T-SQL - 使用 STUFF 连接分组列并删除重复项

T-SQL - using STUFF to concatenate grouped columns and removing duplicates

我有一个 table 看起来像这样:

EmailAddress: nvarchar(255)
MarketingEmailOptIn: nvarchar(50)
NewsletterOptIn: nvarchar(50)
ThoughtLeaderOptIn: nvarchar(50)

下面显示的我的 SQL 语句采用上面的数据并使用逗号作为分隔符连接“订阅类型”:

SELECT  
    EmailAddress,
    STUFF((SELECT ',' + 
              CASE
                 WHEN B.MarketingEmailOptIn = 'TRUE' THEN 'MarketingEmail' 
                 WHEN B.ThoughtLeaderOptIn = 'TRUE' THEN 'ThoughtLeader'
                 WHEN B.NewsletterOptIn = 'TRUE' THEN 'Newsletter'
              END
          FROM UK_AGT_AgentForms_TEST_DE B 
          WHERE ISNULL(B.EmailAddress, '') = ISNULL(A.EmailAddress, '')
          FOR XML PATH('')), 1, 2, '') AS Subscriptions
FROM
    UK_AGT_AgentForms_TEST_DE A
GROUP BY 
    EmailAddress 

运行 这个 SQL 产生以下输出:

但是请注意 MarketingEmail 被列出了两次,因为来源 table 也列出了两次(第一行和第二行)。我需要忽略检测到的任何重复项,这样我得到的 table 看起来像:

我对 STUFF 关键字很陌生。我只是有点迷失了如何在 运行 时间检测重复项 - 任何建议都将不胜感激。谢谢

尝试这样的事情:

DECLARE @Data table (
    EmailAddress nvarchar(255),
    MarketingEmailOptIn nvarchar(50),
    NewsletterOptIn nvarchar(50),
    ThoughtLeaderOptIn nvarchar(50)
);

INSERT INTO @Data VALUES
    ( 'mike@mikemarks.com', 'TRUE', NULL, NULL ),
    ( 'mike@mikemarks.com', 'TRUE', 'TRUE', NULL ),
    ( 'mike@mikemarks.com', 'TRUE', NULL, 'TRUE' );

SELECT
    EmailAddress
    , STUFF ( ( CASE WHEN EOptIn = 'TRUE' THEN ',MarketingEmail' ELSE '' END
        + CASE WHEN NOptIn = 'TRUE' THEN ',Newsletter' ELSE '' END
        + CASE WHEN TOptIn = 'TRUE' THEN ',ThoughtLeader' ELSE '' END 
    ), 1, 1, '' ) AS Subscriptions
FROM (

    SELECT TOP 100 PERCENT
        EmailAddress
        , MAX ( MarketingEmailOptIn ) AS EOptIn
        , MAX ( NewsletterOptIn ) AS NOptIn
        , MAX ( ThoughtLeaderOptIn ) AS TOptIn
    FROM @Data A --UK_AGT_AgentForms_TEST_DE
    GROUP BY EmailAddress
    ORDER BY EmailAddress

) AS x
ORDER BY 
    EmailAddress;

Returns

+--------------------+-----------------------------------------+
|    EmailAddress    |              Subscriptions              |
+--------------------+-----------------------------------------+
| mike@mikemarks.com | MarketingEmail,Newsletter,ThoughtLeader |
+--------------------+-----------------------------------------+

如果您有 Sql Server 2017 或更高版本,您可以使用 String_agg() 来简化此操作:

SELECT   
    EmailAddress,
        STRING_AGG(CASE
                 WHEN MarketingEmailOptIn = 'TRUE' THEN 'MarketingEmail' 
                 WHEN ThoughtLeaderOptIn = 'TRUE' THEN 'ThoughtLeader'
                 WHEN NewsletterOptIn = 'TRUE' THEN 'Newsletter'
              END, ', ') AS Subscriptions
FROM
    UK_AGT_AgentForms_TEST_DE
GROUP BY 
    EmailAddress

如果仍然看到重复项,可以在嵌套查询中使用条件聚合先将其汇总:

SELECT  
    EmailAddress,
          CASE WHEN MarketingEmailOptIn > 0 THEN 'MarketingEmail,' ELSE '' END
        + CASE WHEN ThoughtLeaderOptIn > 0 THEN 'ThoughtLeader,' ELSE '' END
        + CASE WHEN NewsletterOptIn = > 0 THEN 'Newsletter' ELSE '' END
         AS Subscriptions
FROM (
    SELECT EmailAddress
        , SUM(CASE WHEN MarketingEmailOptIn = 'TRUE' THEN 1 ELSE 0 END) MarketingEmailOptIn
        , SUM(CASE WHEN ThoughtLeaderOptIn = 'TRUE' THEN 1 ELSE 0 END) ThoughtLeaderOptIn
        , SUM(CASE WHEN NewsletterOptIn = 'TRUE' THEN 1 ELSE 0 END) NewsletterOptIn
    FROM UK_AGT_AgentForms_TEST_DE
    GROUP BY EmailAddress
) T

呸。我不得不玩这个。也许不是完美的解决方案,但我认为我能够实现您正在尝试的目标。但是它不使用 stuff 函数。它只是连接每个字符串,然后删除最后一个逗号。

SELECT EmailAddress, CASE WHEN LEN(Subscriptions) > 0 THEN LEFT(Subscriptions, LEN(Subscriptions) - 1) ELSE '' END AS Subscriptions
FROM (
    SELECT EmailAddress, CONCAT(
            CASE WHEN SUM(CASE WHEN MarketingEmailOptIn = 'TRUE' THEN 1 ELSE 0 END) > 0 THEN 'MarketingEmail, ' ELSE '' END,
            CASE WHEN SUM(CASE WHEN NewsletterOptIn = 'TRUE' THEN 1 ELSE 0 END) > 0 THEN 'Newsletter, ' ELSE '' END,
            CASE WHEN SUM(CASE WHEN ThoughtLeaderOptIn = 'TRUE' THEN 1 ELSE 0 END) > 0 THEN 'ThoughLeader, ' ELSE '' END
        ) AS Subscriptions
    FROM UK_AGT_AgentForms_TEST_DE 
    GROUP BY EmailAddress
) AS a