Question

我希望在没有使用分析函数的情况下获得user_counts的更改百分比，因为mysql不支持它。只是想知道什么是解决它的最佳方法。

架构： - User_Visits（VisitorID，UserID，Day）

我尝试了什么？

  SELECT Day,
      FROM (
             ( Day_User_Count - LAG( Day_User_Count, 1 ) OVER ( ORDER BY Day ) )/
              Day_User_Count  
            )* 100 AS Percentage_Change     
         ( 
           SELECT Day,
                  COUNT( DISTINCT UserID ) AS Day_User_Count
             FROM User_Visits
            GROUP BY Day
          );

Answer 1

试一试。

SELECT 
    Day, COUNT(DISTINCT UserID) as Day_count, prv.Prev_Day_User_Count,
    (COUNT(DISTINCT UserID) - prv.Prev_Day_User_Count) 
        / prv.Prev_Day_User_Count * 100 AS Percentage_Change
FROM
    User_Visits UV
        INNER JOIN
    (SELECT 
        Day AS prev_day,
            COUNT(DISTINCT UserID) AS Prev_Day_User_Count
    FROM
        User_Visits
    GROUP BY Day) prv ON UV.day = DATE_ADD(prv.prev_day, INTERVAL 1 DAY)
GROUP BY Day;

输入：

UserID Day    
1   2016-10-20
1   2016-10-20
2   2016-10-20
3   2016-10-21
4   2016-10-21
5   2016-10-21
6   2016-10-22
1   2016-10-22
2   2016-10-23
3   2016-10-23
4   2016-10-23
5   2016-10-23
6   2016-10-24
7   2016-10-24

输出：

# Day, prev_day, Day_count, Prev_Day_User_Count, Percentage_Change
'2016-10-21', '2016-10-20', '3', '2', '50.0000'
'2016-10-22', '2016-10-21', '2', '3', '-33.3333'
'2016-10-23', '2016-10-22', '4', '2', '100.0000'
'2016-10-24', '2016-10-23', '2', '4', '-50.0000'

http://sqlfiddle.com/#!9/d3bcb

与没有LAG分析功能的昨天相比，用户的百分比

1 个答案: