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sql - 在 Oracle 中使用增强的聚合报告功能解决问题

问题描述

我正在使用 sqlplus (oracle)。我试图写一个声明来显示基于所有交易价值的账户价值变化

此外，借方交易应为负，贷方交易应保持为正。我需要在 oracle 中使用增强的聚合报告功能来解决它​​。另外，应该有客户总数和总计。

这是我到目前为止得到的。

select wc.first_name "First", wc.surname "Last", wt.account_type "Act Type" 
case when wt.transaction_type is NULL then NULL 
    when wt.transaction_type = 'D'then TO_CHAR(transaction_amount*-1, '$9,999.99') 
    else TO_CHAR(transaction_amount, '$9,999.99') 
    end "Total" 
from wgb_customer wc join wgb_account wa on wa.customer_number = wc.customer_number join wgb_account_type wt on wa.account_type = wt.account_type
left outer join wgb_transaction wt on wt.customer_number = wa.customer_number 
and wt.account_type = wa.account_type
order by 2,3,6;

但是，它不起作用，并且不显示客户总数或总计。请帮忙！

ERD ERD

预期产出 预期产出

标签： sqloraclesqlplus

解决方案

GROUP BY ROLLUP 是要走的路。它允许您创建所有可能的小计和总计。聚合数据后，您只需将连接保留到客户表以检索 first_name 和 last_name。确保进行左连接，以免丢失总计

--SAMPLE TABLES AND TEST DATA 

  CREATE TABLE WGB_TRANSACTION (
    CUSTOMER_NUMBER VARCHAR2(7),
    ACCOUNT_TYPE NUMBER(1),
    TRANSACTION_AMOUNT NUMBER,
    TRANSACTION_TYPE VARCHAR2(1)
  );

  CREATE TABLE WGB_CUSTOMER (
    CUSTOMER_NUMBER VARCHAR2(7),
    FIRST_NAME VARCHAR2(30),
    SURNAME VARCHAR2(30)
   );

INSERT INTO WGB_CUSTOMER VALUES ( '123', 'John','Smith');
INSERT INTO WGB_CUSTOMER VALUES ( '456', 'James','Anderson');
INSERT INTO WGB_TRANSACTION VALUES (123,1,100,'C');
INSERT INTO WGB_TRANSACTION VALUES (123,1,50,'D');
INSERT INTO WGB_TRANSACTION VALUES (123,2,100,'C');
INSERT INTO WGB_TRANSACTION VALUES (456,1,50,'C');
INSERT INTO WGB_TRANSACTION VALUES (456,1,100,'C');
INSERT INTO WGB_TRANSACTION VALUES (456,3,100,'D');
INSERT INTO WGB_TRANSACTION VALUES (456,1,50,'C');

--The Query 
WITH BALANCES AS (
    SELECT CUSTOMER_NUMBER, ACCOUNT_TYPE, 
    SUM(DECODE(TRANSACTION_TYPE,'D',-1,'C',1,0)*TRANSACTION_AMOUNT) BALANCE
    FROM WGB_TRANSACTION WHERE TRANSACTION_TYPE IN ('C','D') 
    GROUP BY ROLLUP ( CUSTOMER_NUMBER, ACCOUNT_TYPE )
)
SELECT C.FIRST_NAME, C.SURNAME, B.ACCOUNT_TYPE, TO_CHAR(B.BALANCE,'$9,999.99')
FROM BALANCES B LEFT JOIN WGB_CUSTOMER C ON C.CUSTOMER_NUMBER=B.CUSTOMER_NUMBER;

--RESULT 
FIRST_NAME  SURNAME ACCOUNT_TYPE    TO_CHAR(B.BALANCE,'$9,999.99')
John    Smith   (null)  $150.00
John    Smith   2   $100.00
John    Smith   1   $50.00
James   Anderson    (null)  $100.00
James   Anderson    3   -$100.00
James   Anderson    1   $200.00
(null)  (null)  (null)  $250.00

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