Question

示例数据

userid         email_address               login_name       name                Title       org            phone_number_com
=============  ==========================  ===============  ==================  ==========  =============  ===================
1192           Steve.Jobs@apple.com        sjobs            Steve Jobs          CEO         Apple          N/A
1274           Steve.Jobs@apple.com        sjobs            Steve Jobs          CFO         Apple          697-4686
1192           Steven.jobs@apple.com       sjobs            Steven jobs         CEO         Apple          604-7126
1885           Bill.Gates@microsoft.com    bgates           Bill Gates          CEO         Microsoft      604-7114
1920           Bill.Gates@microsoft.com    bgates           William Gates       CTR         Microsoft      604-7247
1951           Warren.Buffet@hp.com        wbuffet          Warren Buffet       CEO         HP             614-9141
1954           Warren.Buffet@hp.com        wbuffet          W. Buffet           COO         HP             614-7589
1951           Warren.Buffet@xerox.com     wbuffet          Warren S Buffet     CIO         Xerox          614-8874
1956           Mark.Zuckerberg@fb.com      mzuck            Mark  Zuckerberg    CEO         FB             614-8295

QUERY

SELECT * 
FROM
    (
        SELECT userid, name, login_name, email_address, phone_number_com,
        ROW_NUMBER() OVER(PARTITION BY [login_name] ORDER BY login_name) Num_Duplicates
        FROM web_user
    ) as Rows
WHERE Num_Duplicates > 1

这是我的第一篇文章，希望我遵守所有程序。我得到一个结果集，显示重复的第2行和第3行。我正在尝试GROUP BY login_name，只显示最高Num_Duplicates的行。如果login_name的{{1}}为2和3，则仅显示3行。我希望这有意义！提前感谢您提供的任何指导。

这些结果我想要查询输出：

Num_Duplicates

Answer 1

如果我理解您要正确执行的操作，则首先按登录名分组以获取重复项数：

SELECT login_name, COUNT(*) AS num_duplicates
  FROM web_user
 GROUP BY login_name

在这里你可以使用带有ROW_NUMBER()的子查询（虽然我建议在绑定的情况下使用RANK()）或者你可以在窗口函数中使用聚合：

SELECT login_name, COUNT(*) AS num_duplicates
     , RANK() OVER ( ORDER BY COUNT(*) DESC ) AS rn
  FROM web_user
 GROUP BY login_name;

然后将其放在子查询中以仅获得具有最多重复项的login_name：

SELECT * FROM (
    SELECT login_name, COUNT(*) AS num_duplicates
         , RANK() OVER ( ORDER BY COUNT(*) DESC ) AS rn
      FROM web_user
     GROUP BY login_name
) WHERE rn = 1;

根据OP的评论更新，问题编辑：

SELECT userid, name, login_name, email_address, phone_number_com, num_duplicates
  FROM (
    SELECT userid, name, login_name, email_address, phone_number_com
         , COUNT(*) OVER ( PARTITION BY login_name ) AS num_duplicates
         , ROW_NUMBER() OVER ( PARTITION BY login_name ORDER BY userid ) AS rn
      FROM web_user
) WHERE num_duplicates > 1 AND rn = 1;

我上面所做的是使用COUNT(*)作为窗函数;按login_name分区将获得每个登录名的计数。我也按login_name进行分区以获取ROW_NUMBER()并按userid排序，以便我可以返回最小值（您似乎在所需的输出中执行此操作）。

Answer 2

嗯 - 从你的描述中听起来就像你只是想要这样的东西（在我的头顶）：

SELECT login_name, email_address 
FROM web_user
GROUP BY login_name, email_address
HAVING count(*) > 2

Answer 3

以下内容应该为您提供所需。

ROW_NUMBER窗口函数用于标识login_name的第一行。

COUNT窗口函数用于计算每个login_name的行数。

外部查询然后将结果限制为具有多于1行的login_name，并且仅返回每个login_name的第一行。

DECLARE @users TABLE
(
    userid              int
    , email_address     varchar(100)
    , login_name        varchar(100)
    , name              varchar(100)
    , title             varchar(100)
    , org               varchar(100)
    , phone_number_com  varchar(100)
)

INSERT INTO @users
VALUES
(1192, 'Steve.Jobs@apple.com', 'sjobs', 'Steve Jobs', 'CEO', 'Apple', 'N/A')
, (1274, 'Steve.Jobs@apple.com', 'sjobs', 'Steve Jobs', 'CFO', 'Apple', '697-4686')
, (1192, 'Steven.jobs@apple.com', 'sjobs', 'Steven jobs', 'CEO', 'Apple', '604-7126')
, (1885, 'Bill.Gates@microsoft.com', 'bgates', 'Bill Gates', 'CEO', 'Microsoft', '604-7114')
, (1920, 'Bill.Gates@microsoft.com', 'bgates', 'William Gates', 'CTR', 'Microsoft', '604-7247')
, (1951, 'Warren.Buffet@hp.com', 'wbuffet', 'Warren Buffet', 'CEO', 'HP', '614-9141')
, (1954, 'Warren.Buffet@hp.com', 'wbuffet', 'W. Buffet', 'COO', 'HP', '614-7589')
, (1951, 'Warren.Buffet@xerox.com', 'wbuffet', 'Warren S Buffet', 'CIO', 'Xerox', '614-8874')
, (1956, 'Mark.Zuckerberg@fb.com', 'mzuck', 'Mark Zuckerberg', 'CEO', 'FB', '614-8295')
;

WITH LoginWithWindowFunction AS
(
    SELECT
        *
        , ROW_NUMBER() OVER(PARTITION BY login_name ORDER BY userid) AS LoginOrder
        , COUNT(*) OVER(PARTITION BY login_name) AS Num_Duplicates

    FROM
        @users
)

SELECT
    userid
    , email_address
    , login_name
    , name
    , title
    , org
    , phone_number_com
    , Num_Duplicates

FROM
    LoginWithWindowFunction

WHERE
    LoginOrder = 1
    AND Num_Duplicates > 1

ORDER BY
    userid

组合GROUP BY和ROW_NUMBER（）

3 个答案: