我有一个像这样的表，我通过适度的重新加工this数据+ SQL来生成白天显示，但我想在空的日子里加上右边的运行记录。
╔════════════╦════════╦═════════╦════════╗
║ 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 ║
╚════════════╩════════╩═════════╩════════╝
我不知道如何制作它以便表转换为：
╔════════════╦════════╦═════════╦════════╗
║ 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 ║
╚════════════╩════════╩═════════╩════════╝
答案 0 :(得分：1)
尝试一下（我假设您的表名为tbl）。我使用generate_series生成缺少的行，并使用几个窗口函数在新行中插入相应的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
答案 1 :(得分：1)
数据：
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);
查询：
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)