MySQL查询累计和（案例：当期无订单时，仍为上期的值）

Question

I have order table with order date of course and its amount.

I want to sum the amount of previous period with current period and resulting table like this

my current result

the query I have tried was

    with data as(
select  
        lead_id,
        no_pks,
        customer_name,
        point_name,
        funder_id,
        DATE_FORMAT(order_date,'%Y-%m') as year_and_month_order,
        sum(total_amount) as outstanding
    from 
        mall_order_list_payment
    group by
        lead_id,
        no_pks,
        customer_name,
        point_name,
        funder_id,
        year_and_month_order
)

select 
    *,
    sum(outstanding) over(partition by lead_id,no_pks order by year_and_month_order) as cumulative_outstanding
from
    data
order by lead_id,no_pks

My goals is when on such month there were no order, the amount are 0 while the cumulative amount must be followed the previous month. The result of what I need are

My goals result

Answer 1

You must add one more recursive CTE and generate base dates list (calendar) using min. and max. dates from your data table as generated range borders (or set these borders as parameters) then LEFT JOIN your table to it. In window function you'd use the dates from this base table which will contain all needed dates.

Simple DEMO:

 CREATE TABLE test (the_date DATE, the_value INT); INSERT INTO test SELECT '2022-03-04', 1 UNION ALL SELECT '2022-03-05', 2 UNION ALL SELECT '2022-03-07', 3 ;

 SELECT *, SUM(the_value) OVER (ORDER BY the_date) cumulative FROM test; 

the_date	the_value	cumulative
2022-03-04	1	1
2022-03-05	2	3
2022-03-07	3	6

 WITH RECURSIVE cte AS ( SELECT MIN(the_date) the_date FROM test UNION ALL SELECT the_date + INTERVAL 1 DAY FROM cte WHERE the_date < ( SELECT MAX(the_date) FROM test ) ) SELECT *, SUM(test.the_value) OVER (ORDER BY the_date) cumulative FROM cte LEFT JOIN test USING (the_date); 

the_date	the_value	cumulative
2022-03-04	1	1
2022-03-05	2	3
2022-03-06	null	3
2022-03-07	3	6

db<>fiddle here

---

The variant for different id values

CREATE TABLE test (id INT, the_date DATE, the_value INT); INSERT INTO test SELECT 1, '2022-03-04', 1 UNION ALL SELECT 1, '2022-03-05', 2 UNION ALL SELECT 1, '2022-03-07', 3 UNION ALL SELECT 2, '2022-03-04', 4 UNION ALL SELECT 2, '2022-03-06', 5 UNION ALL SELECT 2, '2022-03-08', 6 ;

 SELECT *, SUM(the_value) OVER (PARTITION BY id ORDER BY the_date) cumulative FROM test;

id | the_date |  the_value | cumulative
-: |  :--------- |  --------: |  ---------: 
 1 |  2022-03-04 |  1 |  1 
 1 |  2022-03-05 |  2 |  3 
 1 |  2022-03-07 |  3 |  6 
 2 |  2022-03-04 |  4 |  4 
 2 |  2022-03-06 |  5 |  9 
 2 |  2022-03-08 |  6 |  15 

WITH RECURSIVE cte1 AS ( SELECT MIN(the_date) the_date FROM test UNION ALL SELECT the_date + INTERVAL 1 DAY FROM cte1 WHERE the_date < ( SELECT MAX(the_date) FROM test ) ), cte2 AS ( SELECT DISTINCT id FROM test ) SELECT *, SUM(test.the_value) OVER (PARTITION BY id ORDER BY the_date) cumulative FROM cte1 CROSS JOIN cte2 LEFT JOIN test USING (id, the_date);

the_date |  id | the_value | cumulative
:--------- |  -: |  --------: |  ---------: 
2022-03-04 |  1 |  1 |  1 
2022-03-05 |  1 |  2 |  3 
2022-03-06 |  1 |  null | 3 
2022-03-07 |  1 |  3 |  6 
2022-03-08 |  1 |  null | 6 
2022-03-04 |  2 |  4 |  4 
2022-03-05 |  2 |  null | 4 
2022-03-06 |  2 |  5 |  9 
2022-03-07 |  2 |  null | 9 
2022-03-08 |  2 |  6 |  15 

db<>fiddle here