Question

我想弄清楚为什么我的代码在取出Account_ID时不会返回任何数据。我希望它运行并找到满足连接条件和我想要返回的字段的每个Account_ID。这是在MySQL，我在这里和谷歌搜索，但没有找到这样的东西或只是没有正确。所以，下面是代码，如果有人可以帮助我，我会非常感激。代码应该每个Account_ID返回1行我不习惯使用MySQL。版本5.6.34使用Toad Data-Point运行sql脚本。谢谢Dan

    SELECT DISTINCT
    pd.ACCOUNT_ID                 
    , do.MEMBER_NAME
    , do.GENDER
    , MaxDate.EarlyDATE AS ACTIVITY_DATE
    , MaxDate.LateDATE  AS LAST_ACTIVITY_DATE
    , pd.NUMBER_WKS        
    ... more fields after this
    Joins below
    FROM ZZ_Program_Details pd 
        LEFT JOIN ZZ_Demographics do 
         ON do.ACCOUNT_ID = pd.ACCOUNT_ID
    INNER JOIN 
        (SELECT DISTINCT cl.ACCOUNT_ID
        , min(cl.activity_date)  AS EarlyDATE
        , cl1.activity_date AS LateDATE
        , cl1.scheduling_selection
        , cl.ATND_STATUS_THIS_INTERVAL         
        , cl.WEIGHT_STATUS_FOR_INTERVAL        
        , TRUNCATE(cl.WEIGHT_WHEN_ENROLLED,2)   AS WEIGHT_WHEN_ENROLLED
        , TRUNCATE(cl.WEIGHT_AT_INTERVAL_END,2) AS WEIGHT_AT_INTERVAL_END
        FROM ZZ_Class_Attendance cl
         INNER JOIN ZZ_Class_Attendance cl1
          ON cl1.account_id = cl.ACCOUNT_ID 
        WHERE cl1.ACTIVITY_DATE > cl.ACTIVITY_DATE
         /*and cl.account_id = '47F74C65BA8BB02DE053BC010B0AF714'*/
        ) AS MaxDate 
    INNER JOIN 
     (SELECT ESRA.healthfleet_member_id
      , ESRA.Sponsor
      , ESRA.uhc_region
      , ESRA.uhc_major_market
      , ESRA.uhc_minor_market AS ESRA_UHC_MINOR_MARKET
      , ESRA.policy_number AS ESRA_POLICY_NBR
      , ESRA.relationship_code
      , sf.Market_Segment
      , sf.UHC_Minor_Market   AS SF_UHC_MINOR_MARKET
      , sf.CLIENT_NAME
      , sf.Client_Owner
      , sf.Payer_Effective_Date
    FROM reporting_adhoc.ZZ_ESRA_People ESRA
     LEFT JOIN reporting_adhoc.ZZ_Salesforce sf
      ON sf.Policy_Number = ESRA.policy_number
    /*WHERE ESRA.healthfleet_member_id='47F74C65BA8BB02DE053BC010B0AF714'*/
    GROUP BY ESRA.healthfleet_member_id
    , ESRA.Sponsor, ESRA.uhc_region
    , ESRA.uhc_major_market
    , ESRA.uhc_minor_market
    , ESRA.relationship_code
    , sf.Market_Segment
    , sf.UHC_Minor_Market
    , sf.CLIENT_NAME
    , sf.Client_Owner, ESRA.policy_number
    , sf.Payer_Effective_Date
    ) ESRA_DATA 

    ON MaxDate.ACCOUNT_ID = pd.ACCOUNT_ID
    AND ESRA_DATA.healthfleet_member_id = pd.ACCOUNT_ID

    /*WHERE pd.account_id = '47F74C65BA8BB02DE053BC010B0AF714'* --with this 
    it works fine */ 
    WHERE pd.account_id > 0 --an attempt with getting all account_ids

    GROUP BY
    pd.ACCOUNT_ID                 
    , do.MEMBER_NAME  
    , do.GENDER
    , MaxDate.EarlyDATE
    , MaxDate.LateDATE
    ... rest of fields
    ORDER BY pd.ACCOUNT_ID

Answer 1

如果无法访问您的表格和数据，我们可能无法提出明确的答案，而且由于您的查询缺少部分，因此它更像是一个猜谜游戏。

让我们从这里开始：

SELECT DISTINCT  ---------- bad bad
      pd.ACCOUNT_ID
    , MaxDate.EarlyDATE AS ACTIVITY_DATE
    , MaxDate.LateDATE  AS LAST_ACTIVITY_DATE

FROM ZZ_Program_Details pd
LEFT JOIN ZZ_Demographics do ON do.ACCOUNT_ID = pd.ACCOUNT_ID
INNER JOIN (
      SELECT DISTINCT  ---------- bad bad
            cl.ACCOUNT_ID
          , MIN(cl.activity_date) AS EarlyDATE
          , cl1.activity_date     AS LateDATE
      FROM ZZ_Class_Attendance cl
      INNER JOIN ZZ_Class_Attendance cl1 ON cl1.account_id = cl.ACCOUNT_ID
      WHERE cl1.ACTIVITY_DATE > cl.ACTIVITY_DATE

      -- no GROUP BY, so how do your get MIN() or MAX() ?

   ) AS MaxDate ON 
                    ( WHAT???? )

首先将您的大查询分成较小的块，就像您在上面看到的那样。 此部分查询是否有效？可能不是因为您尝试使用DISTINCT而不是GROUP BY，并且连接条件也是未知的。

另请注意，您使用了INNER JOIN。如果该子查询MaxDate未能返回任何匹配的行，则整个查询将不会返回行。重新考虑每个INNER JOIN（也许尝试将它们作为LEFT JOINS）。

使部分查询生效后，添加下一个连接并将相关列添加到top select子句中。处理查询的这一部分，直到成功，移到下一个并重复，直到找到有效的查询。

请尽量避免使用＆＃34;选择不同的＆＃34;在涉及多个连接的复杂查询中。这不是一个好习惯。

您可能会发现这些查询将帮助您确定哪些表（或子查询）需要左连接：

SELECT COUNT(distinct pd.ACCOUNT_ID) acct_c, COUNT(*) row_c
FROM ZZ_Program_Details pd 
;

SELECT COUNT(distinct pd.ACCOUNT_ID) acct_c, COUNT(*) row_c
FROM ZZ_Program_Details pd 
INNER JOIN ZZ_Demographics do ON pd.ACCOUNT_ID = do.ACCOUNT_ID
;

SELECT COUNT(distinct pd.ACCOUNT_ID) acct_c, COUNT(*) row_c
FROM ZZ_Program_Details pd 
LEFT JOIN ZZ_Demographics do ON pd.ACCOUNT_ID = do.ACCOUNT_ID
;

SELECT COUNT(distinct pd.ACCOUNT_ID) acct_c, COUNT(*) row_c
FROM ZZ_Program_Details pd 
INNER JOIN ZZ_Class_Attendance cl ON pd.ACCOUNT_ID = cl.ACCOUNT_ID
INNER JOIN ZZ_Class_Attendance cl1 ON cl1.account_id = cl.ACCOUNT_ID 
;

SELECT COUNT(distinct pd.ACCOUNT_ID) acct_c, COUNT(*) row_c
FROM ZZ_Program_Details pd 
LEFT JOIN ZZ_Class_Attendance cl ON pd.ACCOUNT_ID = cl.ACCOUNT_ID
LEFT JOIN ZZ_Class_Attendance cl1 ON cl1.account_id = cl.ACCOUNT_ID 
;

SELECT COUNT(distinct pd.ACCOUNT_ID) acct_c, COUNT(*) row_c
FROM ZZ_Program_Details pd 
INNER JOIN reporting_adhoc.ZZ_ESRA_People ESRA ON pd.ACCOUNT_ID = ESRA.healthfleet_member_id
INNER JOIN reporting_adhoc.ZZ_Salesforce sf ON sf.Policy_Number = ESRA.policy_number
;

SELECT COUNT(distinct pd.ACCOUNT_ID) acct_c, COUNT(*) row_c
FROM ZZ_Program_Details pd 
LEFT JOIN reporting_adhoc.ZZ_ESRA_People ESRA ON pd.ACCOUNT_ID = ESRA.healthfleet_member_id
LEFT JOIN reporting_adhoc.ZZ_Salesforce sf ON sf.Policy_Number = ESRA.policy_number
;

没有使用ID的Where子句时没有结果集

1 个答案: