Question

基本上我需要一个sql查询，它将返回1列的不同值，然后修剪（DateTime）结果以仅显示日期。而我的其他列应该包含当天的项目数。

我尝试了以下内容：

 Select LoggedDate as Date, count(ID) as Amount from Tickets WHERE LoggedDate >=dateadd(day,datediff(day,0,GetDate())- 30,0) AND State = '1' group by LoggedDate

ID   |  Date                   | Amount
---------------------------------------- 
1    | 2016-09-23 11:54:12.000 | 1 
2    | 2016-09-29 09:23:58.000 | 1 
3    | 2016-09-29 09:34:07.000 | 1 
4    | 2016-09-29 09:41:12.000 | 1 
5    | 2016-09-29 09:53:37.000 | 1 
6    | 2016-09-29 09:57:17.000 | 1 
7    | 2016-09-29 10:01:53.000 | 1 
8    | 2016-10-03 21:10:27.000 | 1 
9    | 2016-10-04 03:20:51.000 | 1 
10   | 2016-10-04 03:47:34.000 | 1 
11   | 2016-10-04 03:50:09.000 | 1

我尝试过类似的事情：

Select DISTINCT(LEFT(LoggedDate, 11)) AS Date, count(ID) as Amount from Tickets WHERE LoggedDate >=dateadd(day,datediff(day,0,GetDate())- 30,0) AND State = '1' group by LoggedDate

ID   |  Date        | Amount
----------------------------
1    | OCT 3  2016  |1
2    | OCT 2  2016  |1
3    | SEP 22 2016  |1
4    | SEP 23 2016  |1
5    | SEP 29 2016  |1

我想要下面但不确定如何获取我的数据：

ID   |  Date       | Amount
---------------------------
1    | 2016-09-23  |1
2    | 2016-09-29  |5
3    | 2016-10-03  |1
4    | 2016-10-03  |1
5    | 2016-10-04  |2

Answer 1

如果我做对了，你可以试试这个：

SELECT ROW_NUMBER() Over(ORDER BY Date) [Id], *
FROM 
(
    SELECT CONVERT(VARCHAR(35), LoggedDate, 106) [Date], SUM(Amount) [Total]
    FROM Tickets
    GROUP BY CONVERT(VARCHAR(35), LoggedDate, 106)
) a

您可以通过更改CONVERT函数中的第三个参数来试验日期格式。

您还需要提出任何条件等。

Answer 2

我认为你需要这样的东西：

SELECT DATE(LoggedDate) AS date, COUNT(id) AS Amount 
FROM Tickets 
WHERE state = 1 
GROUP BY DATE(LoggedDate);

Answer 3

尝试使用以下脚本。

SELECT ROW_NUMBER() OVER ( ORDER BY (select 1) ) AS id
       ,CAST(LoggedDate as date) AS Date
       , count(ID) as Amount 
FROM Tickets 
WHERE LoggedDate >=dateadd(day,datediff(day,0,GetDate())- 30,0) AND State = '1' 
GROUP BY  CAST(LoggedDate as date)

Answer 4

我只是猜测你的表结构，但这是我放在一起的示例代码：

DECLARE @t TABLE ( id INT, loggedDate DATETIME ) 

INSERT INTO @t ( id, loggedDate ) VALUES ( 1, '2016-09-23 11:54:12.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 2, '2016-09-29 09:23:58.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 3, '2016-09-29 09:34:07.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 4, '2016-09-29 09:41:12.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 5, '2016-09-29 09:53:37.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 6, '2016-09-29 09:57:17.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 7, '2016-09-29 10:01:53.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 8, '2016-10-03 21:10:27.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 9, '2016-10-04 03:20:51.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 10, '2016-10-04 03:47:34.000' ) 
INSERT INTO @t ( id, loggedDate ) VALUES ( 11, '2016-10-04 03:50:09.000' ) 




SELECT ROW_NUMBER() OVER ( ORDER BY CONVERT(DATE,loggedDate) ) AS id, CONVERT(DATE, loggedDate) AS loggedDate, COUNT(1) AS amount
FROM @t
GROUP BY CONVERT(DATE,loggedDate)

/* output 
id  loggedDate  amount
1   2016-09-23  1
2   2016-09-29  6
3   2016-10-03  1
4   2016-10-04  3
*/

sql获取不同并修剪一列，同时计算另一列中的值的数量

4 个答案: