Question

我的数据格式为

+----------+-------------+
|Customerid|DateofJoining|
+----------+-------------+
|1         |aug 1        |
+----------+-------------+
|2         |aug 1        |
+----------+-------------+
|3         |aug 10       |
+----------+-------------+
|4         |aug 10       |
+----------+-------------+
|5         |aug 10       |
+----------+-------------+
|6         |sept 5       |
+----------+-------------+

所以我想展示结果作为时间进展的客户数量

+--------------+------+
|TotalCustomers|Date  |
+--------------+------+
|2             |Aug 1 |
+--------------+------+
|5             |Aug 10|
+--------------+------+
|6             |Sept 5|
+--------------+------+

我该怎么做。

通过分组，我可以在任何给定日期显示总客户，但不会在此之前显示总和。我该怎么做？

我试过了：

SELECT COUNT(numcustomersjoined), dateofjoining
FROM foo
GROUP BY dateofjoining

但这对我不起作用。

Answer 1

您没有包含sql-server的版本。这适用于sqlserver-2012：

create table foo(Customerid int, DateofJoining date)
insert into foo values 
(1,'2014-08-01'),(2,'2014-08-01'),(3,'2014-08-10'),
(4,'2014-08-10'),(5,'2014-08-10'),(6,'2014-09-05')

SELECT
DateofJoining,
SUM(COUNT(*)) OVER (ORDER BY dateofjoining) CumulativeTotal
FROM foo
GROUP BY DateofJoining

结果：

DateofJoining  CumulativeTotal
2014-08-01     2
2014-08-10     5
2014-09-05     6

Answer 2

您的输出显示您希望在特定日期加入最高customerId。

为此，您可以使用max功能。

select max(Customerid), dateofjoining from foo group by dateofjoining

Answer 3

SELECT MAX(Customerid), dateofjoining FROM foo GROUP BY dateofjoining;

试试这个......

Answer 4

您正在寻找 running total 。让我们来讨论this scenario:

create table t( Customerid int, DateofJoining date );
insert into t values 
(1 ,               '2014-08-01' ),
(12,               '2014-08-01' ),
(53,               '2014-08-10' ),
(14,               '2014-08-10' ),
(5 ,               '2014-08-10' ),
(96,               '2014-09-05' )

然后你可以使用这个非等值的运行总计计数：

;with 
 slots as
(select
 distinct DateofJoining
 from t
)
select
 count(*) as n,
 slots.DateofJoining
from 
 t inner join slots
 on t.DateofJoining <= slots.DateofJoining  --non equijoin
group by
 slots.DateofJoining

<强> Results ：

| N | DATEOFJOINING |
|---|---------------|
| 2 |    2014-08-01 |
| 5 |    2014-08-10 |
| 6 |    2014-09-05 |

按日期间隔分组I.

4 个答案: