使用 SQL 有条件地从贷方和运行余额中借记

Question

我的表格中有如下数据cryptotransactionledger

id	transaction_typeid	transaction_type	amount	totalcoins
1	1	bitcoin-credit	30	30
2	2	ethereum-credit	20	50

如果我花费比特币，我会在同一张表中添加一个如下所示的新条目，其中 transaction_typeid 为 3，对于 ethereum，transaction_typeid 为 4 类似

id	transaction_typeid	transaction_type	amount	totalcoins
1	1	bitcoin-credit	30	30
2	2	etherium-credit	20	50
3	3	bitcoin-debit	-10	40
4	4	etherium-debit	-5	35

假设如果我在表格中的最终数据如下所示，我将拥有 35 个比特币和 20 个以太坊。

id	transaction_typeid	transaction_type	amount	totalcoins
1	1	bitcoin-credit	30	30
2	2	etherium-credit	20	50
3	3	bitcoin-debit	-10	40
4	4	etherium-debit	-5	35
5	1	bitcoin-credit	15	50
6	2	etherium-credit	10	60
7	4	etherium-debit	-5	55

从各个贷方中减少所有借方后，如何使用 SQL 得出以下余额摘要

id	transaction_type	amount	totalcoins
1	bitcoin-credit	35	35
2	etherium-credit	20	55

Answer 1

我会忘记最终结果的“交易类型”，而只关注硬币本身。您的数据模型很奇怪，因为您没有针对硬币类型的单独列。但是您可以从transaction_type 列中提取它。

此外，您的 cmets 建议 totalcoins 应该是列中的最后一个值。对于“比特币”，应该是 50，而不是 35。

无论如何，您都可以使用聚合。我建议：

select regexp_substr(transaction_type, '^[^-]+') as coin,
       sum(amount) as amount,
       max(totalcoins) keep (dense_rank first oder by id desc) as totalcoins
from cryptotransactionledger tl
group by regexp_substr(transaction_type, '^[^-]+');

Here 是一个 dbfiddle。

Answer 2

假设您有表 ledger 和 coin_types：

create table ledger (id varchar2(3) primary key, 
                 coin_type varchar2(2), 
                 amount number(5)
                 );

create table coin_types (id varchar2(2) primary key, 
                         alias varchar2(100));

这些 DML 语句代表您的“最终数据”：

insert into coin_types values (1, 'bitcoin');
insert into coin_types values (2, 'etherium');

insert into ledger values (1, 1, 30);
insert into ledger values (2, 2, 20);
insert into ledger values (3, 1, -10);
insert into ledger values (4, 2, -5);
insert into ledger values (5, 1, 15);
insert into ledger values (6, 2, 10);
insert into ledger values (7, 2, -5);

您的第一步应该是总结金额，按您的硬币类型分组：

  select l.coin_type, 
         c.alias,
         sum(l.amount) amount_coin
    from ledger l, coin_types c
   where l.coin_type = c.id
group by l.coin_type, c.alias

然后您可以使用此选择将您的“totalcoins”作为累积总和，按您的 coin_type (see Oraclee Docs) 排序：

   select coin_type, 
          alias, 
          amount_coin, 
          sum(amount_coin) over (order by coin_type) totalcoins 
      from(
          select l.coin_type, 
                 c.alias,
                 sum(l.amount) amount_coin
            from ledger l, coin_types c
           where l.coin_type = c.id
        group by l.coin_type, c.alias
           );

这是一个有效的 SQL-Fiddle：SQL-Fiddle

Answer 3

我会创建一个 tran 类型表

create table tran_types (
   id int primary key,
   coin_Name varchar2(100),
   op_Name varchar2(100)
   );

insert into tran_types  
select 1, 'bitcoin', 'credit' from dual union all
select 2, 'etherium','credit' from dual union all
select 3, 'bitcoin', 'debit'  from dual union all
select 4, 'etherium','debit'  from dual;

通过这种方式，您可以对交易进行分组以获得总计并根据需要可视化总计行标题

select t.id transaction_typeid, t.coin_Name || '-' || t.op_Name transaction_type, 
    s.amount, s.total_coins
from (
  select t.coin_Name, sum(amount) amount
    , sum(sum(amount)) over(order by t.coin_Name) total_coins
  from ledger r
  join tran_types t on r.tran_type = t.id
  group by t.coin_Name
) s
join tran_types t on t.op_name = 'credit' and t.coin_Name = s.coin_Name
order by t.coin_Name;

Answer 4

感谢所有提供建议和不同解决方案的人。 我终于使用了下面的查询，它为我提供了预期的结果。

WITH cte1
 AS (SELECT a.*,
            CASE
              WHEN ( transaction_typeid = 1
                      OR transaction_typeid = 3 ) THEN 0
              ELSE 1
            END AS category
     FROM   cryptotransactionledger a
     ORDER  BY id),
 cte2
 AS (SELECT CASE
              WHEN ( category = 0 ) THEN 'bitcoin-credit'
              ELSE 'etherium-credit'
            END         AS transaction_type,
            Sum(amount) AS amount
     FROM   cte1 o
     GROUP  BY category),
 cte3
 AS (SELECT Row_number()
              OVER (ORDER BY c.transaction_type) AS id,
            c.*
     FROM   cte2 c) SELECT f.*,
     Sum(amount) OVER(ORDER BY f.id) AS total_coins FROM cte3 f;

Link to DB Fiddle