Question

我正在尝试做一些数据报告，并且不经常做SQL体操以了解我正在寻找什么功能。我称之为“ungroup by”。我有一个SELECT输出，反映了一系列每月订阅项目：创建，关闭以及每月支出的金额：

+----+---------------------------------+------------+------------+------------+
| id | description                     | created_on | closed_on  | monthly    |
+----+---------------------------------+------------+------------+------------+
|  3 | Daily horoscope email           | 2012-01-01 | null       | 10000.0000 |
|  5 | Pet food delivery               | 2012-01-05 | null       |  3500.0000 |
|  6 | Dirty magazine subscription     | 2012-01-09 | null       |  1500.0000 |
|  7 | Stupid nuts posted in a box     | 2012-01-01 | 2012-01-04 |  1500.0000 |
  .... etc ...

我要做的是每天计算出“运行率”。因此，每天都会列出当前每月承诺的总计，即上述数据将映射到：

+------------+----------+
| date       | run_rate |
+------------+----------+
| 2012-01-01 | 11500    |
| 2012-01-02 | 11500    |
| 2012-01-03 | 11500    |
| 2012-01-04 | 10000    |
| 2012-01-05 | 13500    |
| 2012-01-06 | 13500    |
| 2012-01-07 | 13500    |
| 2012-01-08 | 13500    |
| 2012-01-09 | 15000    |

我认为可能是每天创建一行一行的临时表，然后写一个LEFT JOIN / GROUP BY语句引用第一个表来构建我的输出。但我只能看到如何以这种方式创建日常“差异”，而不是运行总计，我需要将第一个表“取消组合”为两个条目，这是创建订阅时的正面条目，以及关闭时的否定条目。

我想坚持使用MySQL，如果可能的话，还要坚持使用一个超级声明。如果那是不可能的，我可以在我的查询框架中添加一些存储过程或临时表。或者我真的必须通过Ruby研磨我的数据？（我确切知道如何，但希望我能将所有逻辑保存在一个地方，并且我正在努力改进当前使用ActiveRecord的缓慢计算）。

Answer 1

尝试这样的事情 - 应该产生你想要的结果：

SET @runtot:=0;
SELECT
   mydates.seeddate,
    (@runtot := @runtot + IFNULL(m.amt,0) - IFNULL(m2.amt,0)) AS rt
FROM
   mydates left join 
    (Select createdon, SUM(monthly) amt
     FROM mytable 
     group by createdon
     ) m on mydates.seeddate = m.createdon
left join 
    (Select closed_on, SUM(monthly) amt
     FROM mytable 
     group by closed_on  
     ) m2 on mydates.seeddate = m2.closed_on

这是SQL Fiddle。

祝你好运。

Answer 2

试试这个：

select date,sum(monthly)
from     
(
   select created_on as date from yourtable 
   union 
   select closed_on from yourtable where closed_on is not null
) as alldates
left outer join yourtable
  on date >= created_on
 and (closed_on is null or date < closed_on)
where date between '2012-1-1' and '2012-1-31'
group by date order by 1

根据您的示例数据，输出为：

+------------+--------------+
| date       | sum(monthly) |
+------------+--------------+
| 2012-01-01 |     11500.00 |
| 2012-01-04 |     10000.00 |
| 2012-01-05 |     13500.00 |
| 2012-01-09 |     15000.00 |
+------------+--------------+
4 rows in set (0.00 sec)

我们可以确定那天没有的日期等于最接近的那一天。“2012-01-02”的run_rate等于'2012-01-01'的run_rate。

假设您已经有一个包含该月所有日期的表格，我们将其称为“mydate”，其中一列为“date”。

mysql> select * from mydate where date >= '2012-1-1' and date <= '2012-1-31';
+------------+
| date       |
+------------+
| 2012-01-01 |
| 2012-01-02 |
| 2012-01-03 |
| 2012-01-04 |
| 2012-01-05 |
| 2012-01-06 |
| 2012-01-07 |
...

然后替换

(
select created_on as date from yourtable 
union 
select closed_on from yourtable where closed_on is not null
) as alldates

mydate

完成！

Answer 3

这是另一种使用它的方法：INFORMATION_SCHEMA.COLLATION_CHARACTER_SET_APPLICABILITY它将在创建临时表的过程中创建日期，而不必创建临时表。 Explan plan可以确认哪种方法最适合您。

SQLFIDDLE DEMO

SET @rrate：= 0;

SELECT X.rdate, (@rrate:=@rrate +
COALESCE(Y.summonthly,0) -
COALESCE(Z.summonthly,0)) as run_rate
FROM(
      SELECT date_add(P.createdon, interval `day` day)
      as rdate
      FROM 
          (SELECT @i:= @i + 1 AS `day`
           FROM   
           INFORMATION_SCHEMA.COLLATION_CHARACTER_SET_APPLICABILITY,
           (SELECT @i:= -1) AS i
           ) As D,
      rategroups P
      GROUP BY rdate
      HAVING rdate <= (SELECT MAX(createdon) FROM rategroups)
      ORDER BY rdate) X
LEFT JOIN 
      (SELECT createdon, sum(monthly) summonthly
       FROM rategroups
       GROUP BY createdon) Y
ON X.rdate = Y.createdon
LEFT JOIN 
      (SELECT closed_on, sum(monthly) summonthly
       FROM rategroups
       GROUP BY closed_on) Z
ON X.rdate = Z.closed_on
GROUP BY X.rdate
;

|                          RDATE | RUN_RATE |
---------------------------------------------
| January, 01 2012 00:00:00+0000 |    11500 |
| January, 02 2012 00:00:00+0000 |    11500 |
| January, 03 2012 00:00:00+0000 |    11500 |
| January, 04 2012 00:00:00+0000 |    10000 |
| January, 05 2012 00:00:00+0000 |    13500 |
| January, 06 2012 00:00:00+0000 |    13500 |
| January, 07 2012 00:00:00+0000 |    13500 |
| January, 08 2012 00:00:00+0000 |    13500 |
| January, 09 2012 00:00:00+0000 |    15000 |

MySQL数据摘要技巧 - 是否有UNGROUP BY？

3 个答案: