使用滞后 function 后如何用以前的值替换当前 null 值

Question

我面临的问题是通过用开放列值减去先前的关闭列值来替换我从滞后 function 获得的差异列上的当前 null 值。 我想跳过当前的 null 值并将所有以前的值从 diff 列提高 1 级

我的查询

SELECT DATE_FORMAT(date, '%m/%d/%Y') as Dates, Open, close, lag(open) OVER (ORDER BY date desc ) - close  as diff FROM example  

当前结果

Dates,         Open,   Close,    Diff
'04/18/2022', '301.07', '298.82', NULL
'04/14/2022', '303.75', '301.86', '-0.79'
'04/13/2022', '307.28', '302.67', '1.08'
'04/12/2022', '306.58', '308.08', '-0.80'
'04/11/2022', '313.09', '308.96', '-2.38'

预期的期望结果

Dates,         Open,   Close,    Diff
'04/18/2022', '301.07', '298.82', '-0.79'
'04/14/2022', '303.75', '301.86', '1.08'
'04/13/2022', '307.28', '302.67', '-0.80'
'04/12/2022', '306.58', '308.08', '-2.38'
'04/11/2022', '313.09', '308.96', 1.40
'04/08/2022', '308.00', '311.69', '0.43'

Answer 1

你为什么不这样做

SELECT 
DATE_FORMAT(date, '%m/%d/%Y') as Dates, 
Open, close, 
Open- lead(close) OVER (ORDER BY date desc )  as diff 
FROM example  

基本上我们从下一行中为每个人提供接近的数据