分组子集对齐后再做差集

【问题】

I have two tables 1)users(id,registerdate) 2)user_answer(userid,answer,updated_date)

I want the count of zero usage per day. How many users are registering but not answering per day. Results will be like this:

Date registedCount notAnsweredCount

15-09-02 20 10
15-09-01 20 10
15-08-31 12 4

Data will be like for user table((1,‘15-09-01’),(2,‘15-09-01’),(3,‘15-09-01’)) for user answer table ((1,0,15-09-01)).. Here you can see three users are registered on the day of sep 01, 2015 but the only one user has answered one question. So, result will be (Date=>15-09-01, registedCount => 3, notAnsweredCount => 2)

有人给出解答，楼主说比较像，但没进一步反馈

SELECT date_range.aDay,
COUNT(DISTINCT users.id) AS registedCount,
SUM(IF(users.id IS NOT NULL AND user_answer.userid IS NULL, 1, 0)) AS notAnsweredCount
FROM
(
SELECT DATE_ADD('2015-09-01', INTERVAL units.aCnt + tens.aCnt * 10 DAY) AS aDay
FROM
(
SELECT 0 AS aCnt UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) units
CROSS JOIN
(
SELECT 0 AS aCnt UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) tens
) date_range
LEFT OUTER JOIN users
ON date_range.aDay = users.registerdate
LEFT OUTER JOIN user_answer
ON users.id = user_answer.userid
GROUP BY date_range.aDay

【回答】

有两个难点要解决：库表按照指定的日期序列分组，而不是库表中的字段；每日注册的 id 如何和每日解答的 userid 进行差集运算。用 SQL 实现会比较难理解，可以用 SPL 来帮助 SQL 实现，代码分步写出很直观：

参数设置：

SPL 脚本

A1、A2：通过 SQL 取表中数据

A3：使用函数 periods，根据参数生成时间序列

A4、A5：使用函数 align，将集合按指定序列对齐

A6：生成新序表，“\”表示差集