Question

问题：
我们有一个交易表。该表中的所有记录均具有以下交易类型之一：钱包存款（付款），钱包提款（销售）和现金返还（用于未来销售的折扣）。我想在显示现金返还余额的每一行中增加一列。现金返还用于新销售的折扣或减少负总余额。

交易表：

customer (int)
transaction_date (date)
amount (int)
transaction_type (varchar(25))

我尝试使用lag函数来获取上一行的值，并将其用于当前行的计算。但这并不总是有效，因为lag函数会回溯到它具体指向的行。

在计算中使用滞后函数：

case when
isnull(lag(balance_cashback) over (partition by client_id order by transaction_date), 0)
+ case when type = "cashback" then amount else 0 end 
+ case when type = "revenu"  and amount < 0 then amount else 0 end 
<= 0 then 0
else
lag(balance_cashback) over (partition by client_id order by transaction_date)
+ case when type = "cashback" then amount else 0 end 
+ case when type = "revenu"  and amount < 0 then amount else 0 end 
end

搜索互联网我认为我应该使用循环还是光标？

想法：
想法是使用事务表并添加两个rownumber列。我要循环通过的事务表中所有行的行号。每个客户所有交易的第二行号。下一步似乎是创建一个空的余额表，其中包含rownumclient，client_id，overall_balance和cashback_balance字段。

行数列的计算：

row_number () over (order by client_id, transaction_date) as rownumber_all
row_number () over (partition by client_id order by client_id, transaction_date) as rownumber_client

具有行号的交易表：

rownumber_all (int)
rownumber_client (int)
client (int)
transaction_date (date)
amount (int)
transaction_type (varchar(25))

余额表：

rownumber_client (int)
client_id (int)
overall_balance (int)
cashback_balance (int)

具有行号的示例交易表：

rownumbwr_all | rownumber_client | client_id | transaction_date | amount | transaction_type  
1           1              123         2018-10-12         10       wallet deposit  
2           2              123         2018-10-27         5        cashback  
3           3              123         2018-11-03         -2,5     wallet withdrawal  
4           4              123         2018-11-13         -5       wallet withdrawal  
5           5              123         2018-12-18         10       wallet deposit  
6           6              123         2018-12-19         20       wallet deposit  
7           7              123         2018-12-21         5        cashback  
8           1              456         2018-10-11         -45      wallet withdrawal  
9           2              456         2018-10-23         5        cashback  
10          3              456         2018-11-01         5        cashback  
11          4              456         2018-11-04         10       wallet deposit  
Etc.

有了额外的行号列和新的余额表，我必须创建一个遍历事务表中所有行的循环。使用rownumber_all列从第一个开始。新创建的余额表用于计算当前行中的现金返还余额。我们将此表与带有行号列的事务表的左连接一起使用。当我们遍历第一行时，余额表为空，但是从第二行开始，上一行具有计算出的现金返还余额。

用于计算当前现金返还余额的选择语句：

select  
 t1.rownumall,
 t1.rownumclient,
 t1.client_id,
 t2.overall_balance + t1.amount as overall_balance,
 case
 when (t2.overall_balance + case when t1.type = 'cashback' then t1.amount else 0 end) < 0 then 0
 when t1.type in (sales, cashback) then amount 
 else null 
 end + t2.cashback_balance as cashback_balance
/*insert into balance*/
from
 transactions as t1
 left join cashback as t2 on t2.client_id = t1.client_id and t2.rownumber_client = t1.rownumber_client-1

对于通过上述select语句的结果循环的每一行，只要有可用的交易记录，都应插入到余额表中。如前所述，现金返还余额要么用于新销售的折扣，要么用于减少负的总余额。也就是说，我要寻找的预期结果如下，其中Cashback_balance是最重要的字段。

带有余额的预期交易表：

client_id | transaction_date | amount | transaction_type | overall_balance | cashback balance  
123         2018-10-12         10       wallet deposit        10                0 
123         2018-10-27         5        cashback              15                5
123         2018-11-03         -2,5     wallet withdrawal     12,5              2,5
123         2018-11-13         -5       wallet withdrawal     7,5               0
123         2018-12-18         10       wallet deposit        17,5              0
123         2018-12-19         20       wallet deposit        37,5              0
123         2018-12-21         5        cashback              42,5              5
456         2018-10-11         -45      wallet withdrawal     -2,5              0
456         2018-10-23         5        cashback              2,5               2,5
456         2018-11-01         5        cashback              7,5               7,5
456         2018-11-04         10       wallet deposit        17,5              7,5
Etc.

我试图尽可能多地解释，希望这一想法和预期结果是明确的。我无法想象我之前没有做过什么，但是我似乎无法在任何地方找到特定的用例。

那么哪位SQL专家会友好地用简单的英语告诉我，如何使用循环，游标或任何其他方式来实现这一目标？任何帮助将不胜感激。如果需要任何澄清，请告诉我。

Answer 1

您是否正在寻找像银行对帐单这样的连续总额。该查询可以做到

SELECT
    client_id,
    Transaction_Date,
    Trans_Type,
    Amount,
    Balance = SUM(CASE WHEN Trans_Type = 'Cash back' Then Amount ELSE 0 END) OVER(ORDER BY RowNumber)

FROM
(
    SELECT 
        client_id,
        Transaction_Date,
        Trans_Type,
        Amount,
        ROW_NUMBER() OVER(ORDER BY Client_id) AS RowN

    FROM temp.dbo.Trans_Table) RC

    GROUP BY client_id, Trans_Type, Transaction_Date, Amount, RowN

样本数据

Answer 2

经过搜索和反复试验，我找到了一种遍历所有行并计算每行正确的现金返还余额的方法。我要感谢所有试图帮助我的人和Jorge E. Hernández的solution来解决我的“循环问题”。

这是我使用的最终代码。

-- declare the start and end variable
declare
    @counter int = 1,
    @max_rownumber int = (select max(rownumber_all) as max_rownumber from dbo.transactions)

-- loop 
while @counter <= @max_rownumber
begin

-- calculate overall_balance and cashback_balance for each row in the transactions table filtered by the rownumber_all field
insert into dbo.transactions_enriched
select  
    t1.rownumber_client as rownumber
  , t1.client_id
  , t1.transaction_date
  , t1.amount
  , t1.transaction_type
  , t1.payment_method
  , isnull(t2.overall_balance ,0) + t1.amount as overall_balance
  , case 
    when t1.transaction_type = 'cashback' and isnull(t2.overall_balance, 0) >= 0 then isnull(t2.cashback_balance, 0) + t1.amount
    when (case when t1.transaction_type = 'revenue' and t1.amount < 0 then t1.amount else 0 end) + isnull(t2.cashback_balance, 0) <= 0 then 0
    else (case when t1.transaction_type = 'revenue' and t1.amount < 0 then t1.amount else 0 end) + isnull(t2.cashback_balance, 0)
    end as cashback_balance
from
    dbo.transactions as t1
    left join dbo.transactions_enriched as t2 on t2.client_id = t1.client_id and t2.rownumber_client = t1.rownumber_client - 1
where 
    t1.rownumber_all = @counter

-- update the counter by adding 1
set @counter = @counter + 1

end

浏览表的每一行，在该行上进行计算，插入临时表，然后将临时表用于下一行

2 个答案: