显示最高PK

Question

以下查询提取相同的数据，但有两个不同的PTACHI_PK。我只需要显示最高的PTACHI_PK。我不确定如何在SQL Server中实现这一点。

SELECT DISTINCT
    PTACHI_PK,
    ACT.TAX_PAYER_ID AS 'AP_OWENER_ID',
    IDM.ID_PK AS 'NCPTS_OWNER_ID',
    '' as REL_PK,
    ADDR.ADDRESS_PK,
    FIRST_NAME, MIDDLE_NAME, LAST_NAME,
    BUSINESS_NAME,
    ADDR1, ADDR2, ADDR3,
    IN_CARE_OF,
    CITY,
    RIGHT(STATE,2) as 'STATE',
    POSTAL_CODE AS 'ZIP',
    POSTAL_CODE_EXT AS 'ZIP4',
    HIS.FIELD_CHANGED,
    HIS.FIELD_OLD_VALUE,
    HIS.FIELD_NEW_VALUE,
    CONVERT(VARCHAR, MODIFIED_TS, 22) AS 'DATE_CHANGED'
FROM
    OWNIDM_ID_MASTER IDM
JOIN 
    OWNREL_OWNER_RELATIONSHIP REL ON IDM.ID_PK = REL.IDM_ID_PK
JOIN 
    OWNACT_ACCOUNT ACT ON REL.ACT_ACCOUNT_PK = ACT.ACCOUNT_PK
JOIN 
    COMADD_COMMON_ADDRESS ADDR ON REL.ADD_ADDRESS_PK = ADDR.ADDRESS_PK
JOIN 
    COMNAM_COMMON_NAME NAME ON REL.NAM_NAME_PK = NAME.NAME_PK
JOIN 
    PTACHI_CHANGE_HISTORY HIS ON IDM.ID_PK = HIS.PARENT_PK
WHERE 
    ACT.TAX_PAYER_ID IS NOT NULL
    AND ACT.TAX_PAYER_ID <> ''
    AND ISDEFAULT_ADDRESS = 'Y'
    AND ADDR.SOURCE_TYPE <> 'ADSRDMV'
    AND HIS.PARENT_TYPE = 'COMADD_COMMON_ADDRESS'
    AND HIS.FIELD_CHANGED = 'OWNER_ADDRESS_UPDATED'
    AND MODIFIED_TS >= DATEADD(day, DATEDIFF(day,0,GETDATE())-1,0)
    AND MODIFIED_TS < DATEADD(day, DATEDIFF(day,0,GETDATE()),0)
    AND ID_PK = 432082
GROUP BY 
    ACT.TAX_PAYER_ID, IDM.ID_PK, ADDR.ADDRESS_PK, 
    FIRST_NAME, MIDDLE_NAME, LAST_NAME,
    BUSINESS_NAME, ADDR1, ADDR2, ADDR3,
    IN_CARE_OF, CITY, STATE,
    POSTAL_CODE, POSTAL_CODE_EXT,
    HIS.FIELD_CHANGED, HIS.FIELD_OLD_VALUE,
    HIS.FIELD_NEW_VALUE, MODIFIED_TS, PTACHI_PK
ORDER BY 
    PTACHI_PK DESC

我正在使用SQL Server。附上一些结果样本。

Answer 1

SELECT TOP 1 

..... 

ORDER BY PTACHI_PK DESC

查看它是否有效并在合理的时间内运行。

Answer 2

您应该能够使用相同的连接和过滤器来确定最大值。例如：

SELECT MAX(PTACHI_PK)
FROM OWNIDM_ID_MASTER IDM
JOIN OWNREL_OWNER_RELATIONSHIP REL ON IDM.ID_PK = REL.IDM_ID_PK
JOIN OWNACT_ACCOUNT ACT ON REL.ACT_ACCOUNT_PK = ACT.ACCOUNT_PK
JOIN COMADD_COMMON_ADDRESS ADDR ON REL.ADD_ADDRESS_PK = ADDR.ADDRESS_PK
JOIN COMNAM_COMMON_NAME NAME ON REL.NAM_NAME_PK = NAME.NAME_PK
JOIN PTACHI_CHANGE_HISTORY HIS ON IDM.ID_PK = HIS.PARENT_PK
WHERE 
    ACT.TAX_PAYER_ID IS NOT NULL
    AND ACT.TAX_PAYER_ID <> ''
    AND ISDEFAULT_ADDRESS = 'Y'
    AND ADDR.SOURCE_TYPE <> 'ADSRDMV'
    AND HIS.PARENT_TYPE = 'COMADD_COMMON_ADDRESS'
    AND HIS.FIELD_CHANGED = 'OWNER_ADDRESS_UPDATED'
    AND MODIFIED_TS >= DATEADD(day, DATEDIFF(day,0,GETDATE())-1,0)
    AND MODIFIED_TS < DATEADD(day, DATEDIFF(day,0,GETDATE()),0)
    AND ID_PK = 432082

知道如何获得最大值，您可以将其作为原始算法的Where中的过滤器：

WHERE 
    ACT.TAX_PAYER_ID IS NOT NULL
    AND ACT.TAX_PAYER_ID <> ''
    AND ISDEFAULT_ADDRESS = 'Y'
    AND ADDR.SOURCE_TYPE <> 'ADSRDMV'
    AND HIS.PARENT_TYPE = 'COMADD_COMMON_ADDRESS'
    AND HIS.FIELD_CHANGED = 'OWNER_ADDRESS_UPDATED'
    AND MODIFIED_TS >= DATEADD(day, DATEDIFF(day,0,GETDATE())-1,0)
    AND MODIFIED_TS < DATEADD(day, DATEDIFF(day,0,GETDATE()),0)
    AND ID_PK = 432082

    --using the code for MAX(PTACHI_PK) to filter PTACHI_PK:
    AND PTACHI_PK = 
          (SELECT MAX(PTACHI_PK)
             FROM OWNIDM_ID_MASTER IDM
             JOIN OWNREL_OWNER_RELATIONSHIP REL ON IDM.ID_PK = REL.IDM_ID_PK
             JOIN OWNACT_ACCOUNT ACT ON REL.ACT_ACCOUNT_PK = ACT.ACCOUNT_PK
             JOIN COMADD_COMMON_ADDRESS ADDR ON REL.ADD_ADDRESS_PK = ADDR.ADDRESS_PK
             JOIN COMNAM_COMMON_NAME NAME ON REL.NAM_NAME_PK = NAME.NAME_PK
             JOIN PTACHI_CHANGE_HISTORY HIS ON IDM.ID_PK = HIS.PARENT_PK
             WHERE 
                   ACT.TAX_PAYER_ID IS NOT NULL
                   AND ACT.TAX_PAYER_ID <> ''
                   AND ISDEFAULT_ADDRESS = 'Y'
                   AND ADDR.SOURCE_TYPE <> 'ADSRDMV'
                   AND HIS.PARENT_TYPE = 'COMADD_COMMON_ADDRESS'
                   AND HIS.FIELD_CHANGED = 'OWNER_ADDRESS_UPDATED'
                   AND MODIFIED_TS >= DATEADD(day, DATEDIFF(day,0,GETDATE())-1,0)
                   AND MODIFIED_TS < DATEADD(day, DATEDIFF(day,0,GETDATE()),0)
                   AND ID_PK = 432082
         )

另外，我会接受Aaron Bertrand的建议而不是distinct加group by，因为它不必要地删除重复次数两次。