从日期开始到当前月/年的 SQL 计数数据

Question

我有从 2019 年 3 月到 2020 年 2 月的客户数据。我正在计算用于购买我们机票的电子邮件总数，然后是每个月，我使用的查询是

  SELECT sub.monthNameYear,sub.monthName,count(*) from
 (SELECT DATE_FORMAT(`reservation_for`,'%Y-%m') as monthNameYear,DATE_FORMAT(`reservation_for`,'%M %Y') as monthName,
         email,
         COUNT(*) AS 'Count'
  FROM `tablename`
  GROUP BY email, DATE_FORMAT(`reservation_for`, '%Y-%m') 
           HAVING COUNT(*) > 1 
  ORDER BY DATE_FORMAT(`reservation_for`, '%Y-%m') ) as sub
GROUP BY sub.monthNameYear;

我想要的结果附在下面。它总是选择日期的开始，然后选择每个月，然后只计算电子邮件总数。

Answer 1

如果您运行的是 MySQL 8.0，则可以进行窗口求和。从您现有的查询开始：

SELECT 
    monthNameYear,
    monthName,
    SUM(COUNT(*)) OVER(ORDER BY monthNameYear) totalPurchase
FROM (
    SELECT 
        DATE_FORMAT(`reservation_for`,'%Y-%m') monthNameYear,
        DATE_FORMAT(`reservation_for`,'%M %Y') monthName,
        email
    FROM `tablename`
    GROUP BY monthNameYear, monthName, email 
    HAVING COUNT(*) > 1 
) as sub
GROUP BY monthNameYear, monthName
ORDER BY monthNameYear;

注意事项：

• 始终枚举group by 子句中的所有非聚合列

• order by 子句应放在外部查询中

• 你是 selecting COUNT(*) in the subquery, but it is not used in the outer query - I removed that column from theselect` 子句

Answer 2

你可以使用窗口函数：

SELECT ed.monthNameYear, ed.monthName,        
       COUNT(*),
       SUM(COUNT(*)) OVER (ORDER BY ed.monthNameYear)
FROM (SELECT DATE_FORMAT(`reservation_for`,'%Y-%m') as monthNameYear,
             DATE_FORMAT(`reservation_for`, '%M %Y') as monthName, 
             email,
             COUNT(*) AS cnt
      FROM `tablename`
      GROUP BY email, DATE_FORMAT(`reservation_for`, '%Y-%m'),
            DATE_FORMAT(`reservation_for`,'%M %Y')
      HAVING COUNT(*) > 1 
     ) ed
GROUP BY ed.monthNameYear, ed.monthName;