Question

我有3个表有外键 ACCOUNT_NO ，第一个表是 CREDIT_LIST ，其中包含从银行取得的所有积分，第二个表是 CUSTOMERS 其中包含客户信息，最后一个是 ACCOUNTS ，其中包含有关帐户本身的所有信息。当我尝试选择

时

SELECT 
    B.CUSTOMER_NO AS CUSTOMER_NO,
    B.CREDIT_TYPE AS CREDIT_TYPE,
    B.ACCOUNT_NO AS CREDIT_ACCOUNT_NUMBER,
    A.BRANCH_CODE AS BRANCH_CODE,
    C.EXTERNAL_ACCOUNT_NO AS EXTERNAL_ACCOUNT_NUMBER
FROM 
    CREDIT_LIST B,
    CUSTOMERS A,
    ACCOUNTS C
WHERE   
    B.STATUS = 'A' -- ACTIVE
    AND A.CUSTOMER_NO = B.CUSTOMER_NO
    AND C.ACCOUNT_NO = B.ACCOUNT_NO
    ;

我的结果为零，因为 ACCOUNTS 中没有EXTERNAL_ACCOUNT_NO且c.account_no = b.account_no。问题是我想显示信息，即使没有EXTERNAL_ACCOUNT_NO并填写为null，例如：

| CUSTOMER_NO | CREDIT_TYPE | CREDIT_ACCOUNT_NUMBER | BRANCH_CODE | EXTERNAL_ACCOUNT_NUMBER
+-------------+-------------+-----------------------+-------------+------------------------
|     1       | some_type   | 123456                | 01          | 
|     2       | some_type   | 654321                | 01          | 111111111111

我觉得这非常愚蠢，但无法弄清楚我到底做错了什么

Answer 1

我认为，这就是你需要的

SELECT 
    B.CUSTOMER_NO AS CUSTOMER_NO,
    B.CREDIT_TYPE AS CREDIT_TYPE,
    B.ACCOUNT_NO AS CREDIT_ACCOUNT_NUMBER,
    A.BRANCH_CODE AS BRANCH_CODE,
    C.EXTERNAL_ACCOUNT_NO AS EXTERNAL_ACCOUNT_NUMBER
FROM 
    CREDIT_LIST B
    JOIN     CUSTOMERS A on A.CUSTOMER_NO = B.CUSTOMER_NO
    LEFT JOIN ACCOUNTS C on C.ACCOUNT_NO = B.ACCOUNT_NO
WHERE   
    B.STATUS = 'A' -- ACTIVE
;

左连接将确保您从A和B获取详细信息，即使C中没有可用数据。

此外AS是多余的，并非真正需要，您只需在查询的第4行中编写B.ACCOUNT_NO CREDIT_ACCOUNT_NUMBER即可。

左外连接 - Reference

不建议，但是，如果有原因，您需要使用旧语法，以下是您需要的

SELECT 
    B.CUSTOMER_NO AS CUSTOMER_NO,
    B.CREDIT_TYPE AS CREDIT_TYPE,
    B.ACCOUNT_NO AS CREDIT_ACCOUNT_NUMBER,
    A.BRANCH_CODE AS BRANCH_CODE,
    C.EXTERNAL_ACCOUNT_NO AS EXTERNAL_ACCOUNT_NUMBER
FROM 
    CREDIT_LIST B,
    CUSTOMERS A,
    ACCOUNTS C
WHERE   
    B.STATUS = 'A' -- ACTIVE
    AND A.CUSTOMER_NO = B.CUSTOMER_NO
    AND C.ACCOUNT_NO(+) = B.ACCOUNT_NO -- (+) will do a outer join for you
    ;

SQL从多个表中进行选择以显示偶数值

1 个答案: