Mysql-如何获取案件数量达到峰值时的最大日期。

Question

我试图在这个问题集上做#8，从sqlzoo (https:/sqlzoo.netwikiWindow_LAG#LAG_using_a_JOIN。).

问题是 "对于每一个在最近一天内有1000个新病例的国家，请说明新病例数量高峰的日期"。

表covid给出了各国每天的covid病例数、死亡数和恢复数，为所以。

+-------------+-------------------------------+-----------+--------+-----------+
|    Name     |              whn              | confirmed | deaths | recovered |
+-------------+-------------------------------+-----------+--------+-----------+
| Afghanistan | Sun, 01 Mar 2020 00:00:00 GMT |         1 |      0 |         0 |
| Albania     | Sun, 01 Mar 2020 00:00:00 GMT |         0 |      0 |         0 |
| Algeria     | Sun, 01 Mar 2020 00:00:00 GMT |         1 |      0 |         0 |
+-------------+-------------------------------+-----------+--------+-----------+

目前我有这个代码：

SELECT c.name, DATE_FORMAT(c.whn,'%Y-%m-%d') as this, d.peak
from ( select tw.name, max(tw.confirmed-lw.confirmed) as peak
FROM covid tw LEFT JOIN covid lw ON 
  DATE_ADD(lw.whn, INTERVAL 1 DAY) = tw.whn
   AND tw.name=lw.name
where tw.confirmed-lw.confirmed > 1000
group by tw.name) d
join covid as c
on d.name = c.name
group by name

这给了我每个国家，日期， 和高峰期的案件数量。然而，日期是显示每个国家的第一天，当案件超过1000。我怎样才能得到案件数量高峰期的日期？

|  Name   |    this    | peak |
|---------|------------|------|
| Austria | 2020-03-26 | 1321 |
| Belarus | 2020-04-20 | 1485 |
| Belgium | 2020-03-26 | 2454 |

Answer 1

你可以计算出 新的 的情况下，通过比较 confirmed 跨越以后的日子；对于这。lag() 很方便。

select 
    t.*,
    confirmed - lag(confirmed, 1, 0) over(partition by name order by whn) new_cases
from mytable t

假设每个国家每天都有一条记录。然后，你可以根据这些记录对每个国家的记录进行排名，并对每个国家排名最高的一天进行过滤。

select *
from (
    select 
        t*, 
        rank() over(partition by name order by new_cases desc) rn
    from (
        select 
            t.*,
            confirmed - lag(confirmed, 1, 0) over(partition by name order by whn) new_cases
        from mytable t
    ) t
    where new_cases > 1000
) t
where rn = 1