MySQL按日期分组和计数，包括缺少的日期

之前，我做了以下工作，从报告表中获取每日计数

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;

---

然而，我希望用0来填充缺失的日期

差不多

---

我的另一次尝试是：

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

---

上述查询的最小数据：

子查询的

所有日期

仅从当前日期（

curdate（）

）开始回溯。如果要包括将来的日期，请将子查询的第一行更改为类似以下内容：

select '2015-01-01' - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date

---

您需要一个

外部联接

在开始和结束之间每天到达，因为如果您使用

内部联接

，则输出将仅限于联接的日期（即报告表中的日期）

此外，当您使用

外部联接时

必须注意

where子句中的条件不会导致
隐式内部联接
；例如，和domain_id=1如果在where子句中使用，将抑制不满足该条件的任何行，但当用作联接条件时，它仅限制报表表的行

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;

通过使用
DATE\u ADD（）
将起点推向未来，我还更改了all_dates派生表，并减小了它的大小。这两个都是选项，可以根据需要进行调整



要获得每一行的域id（如您的问题所示），您需要使用以下内容：；注意，您可以使用特定于MySQL的
IFNULL（）
，但我使用了更通用的SQL的
COALESCE（）
。然而，这里所示的@parameter的使用是特定于MySQL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;

 
如果你愿意，考虑下面简单的两步行动：1。如果您还没有这样做，请提供适当的DDL（和/或SQLFIDLE），以便我们可以更轻松地复制问题。2.如果您还没有这样做，请提供与步骤1中提供的信息相对应的所需结果集。当然，这是包含行的最小表。这只是一个建议，您不必遵循它。太棒了！这就行了：）有可能把一组日子也按周和月分组吗？与我们所有“天”的情况类似，我可以拥有一年中的所有周。很高兴你的回答是这样的-请花一秒钟点击勾号-这表明答案已被接受。可以使用较大的时间单位，如周、月、年，但对于这些时间单位，我们通常使用日期范围（从/到日期对）。
select '2015-01-01' - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date

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;

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;