此查询:
SELECT received_at as sign_up_date,
COUNT(DISTINCT email) AS "count"
FROM seller_v2.users
GROUP BY 1
ORDER BY 1;
创建以下输出:
sign_up_date count
2016-02-18T17:38:51.000Z 1
2016-02-18T21:47:48.000Z 1
每个日期等于用户的注册日期。如何创建一个包含运行总用户数的新列,以便我可以创建时间序列图表?
答案 0 :(得分:1)
如果您使用的是SQL Server(我相信Oracle和其他一些RDBMS的语法相同),您可以使用ROW_NUMBER()
来显示行号,在您的情况下与行号相同:
SELECT received_at as sign_up_date,
ROW_NUMBER() OVER (ORDER BY received_at ASC) AS running_total
FROM seller_v2.users;