如何编写一个按日期接收数据组的查询? [英] How write a query which receives data group by date?

问题描述

我写了一个简单的查询

SELECT date, count(user) as count FROM sessions GROUP BY date;

这样的回答

但是，我想查看每个日期，如果日期不存在，则行必须是这样的

But, I want to see every date, if date is not exist rows must be like this

2018-02-01 | 0
2018-02-02 | 0
2018-02-03 | 0
2018-02-04 | 0
2018-02-05 | 1503582
2018-02-06 | 0
2018-02-06 | 0
2018-02-08 | 5612270
...
...
...
2018-02-31 | 0

我应该如何重写查询以获取飞蛾每天的数据?

How can I should rewrite my query for getting data for every day in the moths ?

推荐答案

您需要先生成一个日历，然后将其与表session的行连接到date列.下面的子查询将根据@START_DATE变量中设置的值生成全部1个月的数据.

You need to generate a calendar first and join it's rows with table session through date column. The subquery below will generate all 1 month based from the value set in the @START_DATE variable.

SET  @START_DATE  = '2018-01-01';  -- set your starting date here

SELECT  a.`DATE`,
        COUNT(b.user) AS `COUNT`
FROM
        (
            SELECT DATE(cal.date) `DATE` 
            FROM ( 
                SELECT @START_DATE + INTERVAL xc DAY AS date 
                FROM ( 
                      SELECT @xi:=@xi+1 as xc from 
                      (SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) xc1, 
                      (SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) xc2, 
                      (SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) xc3, 
                      (SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4) xc4, 
                      (SELECT @xi:=-1) xc0 
                ) xxc1 
            ) cal 
            WHERE cal.date <= DATE_ADD(DATE_ADD(@START_DATE, INTERVAL 1 MONTH), INTERVAL -1 DAY)
        ) a
        LEFT JOIN sessions b
            ON a.`DATE` = b.`DATE`
GROUP   BY a.`DATE`
ORDER   BY a.`DATE`

这是一个演示.

从该

The subquery that generates date was borrowed from this article: Generating a Series of Dates in MySQL and modified a little bit to allow user to input the starting date.

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