Sql 使用给定数据集分组生成结果

有关可用示例，请参见此处的SQL Fiddle：

我有4张桌子，表A、表B、表C和表D

表A数据和结构：

+--------------------+
| ID | Name          |
+--------------------+
| 987 | Mississauga  |
| 454 | Hoboken      |
| 343 | Berkeley     |
+--------------------+

+------------------------------------------------------------+
| City1ID | City1Name     | City2ID | City2Name     | Factor |
+------------------------------------------------------------+
| 343     | Berkeley      | 19      | Oakland       | 0.5    |
| 987     | Mississauga   | 23      | Toronto       | 1.0    |
| 66      | Redmond       | 13      | Seattle       | 1.0    |
| 343     | Berkeley      | 14      | San Jose      | 0.5    |
| 454     | Hoboken       | 55      | New York City | 0.9    |
| 454     | Hoboken       | 44      | Philadelphia  | 0.1    
| 19      | Oakland       | 9       | San Francisco | 1.0    |
...
+------------------------------------------------------------+

+----------------------------------+
| Date | ID | Name |  Hour | Value |
+----------------------------------+

表B数据和结构：

+--------------------+
| ID | Name          |
+--------------------+
| 987 | Mississauga  |
| 454 | Hoboken      |
| 343 | Berkeley     |
+--------------------+

+------------------------------------------------------------+
| City1ID | City1Name     | City2ID | City2Name     | Factor |
+------------------------------------------------------------+
| 343     | Berkeley      | 19      | Oakland       | 0.5    |
| 987     | Mississauga   | 23      | Toronto       | 1.0    |
| 66      | Redmond       | 13      | Seattle       | 1.0    |
| 343     | Berkeley      | 14      | San Jose      | 0.5    |
| 454     | Hoboken       | 55      | New York City | 0.9    |
| 454     | Hoboken       | 44      | Philadelphia  | 0.1    
| 19      | Oakland       | 9       | San Francisco | 1.0    |
...
+------------------------------------------------------------+

+----------------------------------+
| Date | ID | Name |  Hour | Value |
+----------------------------------+

表C：是一个表，其中包含城市一年中每天的小时数据。并非所有城市都在表C中。表a表示表B中存在地图的城市的一小部分。在我的例子中，TableA有439个城市（行），TableB有7000行映射

表C的结构如下：

+------------------------------------------------------------------------+
|Date | CID | Name | Blah1 | Blah2 | Hour1 | Hour2 | Hour3 | ... | Hour24 |
+------------------------------------------------------------------------+

以下是表C中一天的数据示例：

2006-08-01 00:00:00 9 San Francisco Blah1 Blah2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2006-08-01 00:00:00 23 Toronto Blah1 Blah2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 
2006-08-01 00:00:00 13 Seattle Blah1 Blah2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
2006-08-01 00:00:00 14 San Jose Blah1 Blah2 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
2006-08-01 00:00:00 55 New York City Blah1 Blah2 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
2006-08-01 00:00:00 44 Philadelphia Blah1 Blah2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

为了举例说明，奥克兰市不在表C中。然而，伯克利的每小时观测数据是通过求和得出的（圣何塞为0.5倍，奥克兰为0.5倍）

表格结构：

+--------------------+
| ID | Name          |
+--------------------+
| 987 | Mississauga  |
| 454 | Hoboken      |
| 343 | Berkeley     |
+--------------------+

+------------------------------------------------------------+
| City1ID | City1Name     | City2ID | City2Name     | Factor |
+------------------------------------------------------------+
| 343     | Berkeley      | 19      | Oakland       | 0.5    |
| 987     | Mississauga   | 23      | Toronto       | 1.0    |
| 66      | Redmond       | 13      | Seattle       | 1.0    |
| 343     | Berkeley      | 14      | San Jose      | 0.5    |
| 454     | Hoboken       | 55      | New York City | 0.9    |
| 454     | Hoboken       | 44      | Philadelphia  | 0.1    
| 19      | Oakland       | 9       | San Francisco | 1.0    |
...
+------------------------------------------------------------+

+----------------------------------+
| Date | ID | Name |  Hour | Value |
+----------------------------------+

表中为最终结果，应填写表中每个城市每天的小时观测数据

在我的例子中，小时1看起来应该是这样的，米西索加＝2（1×多伦多＝1×2＝2），霍博肯＝5.10（0.9×纽约+ 0.1 x费城＝0.9×5＋0.1 x 0.1）＝伯克利（x x san若泽＋x x奥克兰＝x xα＋x x旧金山＝x xα＋x x＝y）：< /p> 现在，棘手的部分是奥克兰不存在于TabLEC，而是它被映射到旧金山，所以任何被映射到奥克兰的城市都需要参照它所映射的城市。在我的情况下，有很多情况下会发生这种情况，所以我想确保我能正确处理这种情况

我已经将表C解包/规范化并将其与表B连接起来，但不确定如何继续。我知道我需要使用Group By，但不确定如何使用我描述的警告

select * from
(
    select Date, CID, Name, replace(Details, 'Hour', '') as Hour, ObservationValue
    from TableC
    as Result
    unpivot
    (
      ObservationValue for Details in ([Hour1], [Hour2], [Hour3], [Hour4], [Hour5], [Hour6],
      [Hour7],[Hour8],[Hour9],[Hour10],[Hour11],[Hour12],[Hour13],[Hour14],
      [Hour15],[Hour16],[Hour17],[Hour18],[Hour19],[Hour20],[Hour21],[Hour22],
      [Hour23],[Hour24])
)
as UnPvt
) as resultX
join TableB as b on (b.City2ID = resultX.CID)

如果无法访问SQL FIDLE链接，则以下是创建架构的代码：

CREATE TABLE TableA
    ([Id] int, [name] varchar(20));

INSERT INTO TableA
    ([Id], [name])
VALUES
    (987, 'Mississauga'),
    (454, 'Hoboken'),
    (343, 'Berkeley');

CREATE TABLE TableB
    ([City1ID] int, [City1Name] varchar(20), 
     [City2ID] int, [City2Name] varchar(20), [Factor] varchar(20))

INSERT INTO TableB
     ([City1ID], [City1Name],[City2ID], [City2Name], [Factor])
VALUES
     (343, 'Berkeley', 19, 'Oakland', 0.5),
    (987, 'Mississauga', 23, 'Toronto', 1.0),
    (66, 'Redmond', 13, 'Seattle', 1.0),
    (343, 'Berkeley', 14, 'San Jose', 0.5),
    (454, 'Hoboken', 55, 'New York City', 0.9),
    (454, 'Hoboken', 44, 'Philadephia', 0.1),
    (19, 'Oakland', 9, 'San Francisco', 1.0);

CREATE TABLE TableC
    ([date] datetime, [CId] int, [name] varchar(50), [blah1] varchar(10), 
     [blah2] varchar(10), [hour1] int, [hour2] int, [hour3] int, [hour4] int,
    [hour5] int, [hour6] int, [hour7] int, [hour8] int, [hour9] int,
    [hour10] int,[hour11] int, [hour12] int, [hour13] int, [hour14] int,
    [hour15] int, [hour16] int, [hour17] int, [hour18] int, [hour19] int,
    [hour20] int, [hour21] int, [hour22] int, [hour23] int, [hour24] int);

INSERT INTO TableC
    ([date], [CId], [name], [blah1], 
     [blah2], [hour1], [hour2], [hour3], [hour4],
    [hour5], [hour6], [hour7], [hour8], [hour9],
    [hour10],[hour11], [hour12], [hour13], [hour14],
    [hour15], [hour16], [hour17], [hour18], [hour19],
    [hour20], [hour21], [hour22], [hour23], [hour24])

VALUES
('2006-08-01 00:00:00', 9, 'San Francisco', 'Blah1', 'Blah2', 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1),
('2006-08-01 00:00:00', 23,'Toronto', 'Blah1', 'Blah2', 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2),
('2006-08-01 00:00:00', 13,'Seattle', 'Blah1', 'Blah2', 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3),
('2006-08-01 00:00:00', 14,'San Jose', 'Blah1', 'Blah2', 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4),
('2006-08-01 00:00:00', 55,'New York City', 'Blah1', 'Blah2', 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5),
('2006-08-01 00:00:00', 44,'Philadelphia', 'Blah1', 'Blah2', 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6);

这就行了

---

这个问题对我来说太复杂了，请尝试用更简单的数据结构提取更简单的子问题，例如删除不必要的unpivot逻辑并只保存一个小时的列。否则答案会占用太多时间在这种情况下，我的数据结构很重要。您好！谢谢你的帮助。问题是如何获取日期、ID、姓名、小时、值，因为这是表格的结构。它还需要每小时处理一次。哦，你刚刚修改了你的问题。这是同一个问题，因子列应该是十进制类型。所以我更新了。否则，问题中的所有其他内容都是相同的。