Question

我试图为两个问题找到一个优雅的解决方案：我对这两个问题的SELECT部分都有疑问。查询的另一部分很好

从理论上讲，这些问题使没有可用的数据库运行。

第一季度：提取以下报告：显示每天的累计   最近5天（包括5天）内（移动）“点赞”的次数   将显示5月1日至5月5日的总点赞次数。）   美国以外的地区。

列：

DataTimstamp-5天时间范围的上限总计-的数量   在时间范围Region_US中的点赞次数-国家/地区的点赞次数   ='US'Region_rest-国家<>'US'
的点赞次数
第二季度：每天提取创建的用户数和每日更改   与前一天相比。

代码：

SELECT 
       DATEADD(day, 4, se.date) AS DataTimstamp,
       COUNT(se.type_id) AS Total,
       COUNT(CASE WHEN lo.country_3_character_code = 'USA' THEN 1 ELSE NULL END) AS 'Region_USA',
       COUNT(CASE WHEN lo.country_3_character_code != 'USA' THEN 1 ELSE NULL END) AS 'Region_rest'
  FROM  system_events se
  JOIN  location lo ON se.location_id = lo.id
  WHERE se.type = 'like'
GROUP BY 1

SELECT u.creation_date AS 'day', 
      COUNT(IF(day = u.creation_date, u.id, 0)) AS Date_day,
      COUNT(IF(day = u.creation_date - interval 1 day , u.id, 0)) AS Date_before,
      SUM(SUM(Date_day)-SUM(Date_before)) AS daily_change 
FROM user u
GROUP BY 1;

Answer 1

对于第二季度，我认为您可以使用LAG来获取前一天的信息，例如这个简单的例子：

with cte_table as
(
select * from
(values
   ('01-Jan-19',1)
  ,('02-Jan-19',2)
  ,('03-Jan-19',3)
  ,('04-Jan-19',4)
  ,('05-Jan-19',5)
) as t (TheDate,TheValue))

select TheDate, TheValue,
       LAG(TheValue,1,0) OVER(ORDER BY TheDate) as Prev_Value 
from cte_table;

Answer 2

最近5季度的Q1中，我对此进行了讨论，我可能会被SQL Masters吞噬：-o，可能与交叉应用有关，但是在我目前的知识范围之外。

with cte_table as
(
  select * from
    (values
     ('01-Jan-19',10)
    ,('02-Jan-19',20)
    ,('03-Jan-19',10)
    ,('04-Jan-19',5)
    ,('05-Jan-19',10)
    ,('06-Jan-19',20)
    ,('07-Jan-19',10)
    ,('08-Jan-19',10)
    ,('09-Jan-19',10)
    ,('10-Jan-19',5)
) as t (TheDate,TheValue))
, cte_table2 as
(
  select TheDate, TheValue
     ,LAG(TheValue,1,0) OVER(ORDER BY TheDate) as Prev_Value1 
     ,LAG(TheValue,2,0) OVER(ORDER BY TheDate) as Prev_Value2 
     ,LAG(TheValue,3,0) OVER(ORDER BY TheDate) as Prev_Value3 
     ,LAG(TheValue,4,0) OVER(ORDER BY TheDate) as Prev_Value4 
  from cte_table
)

 select TheDate
        ,sum(TheValue) as current_day
        , sum(TheValue) + sum(Prev_Value1) + sum(Prev_Value2) + sum(Prev_Value3) + sum(Prev_Value4) as [last 5 days]
 from cte_table2
 group by TheDate
 order by TheDate
 ;

日期范围内的COUNT CASE函数

2 个答案: