从数据库中获取每日和每月的总销售额

Question

我有一个表，列c_date为datetime，total为mysql中的int类型，我想打印出每天的销售，每个月的总销售，以及每年的总销售，包括日，月，年，其中没有销售。

目前，对于日常销售，我运行以下查询：

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 |
+------------+------------+

对于每月，我得到的是这样的：

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 |
+---------------------+-------+------+-------+

使用MysqL可以做到这一点吗？

Answer 1

由于不可能在MySQL中使用序列(实际上，它们根本不存在)，因此必须首先创建日期范围表。这将是这样的：

CREATE TABLE dates_range (record_date DATE)

然后用sale表中存在的日期填充此表，从日期中的最小值开始，直到最大值。

在此之后，使用SQL LEFT JOIN操作符，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，您需要一个带有日历表的外部连接。

假设有一个日程表，填充如下：

日历

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

然后，您可以编写一个查询，如

         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