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sql - 如何使用 STUFF 和 FOR XML PATH 分隔与其行相关的唯一值

问题描述

我正在运行这个查询：

SELECT A.PeripheralNumber, P.LoginName, P.FirstName, P.LastName, A.EnterpriseName, A.AgentDeskSettingsID, AttributeValues = 
STUFF((SELECT ',' + ATT.EnterpriseName+'='+AT.AttributeValue AS Attribute
FROM Agent_Attribute AT
LEFT JOIN Agent A ON A.SkillTargetID = AT.SkillTargetID
LEFT JOIN Attribute ATT ON ATT.AttributeID = AT.AttributeID
WHERE A.SkillTargetID IN ('26379','24111')
ORDER BY Attribute
FOR XML PATH (''), TYPE ).value('.', 'VARCHAR(MAX)'), 1, 1, '')
FROM Agent A
LEFT JOIN Person P ON P.PersonID = A.PersonID
LEFT JOIN Agent_Desk_Settings ADS ON ADS.AgentDeskSettingsID = A.AgentDeskSettingsID
WHERE A.SkillTargetID IN ('26379','24111')
AND A.EnterpriseName LIKE 'PG_8%'
AND P.FirstName NOT LIKE '%Test%'
GROUP BY A.PeripheralNumber, P.LoginName, P.FirstName, P.LastName, A.EnterpriseName, A.AgentDeskSettingsID

我得到的结果如下：

|PeripheralNumber|LoginName|FirstName|LastName|EnterpriseName|AgentDeskSettingsID|AttributeValues                                                                                                 |
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
|551628          |551628   |John     |Doe     |Doe_John      |5104               |AAH_CSC_EXIST=5,AAH_CSC_FNOL=5,AAH_CSC_ROTL=5,AAH_SERVICE_AUTO=10,AAH_SERVICE_REBRAND=5,AAH_SERVICE_WEB_ONLINE=10|
|558792          |558792   |Jane     |Doe     |Doe_Jane      |5103               |AAH_CSC_EXIST=5,AAH_CSC_FNOL=5,AAH_CSC_ROTL=5,AAH_SERVICE_AUTO=10,AAH_SERVICE_REBRAND=5,AAH_SERVICE_WEB_ONLINE=10|
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

正在发生的事情是正在合并 John Doe 和 Jane Doe 的唯一 AttributeValues。我需要的结果如下：

|PeripheralNumber|LoginName|FirstName|LastName|EnterpriseName|AgentDeskSettingsID|AttributeValues                                                                                                 |
------------------------------------------------------------------------------------------------------------------------------------------------------
|551628          |551628   |John     |Doe     |Doe_John      |5104               |AAH_SERVICE_AUTO=10,AAH_SERVICE_REBRAND=5,AAH_SERVICE_WEB_ONLINE=10|
|558792          |558792   |Jane     |Doe     |Doe_Jane      |5103               |AAH_CSC_EXIST=5,AAH_CSC_FNOL=5,AAH_CSC_ROTL=5                      |
------------------------------------------------------------------------------------------------------------------------------------------------------

帮助！这是 MS SQL Server 2016

标签： sqlsql-serverxmlsql-server-2016stuff

解决方案

你cross apply或outer apply. 像这样的东西：

SELECT DISTINCT A.PeripheralNumber, P.LoginName, P.FirstName, P.LastName, A.EnterpriseName, A.AgentDeskSettingsID, DS.AttributeValues
FROM Agent A
LEFT JOIN Person P
    ON A.PersonID = P.PersonID
LEFT JOIN Agent_Desk_Settings ADS
    ON A.AgentDeskSettingsID = ADS.AgentDeskSettingsID
OUTER APPLY
(
    SELECT STUFF((SELECT ',' + ATT.EnterpriseName+'='+AT.AttributeValue AS Attribute
    FROM Agent_Attribute AT
    INNER JOIN Attribute ATT
        ON ATT.AttributeID = AT.AttributeID
    WHERE A.SkillTargetID = AT.SkillTargetID
    ORDER BY Attribute
    FOR XML PATH (''), TYPE ).value('.', 'VARCHAR(MAX)'), 1, 1, '')
) DS (AttributeValues)
WHERE A.SkillTargetID IN ('26379','24111')
    AND A.EnterpriseName LIKE 'PG_8%'
    AND P.FirstName NOT LIKE '%Test%'

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