Question

我不知道是否可能，但是我想优化我编写的查询，以获取两个日期之间的所有查询组，其中两个日期之间都填充有0个值，用于间隔中的缺失日期。

我正在运行MySQL 5.7

我有一个calendar表，其中包含一年中的所有小时（8760行）

CREATE TABLE calendar (
  date datetime PRIMARY KEY
);

我有一个spents表，其中包含日期，用户，类别和花费的

CREATE TABLE spents (
  date datetime NOT NULL,
  user varchar(24) NOT NULL,
  category enum('food', 'hobbies', 'clothing', 'taxes') NOT NULL,
  spent int(5) unsigned NOT NULL DEFAULT '0',
  UNIQUE KEY hourly_composite (date, user, category)
);

假设spents表包含以下行：

+---------------------+------+----------+-------+
| date                | user | category | spent |
+---------------------+------+----------+-------+
| 2018-10-01 10:00:00 | bob  | food     |    10 |
| 2018-10-01 11:00:00 | bob  | hobbies  |    50 |
| 2018-10-01 11:00:00 | bob  | clothing |    30 |
| 2018-10-01 11:00:00 | bob  | taxes    |     3 |
| 2018-10-01 12:00:00 | bob  | food     |    30 |
| 2018-10-01 15:00:00 | bob  | clothing |    25 |
| 2018-10-01 16:00:00 | bob  | hobbies  |     5 |
+---------------------+------+----------+-------+

例如，我希望在 2018-10-01 时获得 10 和 18 之间的支出总额用户 bob 。
最终结果应如下所示：

+---------------------+------+------------------------+-------------+
| hour                | user | categories             | total_spent |
+---------------------+------+------------------------+-------------+
| 2018-10-01 10:00:00 | bob  | food                   |          10 |
| 2018-10-01 11:00:00 | bob  | clothing,hobbies,taxes |          83 |
| 2018-10-01 12:00:00 | bob  | food                   |          30 |
| 2018-10-01 13:00:00 | bob  |                        |           0 |
| 2018-10-01 14:00:00 | bob  |                        |           0 |
| 2018-10-01 15:00:00 | bob  | clothing               |          25 |
| 2018-10-01 16:00:00 | bob  | hobbies                |           5 |
| 2018-10-01 17:00:00 | bob  |                        |           0 |
| 2018-10-01 18:00:00 | bob  |                        |           0 |
+---------------------+------+------------------------+-------------+

因此查询如下：

-- get the scalar product of unique group and hour
    SELECT hour, user,
      IFNULL(GROUP_CONCAT(DISTINCT IF(hour = DATE_FORMAT(spents.date, "%Y-%m-%d %T") AND spent > 0, category, NULL)), "") AS categories,  
      SUM(IF(hour = DATE_FORMAT(spents.date, "%Y-%m-%d %T"), IFNULL(spent, 0), 0)) AS total_spent
    FROM spents
    CROSS JOIN 
      (
-- get all hours in the time interval
        SELECT DATE_FORMAT(date, "%Y-%m-%d %T") AS hour
        FROM calendar
        WHERE date BETWEEN "2018-10-01 10:00:00" AND "2018-10-01 18:59:59"
        GROUP BY hour
      ) AS interval_units
    WHERE date BETWEEN "2018-10-01 10:00:00" AND "2018-10-01 18:59:59"
    GROUP BY user, hour
    ORDER BY user, hour;

此请求可以完美运行，但我不确定这是执行此操作的最佳方法。
当然，这是spents表的非常简化的版本，想象一下一个唯一的键，该键在一天的每一小时都有8列以上的列表格中有很多行（数百万）。
我使用calendar表的原因是能够获得两个日期之间所有小时的详尽列表。
我还可以按年，月，日，星期几等分组

编辑：
这是EXPLAIN语句：

+----+-------------+------------+------------+-------+------------------+------------------+---------+------+------+----------+-----------------------------------------------------------+
| id | select_type | table      | partitions | type  | possible_keys    | key              | key_len | ref  | rows | filtered | Extra                                                     |
+----+-------------+------------+------------+-------+------------------+------------------+---------+------+------+----------+-----------------------------------------------------------+
|  1 | PRIMARY     | spents     | NULL       | range | hourly_composite | hourly_composite | 5       | NULL |    7 |   100.00 | Using where; Using temporary; Using filesort              |
|  1 | PRIMARY     | <derived2> | NULL       | ALL   | NULL             | NULL             | NULL    | NULL |    9 |   100.00 | Using join buffer (Block Nested Loop)                     |
|  2 | DERIVED     | calendar   | NULL       | range | PRIMARY          | PRIMARY          | 5       | NULL |    9 |   100.00 | Using where; Using index; Using temporary; Using filesort |
+----+-------------+------------+------------+-------+------------------+------------------+---------+------+------+----------+-----------------------------------------------------------+

优化空日期填充查询（使用日历表）

0 个答案: