之前我正在执行以下操作以从报告表中获取每天的计数。

SELECT COUNT(*) AS count_all, tracked_on
 FROM `reports`
 WHERE (domain_id = 939 AND tracked_on >= '2014-01-01' AND tracked_on <= '2014-12-31')
 GROUP BY tracked_on
 ORDER BY tracked_on ASC;

显然，这不会给我 0 计数丢失的日期。

然后终于找到一个optimum solution在给定的日期范围内生成日期序列。 但我面临的下一个挑战是将它与我​​的报告表结合起来，并按日期对计数进行分组。

select count(*), all_dates.Date as the_date, domain_id
from (
    select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date
    from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
    cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
    cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
) all_dates
inner JOIN reports r
    on all_dates.Date >= '2014-01-01'
  and all_dates.Date <= '2014-12-31'
where all_dates.Date between '2014-01-01' and '2014-12-31' AND domain_id = 939 GROUP BY the_date order by the_date ASC ;

得到的结果是

count(*)    the_date    domain_id
46  2014-01-01  939
46  2014-01-02  939
46  2014-01-03  939
46  2014-01-04  939
46  2014-01-05  939
46  2014-01-06  939
46  2014-01-07  939
46  2014-01-08  939
46  2014-01-09  939
46  2014-01-10  939
46  2014-01-11  939
46  2014-01-12  939
46  2014-01-13  939
46  2014-01-14  939
...

填充缺失的日期

有点像

count(*)    the_date    domain_id
12  2014-01-01  939
23  2014-01-02  939
46  2014-01-03  939
0   2014-01-04  939
0   2014-01-05  939
99  2014-01-06  939
1   2014-01-07  939
5   2014-01-08  939
...

select count(*), all_dates.Date as the_date, domain_id
from (
    select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date
    from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
    cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
    cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
) all_dates
inner JOIN reports r
    on all_dates.Date = r.tracked_on
where all_dates.Date between '2014-01-01' and '2014-12-31' AND domain_id = 939 GROUP BY the_date order by the_date ASC ;

结果:

count(*)    the_date    domain_id
38        2014-09-03     939
8         2014-09-04     939

上述查询的最少数据:http://sqlfiddle.com/#!2/dee3e/6

最佳答案

您需要一个 OUTER JOIN 来到达开始和结束之间的每一天，因为如果您使用 INNER JOIN，它会将输出限制为仅包含加入(即只是报告表中的那些日期)。

此外，当您使用OUTER JOIN 时，您必须注意where 子句 中的条件不会导致隐式内部联接;例如 AND domain_id = 1 如果在 where 子句中使用将抑制任何不满足该条件的行，但当用作连接条件时它仅限制报告表的行。

SELECT
      COUNT(r.domain_id)
    , all_dates.Date AS the_date
    , domain_id
FROM (
        SELECT DATE_ADD(curdate(), INTERVAL 2 MONTH) - INTERVAL (a.a + (10 * b.a) ) DAY as Date
        FROM (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
        CROSS JOIN (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
      ) all_dates
      LEFT OUTER JOIN reports r
                  ON all_dates.Date = r.tracked_on
                        AND domain_id = 1
WHERE all_dates.Date BETWEEN '2014-09-01' AND '2014-09-30'
GROUP BY
      the_date
ORDER BY
      the_date ASC;

我还更改了 all_dates 派生表，方法是使用 DATE_ADD() 将起点推到 future ，并且减小了它的大小。这两个都是选项，可以根据需要进行调整。

Demo at SQLfiddle

要为每一行得出一个 domain_id(如您的问题所示)，您需要使用如下内容；请注意，您可以使用特定于 MySQL 的 IFNULL()，但我使用的是更通用的 SQL COALESCE()。然而，这里显示的 @parameter 的使用无论如何都是 MySQL 特定的。

SET @domain := 1;

SELECT
      COUNT(r.domain_id)
    , all_dates.Date AS the_date
    , coalesce(domain_id,@domain) AS domain_id
FROM (
        SELECT DATE_ADD(curdate(), INTERVAL 2 month) - INTERVAL (a.a + (10 * b.a) ) DAY as Date
        FROM (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
        CROSS JOIN (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
      ) all_dates
      LEFT JOIN reports r
                  ON all_dates.Date = r.tracked_on
                        AND domain_id = @domain
WHERE all_dates.Date BETWEEN '2014-09-01' AND '2014-09-30'
GROUP BY
      the_date
ORDER BY
      the_date ASC;

See this at SQLfiddle

关于MySQL 按日期分组并计算包括缺失日期，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25804531/