Question

在SQL Server中，我试图计算自过去5天以来首次观察到与今天相同的天气（今天假设为2018年8月6日）的天数。每个城镇。

以下是数据：

+---------+---------+--------+--------+--------+
| Date    | Toronto | Cairo  | Zagreb | Ankara |
+---------+---------+--------+--------+--------+
| 1.08.18 | Rain    | Sun    | Clouds | Sun    |
| 2.08.18 | Sun     | Sun    | Clouds | Sun    |
| 3.08.18 | Rain    | Sun    | Clouds | Rain   |
| 4.08.18 | Clouds  | Sun    | Clouds | Clouds |
| 5.08.18 | Rain    | Clouds | Rain   | Rain   |
| 6.08.18 | Rain    | Sun    | Sun    | Sun    |
+---------+---------+--------+--------+--------+

这需要表现良好，但到目前为止，我只想针对每个镇进行单个查询（并且将会有数十个镇，而不仅仅是四个镇）。这行得通，但不会扩展。

这是多伦多的那个...

SELECT 
    DATEDIFF(DAY, MIN([Date]), GETDATE()) + 1 
FROM
    (SELECT TOP 5 * 
     FROM Weather 
     WHERE [Date] <= GETDATE()
     ORDER BY [Date] DESC) a
WHERE 
    Toronto = (SELECT TOP 1 Toronto 
               FROM Weather
               WHERE DataDate = GETDATE())

...由于今天有雨，并且在过去5天内第一次下雨是8月3日，因此正确返回4。

但是我想要返回的是一个像这样的表：

+---------+-------+--------+--------+
| Toronto | Cairo | Zagreb | Ankara |
+---------+-------+--------+--------+
| 4       | 5     | 1      | 5      |
+---------+-------+--------+--------+

这怎么可能？

Answer 1

您真的不想尝试对数据透视表执行此操作，尽管您声明数据不是以这种方式存储的，但您没有向我们展示如何以列为中心到达城市的透视图-真可惜。

因此，我已在一个公用表表达式中“取消透视”该样本，然后使用apply operator来计算前5天相同天气的先前发生次数。看来您知道如何旋转，我留给您去旋转最终结果。

with cte as (
        select
              date, city, weather
        FROM (
              SELECT * from mytable
             ) AS cp
        UNPIVOT (
                  Weather FOR City IN (Toronto, Cairo, Zagreb, Ankara)
            ) AS up
    )

select 
        date, city, weather, ca.prior
from cte
cross apply (
    select count(*) as prior
    from cte as prev 
    where prev.city = cte.city
    and prev.date between dateadd(day,-6,cte.date) and dateadd(day,-1,cte.date)
    and prev.weather = cte.weather
    ) ca

使用此示例数据：

CREATE TABLE mytable(
   Date    date  NOT NULL
  ,Toronto VARCHAR(9) NOT NULL
  ,Cairo   VARCHAR(9) NOT NULL
  ,Zagreb  VARCHAR(9) NOT NULL
  ,Ankara  VARCHAR(9) NOT NULL
);
INSERT INTO mytable(Date,Toronto,Cairo,Zagreb,Ankara) VALUES ('20180801','Rain','Sun','Clouds','Sun');
INSERT INTO mytable(Date,Toronto,Cairo,Zagreb,Ankara) VALUES ('20180802','Sun','Sun','Clouds','Sun');
INSERT INTO mytable(Date,Toronto,Cairo,Zagreb,Ankara) VALUES ('20180803','Rain','Sun','Clouds','Rain');
INSERT INTO mytable(Date,Toronto,Cairo,Zagreb,Ankara) VALUES ('20180804','Clouds','Sun','Clouds','Clouds');
INSERT INTO mytable(Date,Toronto,Cairo,Zagreb,Ankara) VALUES ('20180805','Rain','Clouds','Rain','Rain');
INSERT INTO mytable(Date,Toronto,Cairo,Zagreb,Ankara) VALUES ('20180806','Rain','Sun','Sun','Sun');

上面的查询产生了以下结果：

+----+---------------------+---------+---------+-------+
|    |        date         |  city   | weather | prior |
+----+---------------------+---------+---------+-------+
|  1 | 01.08.2018 00:00:00 | Ankara  | Sun     |     0 |
|  2 | 02.08.2018 00:00:00 | Ankara  | Sun     |     1 |
|  3 | 03.08.2018 00:00:00 | Ankara  | Rain    |     0 |
|  4 | 04.08.2018 00:00:00 | Ankara  | Clouds  |     0 |
|  5 | 05.08.2018 00:00:00 | Ankara  | Rain    |     1 |
|  6 | 06.08.2018 00:00:00 | Ankara  | Sun     |     2 |
|  7 | 01.08.2018 00:00:00 | Cairo   | Sun     |     0 |
|  8 | 02.08.2018 00:00:00 | Cairo   | Sun     |     1 |
|  9 | 03.08.2018 00:00:00 | Cairo   | Sun     |     2 |
| 10 | 04.08.2018 00:00:00 | Cairo   | Sun     |     3 |
| 11 | 05.08.2018 00:00:00 | Cairo   | Clouds  |     0 |
| 12 | 06.08.2018 00:00:00 | Cairo   | Sun     |     4 |
| 13 | 01.08.2018 00:00:00 | Toronto | Rain    |     0 |
| 14 | 02.08.2018 00:00:00 | Toronto | Sun     |     0 |
| 15 | 03.08.2018 00:00:00 | Toronto | Rain    |     1 |
| 16 | 04.08.2018 00:00:00 | Toronto | Clouds  |     0 |
| 17 | 05.08.2018 00:00:00 | Toronto | Rain    |     2 |
| 18 | 06.08.2018 00:00:00 | Toronto | Rain    |     3 |
| 19 | 01.08.2018 00:00:00 | Zagreb  | Clouds  |     0 |
| 20 | 02.08.2018 00:00:00 | Zagreb  | Clouds  |     1 |
| 21 | 03.08.2018 00:00:00 | Zagreb  | Clouds  |     2 |
| 22 | 04.08.2018 00:00:00 | Zagreb  | Clouds  |     3 |
| 23 | 05.08.2018 00:00:00 | Zagreb  | Rain    |     0 |
| 24 | 06.08.2018 00:00:00 | Zagreb  | Sun     |     0 |
+----+---------------------+---------+---------+-------+

自首次发生以来（过去5天内）的天数

select 
        date, city, weather, datediff(day,ca.prior,cte.date) as prior
from cte
cross apply (
    select min(prev.date) as prior
    from cte as prev 
    where prev.city = cte.city
    and prev.date between dateadd(day,-6,cte.date) and dateadd(day,-1,cte.date)
    and prev.weather = cte.weather
    ) ca

Answer 2

我想你想要类似的东西

CREATE TABLE T
(
  [Date] DATE,
  Toronto VARCHAR(45),
  Cairo VARCHAR(45),
  Zagreb VARCHAR(45),
  Ankara VARCHAR(45)
);

INSERT INTO T VALUES
('2018-08-01', 'Rain', 'Sun', 'Clouds', 'Sun'),
('2018-08-02', 'Sun', 'Sun', 'Clouds', 'Sun'),
('2018-08-03', 'Rain', 'Sun', 'Clouds', 'Rain'),
('2018-08-04', 'Clouds', 'Sun', 'Clouds', 'Clouds'),
('2018-08-05', 'Rain', 'Clouds', 'Rain', 'Rain'),
('2018-08-06', 'Rain', 'Sun', 'Sun', 'Sun');

SELECT 
  (SELECT MAX(Occ) FROM (SELECT COUNT(Toronto) Occ FROM T WHERE Toronto = (select top 1 toronto from t order by date desc) GROUP BY Toronto) T) Toronto,
  (SELECT MAX(Occ) FROM (SELECT COUNT(Cairo) Occ FROM T WHERE Cairo = (select top 1 Cairo from t order by date desc) GROUP BY Cairo) T) Cairo,
  (SELECT MAX(Occ) FROM (SELECT COUNT(Zagreb) Occ FROM T WHERE Zagreb = (select top 1 Zagreb from t order by date desc)GROUP BY Zagreb) T) Zagreb,
  (SELECT MAX(Occ) FROM (SELECT COUNT(Ankara) Occ FROM T WHERE Ankara = (select top 1 Ankara from t order by date desc)GROUP BY Ankara) T) Ankara

返回

+----+---------+-------+--------+--------+
|    | Toronto | Cairo | Zagreb | Ankara |
+----+---------+-------+--------+--------+
|  1 |       4 |     5 |      1 |      3 |
+----+---------+-------+--------+--------+

Demo

SQL：自首次出现某些值以来的行数

2 个答案: