Question

我有一个数据，可以让用户了解用户何时订阅服务，在城市中何时到期。

如下所示

+------+------------+------------+
| City | Start_Date |  End_Date  |
+------+------------+------------+
| LA   | 2019-06-01 | 2019-06-03 |
| LA   | 2019-06-07 | 2019-06-10 |
| LA   | 2019-06-09 | 2019-06-11 |
| LA   | 2019-06-13 | 2019-06-14 |
| LO   | 2019-06-01 | 2019-06-05 |
| LO   | 2019-06-04 | 2019-06-05 |
| LO   | 2019-06-07 | 2019-06-09 |
| LO   | 2019-06-08 | 2019-06-09 |
+------+------------+------------+

我想要在给定日期的每个城市每天的活跃用户数

输出应与

相似

+------+------------+-------+
| City |    Day     | Count |
+------+------------+-------+
| LA   | 2019-06-01 |     1 |
| LA   | 2019-06-02 |     1 |
| LA   | 2019-06-03 |     1 |
| LA   | 2019-06-04 |     0 |
| LA   | 2019-06-05 |     0 |
| LA   | 2019-06-06 |     0 |
| LA   | 2019-06-07 |     1 |
| LA   | 2019-06-08 |     1 |
| LA   | 2019-06-09 |     2 |
| LA   | 2019-06-10 |     2 |
| LA   | 2019-06-11 |     1 |
| LA   | 2019-06-12 |     0 |
| LA   | 2019-06-13 |     1 |
| LA   | 2019-06-14 |     1 |
| LA   | 2019-06-15 |     0 |
+------+------------+-------+

我显示的输出仅适用于城市LA，并且我还希望表中每个城市的输出类似。

我的输出说明

在City = LA上的Day = 2019-06-01中，有1个用户具有有效的订阅，与Day = 2019-06-02相似。
在City = LA的{{1}}中，有2个用户具有有效的订阅，依此类推。

任何帮助将不胜感激

SQL Fiddle Link

Answer 1

您需要一些日期。数字或理货单非常方便。但是递归CTE也很有帮助。

然后使用CROSS JOIN分配行，使用LEFT JOIN引入所需的值，然后使用GROUP BY获取计数：

with dates as (
      select convert(date, '2019-06-01') as dte
      union all
      select dateadd(day, 1, dte) as dte
      from dates
      where dte < '2019-06-15'
     )
select c.city, d.dte, count(t.city)
from (select distinct city from t) c cross join
     dates d left join
     t
     on t.city = c.city and t.start_date <= d.dte and d.end_date >= t.dte 
group by c.city, d.dte
order by c.city, d.dte;

Here是db <>小提琴-使用SQL Server。

Answer 2

就像这样应该起作用：

select city, '2019-06-11', count(*)
from table1
where end_date >= '2019-06-11'
group by 1

您应将日期替换为变量，然后放入所需的日期。

Answer 3

首先将其设为有用的提示表视图

/****** Object:  View [dbo].[cteTally]    Script Date: 10/06/2019 11:02:06 ******/
SET ANSI_NULLS ON
GO

SET QUOTED_IDENTIFIER ON
GO


CREATE View [dbo].[cteTally] as

WITH
    E1(N) AS (select 1 from (values (1),(1),(1),(1),(1),(1),(1),(1),(1),(1))dt(n)),
    E2(N) AS (SELECT 1 FROM E1 a, E1 b), --10E+2 or 100 rows
    E4(N) AS (SELECT 1 FROM E2 a, E2 b), --10E+4 or 10,000 rows max

    cteTally(N) AS 
    (
        SELECT  ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E4
    )
select N from cteTally


GO

然后，您可以像这样生成您要查询的日期列表（下面的一个查询是演示，而不是完成的编码）

Declare @Start as date = '20170101';
Declare @End as date = '20190601';

SELECT dateadd(day, N-1,@Start) aDate from [dbo].[cteTally] WHERE  dateadd(day, N-1,@Start) <= @End;

现在，我们可以将其扩展为对日期范围匹配的记录进行计数

SELECT Y.City, DQ.aDate, Count(Y.City) as DayCountForCity FROM (SELECT dateadd(day, N-1,@Start) aDate from [dbo].[cteTally] WHERE  dateadd(day, N-1,@Start) <= @End) DQ
        LEFT JOIN YourTable Y ON DQ.adate BETWEEN Y.Start_Date AND Y.End_Date   
        GRoup By Y.City, DQ.adate

尝试使用给定的测试表

Declare @Start as date = '20170101';
Declare @End as date = '20450601';

SELECT * FROM (SELECT Y.City, DQ.aDate, Count(Y.City) as DayCountForCity FROM (SELECT dateadd(day, N-1,@Start) aDate from [dbo].[cteTally] WHERE  dateadd(day, N-1,@Start) <= @End) DQ
        LEFT JOIN Table1 Y ON DQ.adate BETWEEN Y.Start_Date AND Y.End_Date  
        GRoup By Y.City, DQ.adate) Q1 WHERE Q1.city is not null
        order by city, adate

Answer 4

请尝试以下查询。测试数据：

DECLARE @tbl TABLE (City varchar(5),Start_Date date,End_Date date);
INSERT INTO @tbl values
( 'LA','2019-06-01','2019-06-03' ),
( 'LA','2019-06-07','2019-06-10' ),
( 'LA','2019-06-09','2019-06-11' ),
( 'LA','2019-06-13','2019-06-14' ),
( 'LO','2019-06-01','2019-06-05' ),
( 'LO','2019-06-04','2019-06-05' ),
( 'LO','2019-06-07','2019-06-09' ),
( 'LO','2019-06-08','2019-06-09' );

以CTE作为日历功能的实际查询：

DECLARE @start DATE, @end DATE;
SELECT @start = MIN(Start_Date), @end = MAX(End_Date) FROM @tbl;

;WITH cte AS (
    SELECT @start dt
    UNION ALL
    SELECT DATEADD(day, 1, dt) FROM CTE
    WHERE dt < @end
)

SELECT cte.dt, SUM(CASE WHEN t.City IS NULL THEN 0 ELSE 1 END) FROM cte
LEFT JOIN @tbl t 
ON cte.dt BETWEEN t.Start_Date AND t.End_Date AND t.City = 'LA'
GROUP BY cte.dt

Answer 5

您也可以这样做

DECLARE @MN DATETIME = (SELECT MIN(Start_Date) FROM T),
        @MX DATETIME = (SELECT MAX(End_Date) FROM T);

;WITH CTE AS
(
  SELECT @MN D
  UNION ALL
  SELECT DATEADD(Day, 1, D)
  FROM CTE
  WHERE D <= @MX
)
SELECT TT.City,
       CTE.D [Day],
       (
        SELECT COUNT(1) 
        FROM T T1 
        WHERE T1.City = TT.City 
              AND CTE.D BETWEEN T1.Start_Date AND T1.End_Date
       ) Cnt
FROM CTE CROSS JOIN (VALUES('LA')) TT(City);

返回：

+------+---------------------+-----+
| City |         Day         | Cnt |
+------+---------------------+-----+
| LA   | 01/06/2019 00:00:00 |   1 |
| LA   | 02/06/2019 00:00:00 |   1 |
| LA   | 03/06/2019 00:00:00 |   1 |
| LA   | 04/06/2019 00:00:00 |   0 |
| LA   | 05/06/2019 00:00:00 |   0 |
| LA   | 06/06/2019 00:00:00 |   0 |
| LA   | 07/06/2019 00:00:00 |   1 |
| LA   | 08/06/2019 00:00:00 |   1 |
| LA   | 09/06/2019 00:00:00 |   2 |
| LA   | 10/06/2019 00:00:00 |   2 |
| LA   | 11/06/2019 00:00:00 |   1 |
| LA   | 12/06/2019 00:00:00 |   0 |
| LA   | 13/06/2019 00:00:00 |   1 |
| LA   | 14/06/2019 00:00:00 |   1 |
| LA   | 15/06/2019 00:00:00 |   0 |
+------+---------------------+-----+

Live Demo

Answer 6

SELECT date,city,count(id) as total FROM table WHERE STR_TO_DATE("2019-06-01","%Y-%m-%d") = STR_TO_DATE("2019-06-01","%Y-%m-%d") GROUP BY city

获取每天的活跃用户数

6 个答案: