, Count(1) URLCount
from Transactions t
group by t.ClientID, t.Url
) x
) x2
where x2.URLRank = 1 --Set the URL to the highest ranking URL
and Clients.ID = x2.ClientID
如果您只想查看每个客户端最常用的URL，请使用以下命令：
select *
from(
select ClientID, URL, row_number() over (partition by clientID order by URLCount desc) URLRank
from(
select t.ClientID, t.URL, Count(1) URLCount
from Transactions t
group by t.ClientID, t.Url
) x
) x2
where x2.URLRank = 1
答案 1 :(得分：0)
您可以像这样使用2 CTE
;WITH temp as
(
SELECT cl.ID as ClientID, t.URL, COUNT(t.ID) as NumberTransactions
FROM Clients cl
LEFT JOIN [Transactions] t on cl.ID = t.ClientID
GROUP BY cl.ID, t.URL
),
temp1 as
(
SELECT *,
ROW_NUMBER() OVER(PARTITION BY t.ClientID ORDER BY t.NumberTransactions desc) as Rn
FROM temp t
)
SELECT t.ClientID, t.URL
FROM temp1 t
WHERE t.Rn = 1
答案 2 :(得分：0)
鉴于此数据：
IF OBJECT_ID('tempdb..#client') IS NOT NULL DROP TABLE #client
IF OBJECT_ID('tempdb..#transactions') IS NOT NULL DROP TABLE #transactions
CREATE TABLE #client (id INT, name VARCHAR(100), url VARCHAR(100));
CREATE TABLE #transactions (id INT IDENTITY(1, 1), clientID INT, url VARCHAR(100));
INSERT #client
( id, name, url )
VALUES ( 9, 'a', null ),
( 8, 'b', null ),
( 7, 'c', null ),
( 6, 'd', null ),
( 5, 'e', null );
INSERT #transactions
( clientID, url )
VALUES (9, 'http://www.google.com' ),
(9, 'http://www.yahoo.com' ),
(9, 'http://www.google.com' ),
(8, 'http://www.stackoverflow.com' ),
(8, 'http://www.stackoverflow.com' ),
(8, 'http://www.slack.com' ),
(7, 'http://www.dotnetpearls.com' ),
(7, 'http://www.c-sharpcorner.com' ),
(6, 'http://www.roastmymealdeal.co.uk' ),
(6, 'http://www.roastmymealdeal.co.uk' ),
(6, 'http://www.roastmymealdeal.co.uk' ),
(6, NULL ),
(5, NULL ),
(5, 'http://www.w3schools.com' ),
(5, 'http://www.w3schools.com' );
此查询将获取每个clientID的每个网址的计数。但是，我们确实每次交易获得一行
SELECT
urlCount = COUNT(*) OVER (PARTITION BY clientID, url)
, transactionURL = t.URL
, t.clientID
FROM #transactions t
从该查询中我们只需要最高urlCount，我们可以通过在urlCount DESC上订购并获得TOP 1.我们确实想要获得每个客户端这个最重要的网址
我们可以通过使用交叉应用来执行此操作，以便为每个客户端运行此计数和前1。交叉应用将在客户端表中的每一行运行一次内部查询。在client.ID上过滤内部查询以获取该客户端的urlCount。
内部查询按urlCount DESC排序，以获得最高计数的URL（每个客户端）。作为一个打破平局，它也在transactions.ID上排序 - 这样做是为了每次运行时产生相同的结果。
SELECT c.url, transURLs.transactionURL, c.id
FROM #client c
CROSS APPLY (
SELECT TOP 1
urlCount = COUNT(*) OVER (PARTITION BY clientID, url)
, transactionURL = t.URL
FROM #transactions t
WHERE t.clientID = c.ID
ORDER BY urlCount DESC, t.id
) transURLs
要完成它，我们只需更新
WITH baseData AS (
SELECT c.url, transURLs.transactionURL, c.id
FROM #client c
CROSS APPLY (
SELECT TOP 1
urlCount = COUNT(*) OVER (PARTITION BY clientID, url)
, transactionURL = t.URL
FROM #transactions t
WHERE t.clientID = c.ID
ORDER BY urlCount DESC, t.id
) transURLs
)
UPDATE baseData
SET url = transactionURL;
SELECT * FROM #client
答案 3 :(得分：0)
with ctetbl (clientid,url,cnt,rowid) As
(
Select t.clientid, t.url,t.cnt,
ROW_NUMBER () over (partition by clientid order by t.cnt desc)as RowId
from (select clientid,url, COUNT(1)as cnt from transactions group by clientid,url)t
)
Update c
set url=ct.url
from clients c
inner join
ctetbl ct on c.id=ct.clientid where rowid=1
答案 4 :(得分：0)
/*************Script to recreate the scenario *************/
Create table [Clients]
(
Id bigint PRIMARY KEY,
Name nvarchar(100),
URL nvarchar(1000)
)
Insert into Clients VALUES
(999999999,'Testing Client 1',NULL),
(999999998,'Testing Client 2',NULL),
(999999997,'Testing Client 3',NULL),
(999999996,'Testing Client 4',NULL),
(999999995,'Testing Client 5',NULL)
Create table Transactions
(
ID bigint,
ClientID bigint FOREIGN KEY REFERENCES Clients(ID),
URL nvarchar(1000)
)
Insert into Transactions VALUES
(73611, 999999999,'http://www.google.com'),
(73612, 999999999,'http://www.yahoo.com'),
(73613, 999999999,'http://www.google.com'),
(73626, 999999998,'http://www.stackoverflow.com'),
(73627, 999999998,'http://www.stackoverflow.com'),
(73628, 999999998,'http://www.slack.com'),
(73629, 999999997,'http://www.dotnetpearls.com'),
(73630, 999999997,'http://www.c-sharpcorner.com'),
(73631, 999999996,'http://www.roastmymealdeal.co.uk'),
(73632, 999999996,'http://www.roastmymealdeal.co.uk'),
(73633, 999999996,'http://www.roastmymealdeal.co.uk'),
(73634, 999999996,NULL),
(73635, 999999995,NULL),
(73636, 999999995,'http://www.w3schools.com'),
(73637, 999999995,'http://www.w3schools.com')
/***************List the tables *****************/
Select * from dbo.Clients
Select * from dbo.Transactions
/***************************************************************************************
cte_grp1 --
Using window functions find the ClientID and URL and number of times it is found in Transaction
cte_grp2
In second step, just find out the ones with maximum count and rank them accordingly
Store the result into another table
**************************************************************************************/
;with cte_grp1
as
(
SELECT
ClientID,
URL,
Count(URL) as CountOfURL
FROM Transactions
WHERE URL IS NOT NULL
GROUP BY ClientID,URL
-- Order by ClientID ASC,CountOfURL DESC
),
cte_grp2
as
(
SELECT y.ClientID,
y.URL,
x.MaxCount
FROM cte_grp1 y
INNER JOIN
(Select ClientID,URL,Max(CountOfURL) as MaxCount
FROM cte_grp1
Group by ClientID,URL ) x
ON x.ClientId=y.ClientID and x.URL=y.URL
--Order by y.ClientID,x.MaxCount DESC
)
Select ClientID,URL,MaxCount,
DENSE_RANK() OVER (PARTITION BY ClientId ORDER BY MaxCount DESC) as Rnk
INTO #Temp_Resultant
from cte_grp2
/*******************************************************************
Using this temp table we will be using FOR XML clause for those links
where a Client has visited the link one time each as they both quality
to be updated in the Clients table
At last.. Update based on ClientID
************************************************************************/
;with resultant
as
(
Select distinct t2.ClientID,STUFF((SELECT ','+ t1.URL
FROM #Temp_Resultant t1
WHERE Rnk=1
AND t1.ClientID=t2.ClientID
FOR XML PATH('')),1,1,'') as CommonURL
From #Temp_Resultant t2
)
Update A
SET A.URL=B.CommonURL
FROM Clients A INNER JOIN resultant B
ON A.Id=B.ClientID
---Check the results
Select * from Clients
Select * from Transactions