根据另一列的前一行计算一列

Question

我有一个商店和一个产品的以下数据，我需要根据另一列计算一列。

初始数据集：

store product tran_date  audit_date audit_bal inv_value
10001 323232  2020-01-01 null       null      5
10001 323232  2020-01-02 2020-01-02 20        31
10001 323232  2020-01-03 null       null      13
10001 323232  2020-01-04 null       null      6
10001 323232  2020-01-05 null       null      21
10001 323232  2020-01-06 null       null      17
10001 323232  2020-01-07 null       null      6
10001 323232  2020-01-08 null       null      34
10001 323232  2020-01-09 null       null      35
10001 323232  2020-01-10 2020-01-10 60        17
10001 323232  2020-01-12 null       null      6
10001 323232  2020-01-13 null       null      9
10001 323232  2020-01-14 null       null      5
10001 323232  2020-01-15 null       null      29

逻辑： start_stock audit_date 旁边的那一天应该是 audit_date 的 audit_bal 对于剩余的天数，应该是前几天 end_stock

end_stock audit_date 旁边的那一天应该是 audit_date + inv_value 的 audit_bal 对于剩余的天数，应该是 start_stock(前几天 end_stock) + inv_value

最终数据集应该是

store product tran_date  audit_date audit_bal inv_value start_stock end_stock
10001 323232  2020-01-01 null       null      5         6           11  
10001 323232  2020-01-02 2020-01-02 20        31        11          42
10001 323232  2020-01-03 null       null      13        20          33
10001 323232  2020-01-04 null       null      6         33          39
10001 323232  2020-01-05 null       null      21        39          60
10001 323232  2020-01-06 null       null      17        60          77
10001 323232  2020-01-07 null       null      6         77          83 
10001 323232  2020-01-08 null       null      34        83          117
10001 323232  2020-01-09 null       null      35        117         152
10001 323232  2020-01-10 2020-01-10 120       17        152         169
10001 323232  2020-01-12 null       null      6         120         126
10001 323232  2020-01-13 null       null      9         126         135 
10001 323232  2020-01-14 null       null      5         135         140
10001 323232  2020-01-15 null       null      29        140         169

我使用了下面的查询但没有得到正确的结果

WITH Inv AS (
    SELECT      *,case when tran_date = date_add(lag_audit_date,1)   then lag_audit_bal + inv_value  
                       when date_add(lag_audit_date,1) != tran_date  then lag_audit_bal + inv_value
                       else SUM(inv_value) OVER (partition by store,product ORDER BY tran_date ASC ROWS UNBOUNDED PRECEDING) end as end_stock 
    FROM        
                 basedata
)
SELECT          tran_date,audit_date,audit_bal,lag_audit_date,lag_audit_bal,
                 case when tran_date = date_add(lag_audit_date,1)  then lag_audit_bal 
                      when date_add(lag_audit_date,1) != tran_date  then lag_audit_bal 
                      else LAG(end_stock,1,0) OVER (partition by store,product ORDER BY transaction_date ASC) end as start_stock, 
                inv_val,
                end_stock
FROM            Inv;

谁能帮帮我。

Answer 1

在这个dba问题中，提出了以下逻辑：

https://dba.stackexchange.com/questions/94545/calculate-row-value-based-on-previous-and-actual-row-values

SELECT 
    s.stmnt_date, s.debit, s.credit,
    SUM(s.debit - s.credit) OVER (ORDER BY s.stmnt_date
                                  ROWS BETWEEN UNBOUNDED PRECEDING
                                           AND CURRENT ROW)
        AS balance
FROM
    statements AS s
ORDER BY
    stmnt_date ;

另请注意关于（仍然可能？）MySQL 分析函数限制的附加讨论（如果适用于您的情况）。

如果您的数据库不支持上述语法，该帖子还提出了一个可能的（效率较低的）解决方案（注意：我还没有测试过这些解决方案中的任何一个）：

SELECT 
    s.stmnt_date, s.debit, s.credit,
    @b := @b + s.debit - s.credit AS balance
FROM
    (SELECT @b := 0.0) AS dummy 
  CROSS JOIN
    statements AS s
ORDER BY
    stmnt_date ;

如果您正在使用 MySQL，您可能还会发现感兴趣的 LAG() 函数

注意本教程中的示例：

https://www.mysqltutorial.org/mysql-window-functions/mysql-lag-function/

https://dev.mysql.com/doc/refman/8.0/en/window-function-descriptions.html#function_lag

SELECT
         t, val,
         LAG(val)        OVER w AS 'lag',
         LEAD(val)       OVER w AS 'lead',
         val - LAG(val)  OVER w AS 'lag diff',
         val - LEAD(val) OVER w AS 'lead diff'
       FROM series
       WINDOW w AS (ORDER BY t);

LAG(expr [ N[ default]]) [null_treatment] over_clause

Returns 行中 expr 的值，该行在其分区内滞后（先于）当前行 N 行。如果没有这样的行，则 return 值为默认值。例如，如果 N 为 3，则前两行的默认值为 return。如果缺少 N 或默认值，则默认值分别为 1 和 NULL。

N 必须是文字非负整数。如果 N 为 0，则对当前行计算 expr。

从MySQL8.0.22开始，N不能为NULL。此外，它现在必须是 1 到 263 范围内的整数，包括以下任何形式：

an unsigned integer constant literal
a positional parameter marker (?)
a user-defined variable
a local variable in a stored routine

Answer 2

..您的查询，稍作调整..

declare @t table (store int, product int, tran_date date, audit_date date, audit_bal int, inv_value int);

insert into @t
values
--store product tran_date  audit_date audit_bal inv_value
(10001, 323232,  '20200101', null,       null,      5),
(10001, 323232,  '20200102', '20200102', 20,        31),
(10001, 323232,  '20200103', null,       null,      13),
(10001, 323232,  '20200104', null,       null,      6),
(10001, 323232,  '20200105', null,       null,      21),
(10001, 323232,  '20200106', null,       null,      17),
(10001, 323232,  '20200107', null,       null,      6),
(10001, 323232,  '20200108', null,       null,      34),
(10001, 323232,  '20200109', null,       null,      35),
(10001, 323232,  '20200110', '20200110', 120,       17),
(10001, 323232,  '20200112', null,       null,      6),
(10001, 323232,  '20200113', null,       null,      9),
(10001, 323232,  '20200114', null,       null,      5),
(10001, 323232,  '20200115', null,       null,      29);

select 
    store, product, tran_date, audit_date, audit_bal, inv_value,
    --start_stock = end_stock-inv_value 
    end_stock-inv_value as start_stock, end_stock
from
(
    --calculate end_stock
    select *,
        max(grp_audit_bal) over(partition by store, product, grp_audit_date order by tran_date)
        + 
        sum(inv_value) over(partition by store, product, grp_audit_date order by tran_date) as end_stock
    from
    (
        --groups are defined by latest audit_date 
        --also get the audit_bal per group (audit_bal is assigned only to the first member of the group, lag() is used)
        select *, 
            max(audit_date) over(partition by store, product order by tran_date ROWS between UNBOUNDED PRECEDING and 1 PRECEDING) as grp_audit_date,
            lag(audit_bal) over(partition by store, product order by tran_date) as grp_audit_bal
        from @t
    ) as xyz
) as src;

根据另一列的前一行计算一列

Calculate one column based on another column's previous row

tsql

sql-server

recursive-query