Question

我有三个查询，查询1和查询2必须在Customer_number上加入内部 query1和query2的结果输出将进一步内连接到3个键customer_number，account_number，bank_number。

任何人都可以帮我写这些查询。我无法推导出来。

提前致谢，

巴斯卡尔

查询1：

SELECT
CUST.BANK_NUMBER as CUSTBANKNUMBER,
CUST.CUSTOMER_NUMBER,
CUST.DATE_DECEASE,
FROM
SIGN_LND.T_CUSTOMER_MASTER_FILE_LND CUST
LEFT OUTER JOIN
SIGN_LND.T_CUST_USER_DEF_FIELDS_LND USER_DEF
ON (CUST.CUSTOMER_NUMBER = USER_DEF.CUSTOMER_NUMBER AND CUST.EFF_DT = USER_DEF.EFF_DT)
WHERE
CUST.EFF_DT = TO_DATE('2015-01-22','YYYY-MM-DD')

查询2：

SELECT
S.CUSTOMER_NUMBER,
S.ALIAS_ACCOUNT_NBR,
M.BANK_NUMBER,
M.ACCOUNT_NUMBER,
M.PRODUCT_CODE,
M.APPLICATION_NUMBER,
M.APPL_ID
FROM
(
SELECT T1.BANK_NUMBER,
   T1.ACCOUNT_NUMBER,
   T1.PRODUCT_CODE,
   CAST (NULL AS VARCHAR(5))              AS SUB_PRODUCT_CODE,
   T1.APPLICATION_NUMBER,
   T1.APPL_ID
   FROM   SIGN_LND.T_LOAN_MASTER_FILE_LND T1
   INNER JOIN SIGN_LND.T_LOAN_PRODUCT_DEFINITION_LND T2
              ON  T1.BANK_NUMBER = T2.BANK_NUMBER
              AND T1.PRODUCT_CODE = T2.PRODUCT_CODE
                      AND T1.EFF_DT=TO_DATE('2015-01-22','YYYY-MM-DD')
              AND T2.EFF_DT=TO_DATE('2015-01-22','YYYY-MM-DD')
   WHERE          T1.PRODUCT_CODE IS NOT NULL
              AND T1.DATE_OPEN IS NOT NULL
   UNION ALL
   SELECT T1.BANK_NUMBER,
   T1.ACCOUNT_NUMBER,
   T1.PRODUCT_CODE,
   T1.PRODUCT_CODE                   AS SUB_PRODUCT_CODE,
   T1.APPLICATION_NUMBER,
   T1.APPL_ID

   FROM   SIGN_LND.T_TA_MASTER_FILE_LND T1
   INNER JOIN SIGN_LND.T_TA_PRODUCT_DEFINITION_LND T2
              ON  T1.BANK_NUMBER = T2.BANK_NUMBER
              AND T1.PRODUCT_CODE = T2.PRODUCT_CODE
                      AND T1.ACCOUNT_TYPE = T2.ACCOUNT_TYPE
                      AND T1.EFF_DT=TO_DATE('2015-01-22','YYYY-MM-DD')
              AND T2.EFF_DT=TO_DATE('2015-01-22','YYYY-MM-DD')
   LEFT JOIN SIGN_LND.T_BONUS_INT_MASTER_FILE_LND T3
             ON  T1.ACCOUNT_NUMBER = T3.ACCOUNT_NUMBER
             AND T1.BANK_NUMBER = T3.BANK_NUMBER
                     AND T3.EFF_DT=TO_DATE('2015-01-22','YYYY-MM-DD')
   WHERE         T1.PRODUCT_CODE IS NOT NULL
             AND T1.ACCOUNT_STATUS_CODE IS NOT NULL
             AND T1.DATE_OPEN IS NOT NULL
   UNION ALL
   SELECT T1.BANK_NUMBER,
   T1.ACCOUNT_NUMBER,
   T1.PRODUCT_CODE,
   T1.PRODUCT_CODE                     AS SUB_PRODUCT_CODE,
   T1.APPLICATION_NUMBER,
   T1.APPL_ID,

   FROM              SIGN_LND.T_TIME_MASTER_FILE_LND T1
   INNER JOIN SIGN_LND.T_TM_PRODUCT_DEFINITION_LND T2
              ON  T1.BANK_NUMBER = T2.BANK_NUMBER
              AND T1.PRODUCT_CODE = T2.PRODUCT_CODE
                      AND T1.EFF_DT=TO_DATE('2015-01-22','YYYY-MM-DD')
              AND T2.EFF_DT=TO_DATE('2015-01-22','YYYY-MM-DD')
   LEFT JOIN SIGN_LND.T_BONUS_INT_MASTER_FILE_LND T3
             ON  T1.ACCOUNT_NUMBER = T3.ACCOUNT_NUMBER
             AND T1.BANK_NUMBER = T3.BANK_NUMBER
                     AND T3.EFF_DT=TO_DATE('2015-01-22','YYYY-MM-DD')
   WHERE         T1.PRODUCT_CODE IS NOT NULL
             AND T1.ACCOUNT_STATUS_CODE IS NOT NULL
             AND T1.DATE_OPEN IS NOT NULL
                       )M
   INNER JOIN
   ( 
   SELECT
    T1.ACCOUNT_NUMBER,
    T1.CUSTOMER_NUMBER,
    T1.RELATIONSHIP ,
    T2.ALIAS_ACCOUNT_NBR,
    T2.APPLICATION_NUMBER
    FROM SIGN_LND.T_CUSTOMER_CROSS_REF_LND T1
    INNER JOIN
     SIGN_LND.T_ALIAS_ACCT_NBR_CRSS_REF_LND T2
     ON  T1.ACCOUNT_NUMBER=T2.ACCOUNT_NUMBER
      WHERE   ALIAS_ACCOUNT_NBR_TYPE='UK'
    GROUP BY  T1.ACCOUNT_NUMBER,
     T1.CUSTOMER_NUMBER,
      T1.RELATIONSHIP ,
     T2.ALIAS_ACCOUNT_NBR,
    T2.APPLICATION_NUMBER
    )S
    ON  M.ACCOUNT_NUMBER=S.ACCOUNT_NUMBER
    AND
     M.APPLICATION_NUMBER=S.APPLICATION_NUMBER
      )

QUERY3：

    SELECT
    SPLIT.BANK_NUMBER,
    SPLIT.CUSTOMER_NUMBER,
    SPLIT.DATE_MOVING_IN,
    SPLIT.TITLE,
    SPLIT.FIRST_NAME,
    SPLIT.SECOND_NAME,
    SPLIT.SURNAME,
    SPLIT.CURR_APT_NUMBER_UPS,
    SPLIT.CURR_HOUSE_NAME_UPS,
    SPLIT.CURR_HOUSE_NBR,
    SPLIT.CURR_STREET_UPS,
    SPLIT.CURR_DISTRICT_UPS,
    SPLIT.CURR_POST_TOWN_UPS,
    SPLIT.POSTAL_CODE,
    SPLIT.CURR_COUNTRY,
    CUST_EXT.LOYALTY_CARD_NBR,
    XREF.ACCOUNT_NUMBER,
    XREF.RELATIONSHIP as RELATIONSHIP_RB,
     RB_ALIAS_XREF.APPL_ID as RBXREFAPPID,RB_ALIAS_XREF.ALIAS_ACCOUNT_NBR AS ALIAS_ACCOUNT_NBR_RB,
     PROD_OFFER.DATE_CLOSE
   FROM
     (
       SIGN_LND.T_CIF_SPLIT_NAME_ADDR_UPS_LND SPLIT
       LEFT OUTER JOIN
       SIGN_LND.T_CUST_MASTER_EXTENSION_LND CUST_EXT
          ON (SPLIT.CUSTOMER_NUMBER = CUST_EXT.CUSTOMER_NUMBER AND               SPLIT.EFF_DT = CUST_EXT.EFF_DT)
          )
         LEFT OUTER JOIN
        (
         SIGN_LND.T_CUSTOMER_CROSS_REF_LND XREF
         LEFT OUTER JOIN
        (
         SELECT * FROM SIGN_LND.T_ALIAS_ACCT_NBR_CRSS_REF_LND WHERE  EFF_DT = TO_DATE('2015-01-22','YYYY-MM-DD') AND ALIAS_ACCOUNT_NBR_TYPE = 'RB'
        ) RB_ALIAS_XREF
        ON (XREF.ACCOUNT_NUMBER = RB_ALIAS_XREF.ACCOUNT_NUMBER AND  XREF.EFF_DT = RB_ALIAS_XREF.EFF_DT)
        LEFT OUTER JOIN
        SIGN_LND.T_PRODUCT_OFFER_PURCHASE_LND PROD_OFFER
        ON (XREF.BANK_NUMBER = PROD_OFFER.BANK_NUMBER AND XREF.ACCOUNT_NUMBER = PROD_OFFER.ACCOUNT_NUMBER AND XREF.EFF_DT = PROD_OFFER.EFF_DT)
          )
         ON (SPLIT.CUSTOMER_NUMBER = XREF.CUSTOMER_NUMBER AND SPLIT.EFF_DT = XREF.EFF_DT)
         WHERE SPLIT.EFF_DT = TO_DATE('2015-01-22','YYYY-MM-DD')``

Answer 1

这应该有效

WITH query1 as
(your query 1 code),
query2 as
(you query 2 code),
query3 as
(your query 3 code)

select * from query1 
join query2 on query1.Customer_number=query2.Customer_number
join query3 on query3.customer_number=query2.customer_number
            and query3.account_number=query2.account_number
            and query3.bank_number=query2.bank_number

Oracle复合加入

1 个答案: