-1

我有一个看起来像这样的表：

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

在此处输入图像描述

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

运行此 SQL 会产生以下输出：

在此处输入图像描述

但是请注意，MarketingEmail它列出了两次，因为源表也列出了两次（第 1 行和第 2 行）。我需要省略检测到的任何重复项，以便生成的表如下所示：

在此处输入图像描述

我对STUFF关键字很陌生。我只是有点迷失如何在运行时检测重复项 - 任何建议都值得赞赏。谢谢

4

3 回答 3

2

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;

退货

+--------------------+-----------------------------------------+
|    EmailAddress    |              Subscriptions              |
+--------------------+-----------------------------------------+
| mike@mikemarks.com | MarketingEmail,Newsletter,ThoughtLeader |
+--------------------+-----------------------------------------+
于 2021-04-05T18:29:17.747 回答
1

如果你有 Sql Server 2017 或更高版本，你可以使用它String_agg()来简化：

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

如果您仍然看到重复项，您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
于 2021-04-05T18:37:03.533 回答
1

皮尤。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。它虽然不使用 stuff 功能。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
于 2021-04-05T18:37:19.383 回答