Question

我写了一个查询来显示客户ID，登录ID和客户信用债务金额之和。如果要从中扣除客户到期金额，则必须创建虚拟列，并从客户其他客户处显示为“到期”。请帮我解决一下这个问题。

我将表格结构和值与我的查询一起附加。

- 客户表

CREATE TABLE isbs_customer_mst
  (
     cust_id   VARCHAR2(30) NOT NULL,
     login_id  VARCHAR2(30) NOT NULL,
     cust_nm   VARCHAR2(30),
     cust_addr VARCHAR2(300),
     CONSTRAINT isbs_customer_mst_pk PRIMARY KEY (cust_id)
  );

- 值

INSERT INTO ISBS_CUSTOMER_MST (CUST_ID, LOGIN_ID, CUST_NM, CUST_ADDR) 
VALUES ('CUST0000000001', 'USER1', 'User Login ID 1', '143/1 Uthamar Gandhi Salai, Nungambakkam, Chennai - 34');

INSERT INTO ISBS_CUSTOMER_MST (CUST_ID, LOGIN_ID, CUST_NM, CUST_ADDR)
VALUES ('CUST0000000002', 'USER2', 'User Login ID 2', '143/2 Uthamar Gandhi Salai, Nungambakkam, Chennai - 34');

INSERT INTO ISBS_CUSTOMER_MST (CUST_ID, LOGIN_ID, CUST_NM, CUST_ADDR) 
VALUES ('CUST0000000003', 'USER3', 'User Login ID 3', '143/3 Uthamar Gandhi Salai, Nungambakkam, Chennai - 34');

INSERT INTO ISBS_CUSTOMER_MST (CUST_ID, LOGIN_ID, CUST_NM, CUST_ADDR) 
VALUES ('CUST0000000004', 'USER4', 'User Login ID 4', '143/4 Uthamar Gandhi Salai, Nungambakkam, Chennai - 34');

- 信用借记表

CREATE TABLE isbs_acct_ledger_det
  (
     acct_ledger_id    VARCHAR2(30),
     cust_id           VARCHAR2(30),
     credit_debit_amt  VARCHAR2(30) NOT NULL,
     credit_debit_dttm TIMESTAMP NOT NULL,
     CONSTRAINT isbs_acct_ledger_det_pk PRIMARY KEY (acct_ledger_id),
     CONSTRAINT isbs_acct_ledger_det_fk FOREIGN KEY (cust_id) REFERENCES
     isbs_customer_mst (cust_id)
  );

- 值

INSERT INTO ISBS_ACCT_LEDGER_DET (ACCT_LEDGER_ID, CUST_ID, CREDIT_DEBIT_AMT, CREDIT_DEBIT_DTTM) 
VALUES ('ACC0000000001', 'CUST0000000001', -1000.25, TO_DATE('01-10-2008 11:00:00', 'DD-MM-YYYY HH24:MI:SS'));

INSERT INTO ISBS_ACCT_LEDGER_DET (ACCT_LEDGER_ID, CUST_ID, CREDIT_DEBIT_AMT, CREDIT_DEBIT_DTTM) 
VALUES ('ACC0000000002', 'CUST0000000002', -256.75, TO_DATE('01-10-2008 11:00:00', 'DD-MM-YYYY HH24:MI:SS'));

INSERT INTO ISBS_ACCT_LEDGER_DET (ACCT_LEDGER_ID, CUST_ID, CREDIT_DEBIT_AMT, CREDIT_DEBIT_DTTM) 
VALUES ('ACC0000000003', 'CUST0000000002', 100.25, TO_DATE('05-10-2008 11:00:00', 'DD-MM-YYYY HH24:MI:SS'));

- 查询

SELECT c.CUST_NM
       , c.LOGIN_ID
       , SUM(a.CREDIT_DEBIT_AMT) "Outstanding Amt"
       , CASE WHEN a.CREDIT_DEBIT_AMT <= -9999.99 
             THEN 'Due to Cust' 
          ELSE 'Due from Cust' END "Due"
FROM ISBS_CUSTOMER_MST c
JOIN ISBS_ACCT_LEDGER_DET a
     ON c.CUST_ID = a.CUST_ID
GROUP BY c.CUST_NM, c.LOGIN_ID, a.CREDIT_DEBIT_AMT;

提前致谢

Answer 1

您可能需要从a.CREDIT_DEBIT_AMT移除group by并在CASE上使用SUM()表达。

SELECT c.cust_nm,
       c.login_id,
       SUM(a.credit_debit_amt) "Outstanding Amt",
       CASE
         WHEN SUM(a.credit_debit_amt) <= -9999.99 THEN 'Due to Cust'
         ELSE 'Due from Cust'
       END                     "Due"
FROM   isbs_customer_mst c
       JOIN isbs_acct_ledger_det a
         ON c.cust_id = a.cust_id
GROUP  BY c.cust_nm,
          c.login_id ;

Answer 2

您需要在group by子句中添加case语句：

SELECT c.CUST_NM , c.LOGIN_ID , SUM(a.CREDIT_DEBIT_AMT) "Outstanding Amt" , CASE WHEN a.CREDIT_DEBIT_AMT <= -9999.99 THEN 'Due to Cust' ELSE 'Due from Cust' END "Due" FROM ISBS_CUSTOMER_MST c JOIN ISBS_ACCT_LEDGER_DET a ON c.CUST_ID = a.CUST_ID GROUP BY c.CUST_NM, c.LOGIN_ID, CASE WHEN a.CREDIT_DEBIT_AMT <= -9999.99 THEN 'Due to Cust' ELSE 'Due from Cust' END "Due";

因为聚合函数的规则是在group by子句中包含所有其他所选属性

Answer 3

为什么在处理总和时会使用sin函数。 Sin给出参数的sin值而abs则返回正数，同样

Select abs(-5) from dual ;

将输出5作为输出。

我认为你的要求不需要这两种功能。

案例和解码

3 个答案: