使用mysql或更好的algothrim匹配借记卡和每笔交易的贷方?
作者:互联网
首先让我解释一下表结构和我所需的输出.
userid date amount
123 2017-01-01 5
123 2017-01-03 2
124 2017-01-04 2
124 2017-01-04 3
123 2017-01-05 -2
借方交易是负面的,正面交易是会员的信用交易.
我们可以通过此查询轻松完成帐户对帐单
select date as BookingDate,
if(amount<0,amount,0) as Debit_Amount,
if(amount>0,amount,0) as Credit_Amount,
(@runtot := amount + @runtot) AS Balance
from transactions,
(SELECT @runtot:=0) c
where userid=123
BookingDate Debit_Amount Credit_Amount Balance
2017-01-01 0 5 5
2017-01-03 0 2 7
2017-01-05 -2 0 5
我的要求是使用FIFO方法标记每笔支付或部分支付的交易.像这样.这可能通过mysql查询或更好的算法?
userid date amount status
123 2017-01-01 5 partial_paid(-2)
123 2017-01-03 2
124 2017-01-04 2
124 2017-01-04 3
123 2017-01-05 -2
谢谢
解决方法:
MariaDB [sandbox]> create table t(userid int,dt date, amount int);
Query OK, 0 rows affected (0.28 sec)
MariaDB [sandbox]> truncate table t;
Query OK, 0 rows affected (0.20 sec)
MariaDB [sandbox]> insert into t values
-> (123 , '2017-01-01' , 5) ,
-> (123 , '2017-01-03' , 2 ),
-> (124 , '2017-01-04' , 2) ,
-> (124 , '2017-01-04' , 3 ),
-> (123 , '2017-01-05' , -2),
-> (125 , '2017-01-01' , 5) ,
-> (125 , '2017-01-03' , 2 ),
-> (125 , '2017-01-05' , -6),
-> (126 , '2017-01-01' , 5) ,
-> (126 , '2017-01-02' , -10),
-> (126 , '2017-01-03' , 2 ),
-> (126 , '2017-01-05' , -10),
-> (126 , '2017-01-06' , 13);
Query OK, 13 rows affected (0.06 sec)
Records: 13 Duplicates: 0 Warnings: 0
MariaDB [sandbox]>
MariaDB [sandbox]>
MariaDB [sandbox]>
MariaDB [sandbox]> select s.userid,s.dt,s.amount,
-> case when s.crs is null then 0 else s.crs end crs,
-> case when s.exhaust is null then 0 else s.exhaust end exhaust,
-> case when s.amount > 0 and s.amount <= s.crs and s.crs > 0 then 'Fully paid'
-> when s.amount > 0 and s.amount > s.crs and s.crs > 0 then concat('Part paid -' ,s.crs)
-> else ''
-> end msg
-> from
-> (
-> select t1.*,
-> if(t1.userid <> @p ,
-> @crs:=(select sum(t2.amount) * - 1 from t t2 where t2.userid = t1.userid and t2.amount < 0)
-> ,@crs:=@crs) crs,
-> if(t1.amount < 0 ,@crs:=@crs,if (t1.amount > @crs , @crs:=0,@crs:=@crs - t1.amount)) exhaust,
-> @p:=t1.userid p
->
-> from (select @p:=0,@crs:=0) p ,t t1
-> order by t1.userid, t1.dt
-> ) s
-> ;
+--------+------------+--------+------+---------+--------------+
| userid | dt | amount | crs | exhaust | msg |
+--------+------------+--------+------+---------+--------------+
| 123 | 2017-01-01 | 5 | 2 | 0 | Part paid -2 |
| 123 | 2017-01-03 | 2 | 0 | 0 | |
| 123 | 2017-01-05 | -2 | 0 | 0 | |
| 124 | 2017-01-04 | 2 | 0 | 0 | |
| 124 | 2017-01-04 | 3 | 0 | 0 | |
| 125 | 2017-01-01 | 5 | 6 | 1 | Fully paid |
| 125 | 2017-01-03 | 2 | 1 | 0 | Part paid -1 |
| 125 | 2017-01-05 | -6 | 0 | 0 | |
| 126 | 2017-01-01 | 5 | 20 | 15 | Fully paid |
| 126 | 2017-01-02 | -10 | 15 | 15 | |
| 126 | 2017-01-03 | 2 | 15 | 13 | Fully paid |
| 126 | 2017-01-05 | -10 | 13 | 13 | |
| 126 | 2017-01-06 | 13 | 13 | 0 | Fully paid |
+--------+------------+--------+------+---------+--------------+
13 rows in set (0.03 sec)
当心: – 我还没有完全测试过这个!!
标签:algorithm,mysql,sql,user-accounts,banking 来源: https://codeday.me/bug/20190701/1351083.html