Question

我已经搜索了论坛，但要么不能正确地提出问题，要么不理解答案，需要有人带我一步一步。

问题是：我的数据库中有一个表，用户。根据电子邮件比较，有一些重复。现在，根据注册日期，其中一些具有更高的优先级（我们将省略具有较早注册日期的那些），但是一些具有较低优先级日期的记录具有更多信息填充（例如性别，地址，电话等）上）。

我想得到的流程是： - ＆GT;根据电子邮件查找重复项 - ＆GT;使用最新注册日期确定行的优先级 - ＆GT;如果此行中的单元格为空，请使用优先级较低的行中的数据填充

P.S。问题是，最多可能有三个重复的帐户使用相同的电子邮件。

CREATE TABLE [dbo].[Person](
    [userID] [nvarchar] PRIMARY KEY,
    [email] [nvarchar] (50),
    [priority] [nvarchar](2),
    [FirstName] [nvarchar](50),
    [LastName] [nvarchar](50)
)

GO

INSERT INTO Person VALUES (1,'a@a.com','1','','');
INSERT INTO Person VALUES (2,'a@a.com','2','Dennis','Li');
INSERT INTO Person VALUES (3,'b@b.com','1','Brent','Li');
INSERT INTO Person VALUES (4,'c@c.com','1','','');
INSERT INTO Person VALUES (5,'c@c.com','2','','Raji');
INSERT INTO Person VALUES (6,'c@c.com','3','Ben','Raji');
GO

Answer 1

使用下面的脚本，我们将注册表保留在最近的regDate中，并使用之前的注册表填充NULL值以用于同一封电子邮件。

但是，如果您有三个或更多用户使用相同的电子邮件，则会忽略较旧的行，我们只会将最新的行与最新的行合并：

INSERT INTO Users ([email],[firstName],[lastName],[street],[city],[code],[country],[phone],[regDate])
VALUES ('a@a.com', 'Andrew', null, null, null, null, null, null, '2018-03-09 00:00:00');

INSERT INTO Users ([email],[firstName],[lastName],[street],[city],[code],[country],[phone],[regDate])
VALUES ('a@a.com', 'ANDREEW', 'Lopez', null, 'Santos', null, null, null, '2018-03-08 00:00:00');

INSERT INTO Users ([email],[firstName],[lastName],[street],[city],[code],[country],[phone],[regDate])
VALUES ('b@b.com', 'Bob', 'Wilk', null, null, null, null, null, '2018-03-10 00:00:00');

INSERT INTO Users ([email],[firstName],[lastName],[street],[city],[code],[country],[phone],[regDate])
VALUES ('b@b.com', 'Robert', null, 'Sandiego Street', 'Santos', null, null, '456 123 789', '2018-03-05 00:00:00');

SELECT * FROM Users;

INSERT INTO Users ([email],[firstName],[lastName],[street],[city],[code],[country],[phone],[regDate])
SELECT
u.[email],
ISNULL(u.firstName,old.firstName),
ISNULL(u.lastName,old.lastName),
ISNULL(u.street,old.street),
ISNULL(u.city,old.city),
ISNULL(u.code,old.code),
ISNULL(u.country,old.country),
ISNULL(u.phone,old.phone),
u.regDate
FROM Users u
INNER JOIN Users old ON old.Id = (SELECT TOP 1 Id FROM Users oldMax WHERE oldMax.email = u.email AND oldMax.Id <> u.Id ORDER BY oldMax.regDate DESC)
WHERE u.Id = (SELECT TOP 1 new.Id From Users new WHERE new.email = u.email ORDER BY new.regDate DESC);

DELETE FROM Users WHERE Id NOT IN (SELECT MAX(Id) FROM Users GROUP BY email);

SELECT * FROM Users;

Here你是一个工作小提琴。

Answer 2

这应该这样做

declare @T TABLE (
    [userID] int PRIMARY KEY,
    [email] [nvarchar] (50),
    [priority] tinyint,
    [FirstName] [nvarchar](50),
    [LastName] [nvarchar](50)
);
INSERT INTO @T VALUES 
       (1,'a@a.com', 1, null, null)
     , (2,'a@a.com', 2, 'Dennis','Li')
     , (3,'b@b.com', 1, 'Brent','Li')
     , (4,'c@c.com', 1, null,null)
     , (5,'c@c.com', 2, null,'Raji')
     , (6,'c@c.com', 3, 'Ben','Raji'); 
select t1.email 
    , (select top 1 tt.FirstName from @T tt where tt.FirstName is not null and tt.email = t1.email order by tt.priority asc) as FN
    , (select top 1 tt.LastName  from @T tt where tt.LastName  is not null and tt.email = t1.email order by tt.priority asc) as LN
from @T t1  
group by t1.email
order by t1.email;

Answer 3

下一个CTE只显示重复的电子邮件数据。如果您需要一个适用于重复电子邮件和非重复电子邮件的查询，您应该删除第一个CTE并完成！

;WITH DuplicatedEmails AS
(
    SELECT
        P.Email
    FROM
        Person AS p
    GROUP BY
        P.Email
    HAVING
        COUNT(1) > 1
),
DuplicatedEmailUserData AS
(
  SELECT
      P.*,
      EmailRanking = ROW_NUMBER() OVER (PARTITION BY Email ORDER BY Priority DESC) -- Assuming a higher priority comes first
  FROM
      Persons AS P
      INNER JOIN DuplicatedEmails AS E ON P.Email = E.Email
)
SELECT
    D1.UserID,
    D1.Email,
    D1.Priority,
    FirstName = COALESCE(D1.FirstName, D2.FirstName, D3.Firstname), -- Use COALESCE for the columns that might be NULL on 1st record
    LastName = COALESCE(D1.LastName, D2.LastName, D3.Lastname)
FROM
    DuplicatedEmailUserData AS D1
    LEFT JOIN DuplicatedEmailUserData AS D2 ON 
        D1.Email = D2.Email AND
        D1.EmailRanking + 1 = D2.EmailRanking
    LEFT JOIN DuplicatedEmailUserData AS D3 ON 
        D1.Email = D3.Email AND
        D2.EmailRanking + 1 = D3.EmailRanking
WHERE
    D1.EmailRanking = 1

使用此方法，您可能需要LEFT JOIN与重复的电子邮件一样多次。

如果first为null，SQL Server将比较两行并组合第二行中的单元格

3 个答案: