爆炸一個SQL表，並插入一行，每天不存在

Question

我有一個像這樣的表，我通過適度返工this數據+ SQL來顯示白天，但我想添加空天以及與在右手邊運行理貨。

╔════════════╦════════╦═════════╦════════╗ 
║ day  ║ num1 ║ num2 ║ tally ║ 
╠════════════╬════════╬═════════╬════════╣ 
║ 2016-06-10 ║ 9.99 ║   ║ 9.99 ║ 
║ 2016-06-12 ║ 136.00 ║ 9.99 ║ 145.99 ║ 
║ 2016-06-14 ║  ║ 145.99 ║ 145.99 ║ 
║ 2016-06-18 ║ 9.99 ║ 145.99 ║ 155.98 ║ 
║ 2016-06-19 ║ 210.00 ║ 145.99 ║ 365.98 ║ 
║ 2016-06-22 ║ 50.00 ║ 9.99 ║ 279.98 ║ 
║ 2016-06-28 ║ 69.99 ║ 59.99 ║ 349.97 ║ 
╚════════════╩════════╩═════════╩════════╝ 

我不知道如何讓這個錶轉換爲：

╔════════════╦════════╦═════════╦════════╗ 
║ day  ║ num1 ║ num2 ║ tally ║ 
╠════════════╬════════╬═════════╬════════╣ 
║ 2016-06-10 ║ 9.99 ║   ║ 9.99 ║ 
║ 2016-06-11 ║  ║   ║ 9.99 ║ <- new row with previous value 
║ 2016-06-12 ║ 136.00 ║ 9.99 ║ 145.99 ║ 
║ 2016-06-13 ║  ║   ║ 145.99 ║ <- new row with previous value 
║ 2016-06-14 ║  ║ 145.99 ║ 145.99 ║ 
║ 2016-06-15 ║  ║   ║ 145.99 ║ <- new row with previous value 
║ 2016-06-16 ║  ║   ║ 145.99 ║ <- new row with previous value 
║ 2016-06-17 ║  ║   ║ 145.99 ║ <- new row with previous value 
║ 2016-06-18 ║ 9.99 ║ 145.99 ║ 155.98 ║ 
║ 2016-06-19 ║ 210.00 ║ 145.99 ║ 365.98 ║ 
║ 2016-06-20 ║  ║   ║ 365.98 ║ <- new row with previous value 
║ 2016-06-21 ║  ║   ║ 365.98 ║ <- new row with previous value 
║ 2016-06-22 ║ 50.00 ║ 9.99 ║ 279.98 ║ 
║ 2016-06-23 ║  ║   ║ 279.98 ║ <- new row with previous value 
║ 2016-06-24 ║  ║   ║ 279.98 ║ <- new row with previous value 
║ 2016-06-25 ║  ║   ║ 279.98 ║ <- new row with previous value 
║ 2016-06-26 ║  ║   ║ 279.98 ║ <- new row with previous value 
║ 2016-06-27 ║  ║   ║ 279.98 ║ <- new row with previous value 
║ 2016-06-28 ║ 69.99 ║ 59.99 ║ 349.97 ║ 
╚════════════╩════════╩═════════╩════════╝ 

Answer 1

試試這個（我假設你的表稱爲tbl）。我使用generate_series來生成缺失的行，並使用幾個窗口函數在新行中插入適當的tally值。

with all_dates as (
    SELECT day 
    FROM generate_series 
     ('2016-06-10'::date, 
     '2016-06-28'::date, 
     '1 day'::interval) day 
), partitioned_data as (
    select d.day, t.num1, t.num2, t.tally, 
     sum(case when t.tally is not null then 1 else 0 end) over (order by d.day) as partition_id 
    from all_dates d 
    left join tbl t 
     on t.day = d.day 
) 
select t.day, t.num1, t.num2, 
     first_value(t.tally) over (partition by t.partition_id) as tally 
    from partitioned_data t 
order by t.day 

Answer 2

數據：

create table the_data(day date, num1 numeric, num2 numeric, tally numeric); 
insert into the_data values 
('2016-06-10', 9.99, null, 9.99), 
('2016-06-12', 136.00, 9.99, 145.99), 
('2016-06-14', null, 145.99, 145.99), 
('2016-06-18', 9.99, 145.99, 155.98), 
('2016-06-19', 210.00, 145.99, 365.98), 
('2016-06-22', 50.00, 9.99, 279.98), 
('2016-06-28', 69.99, 59.99, 349.97); 

查詢：

with the_data as ( 
    select d::date as day, num1, num2, tally 
    from generate_series('2016-06-10'::date, '2016-06-28', '1d') d 
    left join the_data on d = day 
    ) 
select distinct on (a.day) a.day, a.num1, a.num2, b.tally 
from the_data a 
join the_data b 
on a.day >= b.day and b.tally is not null 
order by a.day, b.day desc; 

    day  | num1 | num2 | tally 
------------+--------+--------+-------- 
2016-06-10 | 9.99 |  | 9.99 
2016-06-11 |  |  | 9.99 
2016-06-12 | 136.00 | 9.99 | 145.99 
2016-06-13 |  |  | 145.99 
2016-06-14 |  | 145.99 | 145.99 
2016-06-15 |  |  | 145.99 
2016-06-16 |  |  | 145.99 
2016-06-17 |  |  | 145.99 
2016-06-18 | 9.99 | 145.99 | 155.98 
2016-06-19 | 210.00 | 145.99 | 365.98 
2016-06-20 |  |  | 365.98 
2016-06-21 |  |  | 365.98 
2016-06-22 | 50.00 | 9.99 | 279.98 
2016-06-23 |  |  | 279.98 
2016-06-24 |  |  | 279.98 
2016-06-25 |  |  | 279.98 
2016-06-26 |  |  | 279.98 
2016-06-27 |  |  | 279.98 
2016-06-28 | 69.99 | 59.99 | 349.97 
(19 rows)