从数据库中获取每天和每月的总销售额

Question

我有c_date列为datetime表， total为mysql中的int类型，我想打印出每一天的销售额，以及每月的总销售额，以及每年的总销售额，包括没有销售额的日，月，年。

当前用于日常销售，我正在运行以下查询：

mysql> select date(c_date) as date, sum(total) as total_sale from sale group by date;
+------------+------------+
| date       | total_sale |
+------------+------------+
| 2013-10-3  |        798 |
| 2013-10-6  |        114 |
+------------+------------+

但是，我想要这样的东西：

mysql> select date(c_date) as date, sum(total) as total_sale from sale group by date;
+------------+------------+
| date       | total_sale |
+------------+------------+
| 2013-10-1  |          0 |
| 2013-10-2  |          0 |
| 2013-10-3  |        798 |
| 2013-10-4  |          0 |
| 2013-10-5  |          0 |
| 2013-10-6  |        114 |
+------------+------------+

而对于每月，我得到这个：

mysql> select c_date, month(c_date) as month, year(c_date) as year, sum(total) as total from sale group by c_date order by c_date;
+---------------------+-------+------+-------+
| c_date              | month | year | total |
+---------------------+-------+------+-------+
| 2013-10-3 02:40:06  |    10 | 2013 |   228 |
| 2013-10-3 02:41:58  |    10 | 2013 |   114 |
| 2013-10-3 02:44:36  |    10 | 2013 |   114 |
| 2013-10-3 02:46:40  |    10 | 2013 |   114 |
| 2013-10-3 02:49:15  |    10 | 2013 |   114 |
| 2013-10-3 02:53:36  |    10 | 2013 |   114 |
| 2013-10-6 07:43:27  |    10 | 2013 |   114 |
+---------------------+-------+------+-------+

但是我想要这样的东西：

mysql> select c_date, month(c_date) as month, year(c_date) as year, sum(total) as total from sale group by c_date order by c_date;
+---------------------+-------+------+-------+
| c_date              | month | year | total |
+---------------------+-------+------+-------+
| 2013-1-3 02:40:06   |     1 | 2013 |     0 |
| 2013-2-3 02:41:58   |     2 | 2013 |     0 |
| 2013-3-3 02:44:36   |     3 | 2013 |     0 |
| 2013-4-3 02:46:40   |     4 | 2013 |     0 |
| 2013-5-3 02:49:15   |     5 | 2013 |     0 |
| 2013-6-3 02:53:36   |     6 | 2013 |     0 |
| 2013-7-6 07:43:27   |     7 | 2013 |     0 |
| 2013-8-3 02:44:36   |     8 | 2013 |     0 |
| 2013-9-3 02:46:40   |     9 | 2013 |     0 |
| 2013-10-3 02:49:15  |    10 | 2013 |   912 |
| 2013-11-3 02:53:36  |    11 | 2013 |     0 |
| 2013-12-6 07:43:27  |    12 | 2013 |     0 |
+---------------------+-------+------+-------+

MysqL有可能吗？

Answer 1

由于不可能在MySQL中使用序列 （实际上， 序列根本不存在），因此您必须首先创建日期范围表。 就像：

CREATE TABLE dates_range (record_date DATE)

然后在此表中填充日期，从sale表中存在的最小日期开始直到最大为止。

之后，使用SQL LEFT JOIN运算符，您将能够像这样聚合数据：

SELECT
  YEAR(dates_range.record_date),
  MONTH(dates_range.record_date),
  DAY(dates_range.record_date),
  COALESCE(SUM(sale.total), 0) AS total_sum
FROM
  dates_range
    LEFT JOIN sale
      ON dates_range.record_date=DATE(sale.c_date)
GROUP BY
  YEAR(dates_range.record_date),
  MONTH(dates_range.record_date),
  DAY(dates_range.record_date)

Answer 2

在我看来，您需要与日历表进行外部联接。

假设日历表的填充方式如下：

日历

Year Month   Day
2013 201310  2013-10-1    
2013 201310  2013-10-2
...

然后您可以编写查询

         select date(c_day) as date, 
                sum(total) as total_sale 
           from calendar c 
left outer join sale s 
             on c.day = s.c_date
          where c.month = 201310
       group by c_day
         having c_day <= max(s.c_date); -- this is to avoid to show all 
                                        -- days for October