日期:2014-05-17  浏览次数:20848 次

一条记录拆分成多条插入
有表:
create table ORIGIN_TRAN
(
  TRANID NUMBER(18) not null,
  INACC VARCHAR2(20) not null,
  TOTALAMT NUMBER(18) not null,
  OUTACC VARCHAR2(20) not null,
  STATUS CHAR(1) default '0' not null,
  CHANNID VARCHAR2(4) not null,
)

insert into origin_tran values(1001,'888881',10000,'999999','0','0001');
insert into origin_tran values(1002,'888882',20000,'999999','0','0001');
insert into origin_tran values(1003,'888883',30000,'999999','0','0001');
insert into origin_tran values(1004,'888884',40000,'999999','0','0001');
insert into origin_tran values(1005,'888885',50000,'999999','0','0001');

create table SPLIT_TRAN
(
  AUTOID NUMBER(16) not null,
  ORGINTRANID NUMBER(18) not null,
  AMOUNT NUMBER(18) not null,
  STATUS CHAR(1) not null
)

create table PAYCHANNEL
(
  ID VARCHAR2(4) not null,
  NAME VARCHAR2(30) not null,
  MAXTRANAMT NUMBER(18)
)
insert into paychannel values('0001','A通道',10000);
insert into paychannel values('0002','B通道',3000);

从origin_tran中,查找出CHANNID='0001'的,再从paychannel中,查找出MAXTRANAMT,根据它来拆分记录 写入到split_tran 中去。

希望最后出来的split_tran中的结果是这样:
1 1001 10000 '0',


2 1002 10000 '0',
3 1002 10000 '0',

4 1003 10000 '0',
5 1003 10000 '0',
6 1003 10000 '0',



7 1004 10000 '0',
8 1004 10000 '0',
9 1004 10000 '0',
10 1004 10000 '0',


11 1005 10000 '0',
12 1005 10000 '0',
13 1005 10000 '0',
14 1005 10000 '0',
15 1005 10000 '0'


有没有高效一点的方法???


------解决方案--------------------
搞个字段关联查询
------解决方案--------------------
没法一句SQL搞定,坐等大牛。

SQL code

---- 测试数据       
  WITH tmp AS
  (
    SELECT 1 ID, 21000 TOTAL_VALUES , 10000 MAX_VALUES FROM dual UNION ALL
    SELECT 2 ID, 20000 TOTAL_VALUES , 10000 MAX_VALUES FROM dual UNION ALL
    SELECT 3 ID, 5000 TOTAL_VALUES , 10000 MAX_VALUES FROM dual
  )


---- 查询SQL

  SELECT ID,"VALUES"
    FROM (
        -- 抽取一 整除的部分(21000 = 10000,10000) 剩余 1000 留在抽取二 中查询 + 合并
        SELECT DISTINCT ID
               , CASE WHEN MAX_VALUES > TOTAL_VALUES
                           THEN TOTAL_VALUES
                      ELSE MAX_VALUES END "VALUES"
               , LEVEL
             FROM tmp
             WHERE TOTAL_VALUES > MAX_VALUES
            CONNECT BY LEVEL <= trunc(TOTAL_VALUES / MAX_VALUES) 

        UNION ALL

        -- 抽取二 不能整除的部分(21000 = 1000)、以及 不足额(5000 < 分隔标准 10000)
        SELECT ID
               , MOD(TOTAL_VALUES , MAX_VALUES) "VALUES"
               ,-9
             FROM tmp
             WHERE MOD(TOTAL_VALUES , MAX_VALUES) > 0 -- 不能整除的部分(取余)
                   OR MAX_VALUES > TOTAL_VALUES   -- 不足额的部分
                   )
      ORDER BY 1,2;