PostgreSQL 用户和余额 - 如果缺少值，则在时间序列中为用户使用以前的余额答案

【问题标题】：PostgreSQL users and balances - use previous balance for user in time series if value is missingPostgreSQL 用户和余额 - 如果缺少值，则在时间序列中为用户使用以前的余额
【发布时间】：2022-10-13 03:42:15
【问题描述】：

给定下表：

用户：

name
alice
bob

余额：

id	user_name	date	balance
1	alice	2022-01-01	100
2	alice	2022-01-03	200
3	alice	2022-01-04	300
4	bob	2022-01-01	400
5	bob	2022-01-02	500
6	bob	2022-01-05	600

我想获得所有用户从第一天到最后一天的完整列表，用该用户的最后可用余额替换 NULL 余额。

这是我到目前为止所拥有的：

select u.name, s.day, b.balance
from users u
cross join (select generate_series(min(day)::date, max(day)::date, interval '1 day')::date as day from balances) s
left join balances b on b.user_name = u.name and s.day = b.day
order by u.name, s.day 
;

SQL Fiddle Here

我已经尝试过LAG() 和此处找到的其他一些示例，但它们似乎都没有为用户获得正确的最后平衡。

【问题讨论】：

标签： sql postgresql left-join

【解决方案1】：

我们要使用 running_total：sum() over()

select      u.name
           ,s.day
           ,sum(b.balance) over(partition by u.name order by s.day) as balance
from        users u
cross join (select generate_series(min(day)::date, max(day)::date, interval '1 day')::date as day from balances) s
left join   balances b on b.user_name = u.name and s.day = b.day
order by    u.name, s.day

name	day	balance
alice	2022-01-01	100
alice	2022-01-02	100
alice	2022-01-03	300
alice	2022-01-04	600
alice	2022-01-05	600
bob	2022-01-01	400
bob	2022-01-02	900
bob	2022-01-03	900
bob	2022-01-04	900
bob	2022-01-05	1500

Fiddle

【讨论】：

此解决方案将所有先前余额的总和相加。我需要它来添加最后一个余额值。例如。 bob 的 2022-01-05 余额应该是 600

【解决方案2】：

基于How do I efficiently select the previous non-null value?，我最终通过以下查询获得了成功的结果：

select
  name, 
  day, 
  first_value(balance) over (partition by x.name, value_partition order by day) as balance
from (
  select 
    u.name as name, 
    s.day as day, 
    b.balance as balance,
    sum(case when b.balance is null then 0 else 1 end) over (partition by u.name order by s.day) as value_partition
  from users u
  cross join (select generate_series(min(day)::date, max(day)::date, interval '1 day')::date as day from balances) s
  left join balances b on b.user_name = u.name and s.day = b.day
) x
order by x.name, x.day

DB Fiddle

【讨论】：