我有一张表格如下
order no orderdate
01 2012-08-01 00:00
02 2012-08-01 00:10
03 2012-08-01 00:15
04 2012-08-01 00:30
05 2012-08-01 10:00
06 2012-08-01 10:35
07 2012-08-01 14:00
08 2012-08-02 20:30
09 2012-08-02 20:35
10 2012-08-02 23:00
我想按时间间隔对这些值进行分组（如00 -01,01-02 ...... 22- 23,23-24）（不包括日期值）
如何编写查询以按时间对此进行分组？
我希望获得24行（24个间隔）的订单数量
答案 0 :(得分：1)
您可以使用 GROUP BY DATEPART(HOUR, orderdate) 。
查询：
SELECT DATEPART(HOUR, orderdate) AS HOUR,
COUNT(*) AS ORDERS_PER_HOUR
FROM YourTable
WHERE CAST(orderdate AS DATE) = @someDate
GROUP BY DATEPART(HOUR, orderdate)
请注意，如果orderdate中不存在每小时，则上述查询的结果将少于24个。
如果需要表示所有小时数，那么您可以使用递归公用表表达式（CTE）来首先生成所有小时数（0-23），并且然后离开你的桌子与CTE：
;WITH hours AS
(
SELECT 0 AS h
UNION ALL
SELECT h + 1 FROM hours WHERE h < 23
)
SELECT hours.h AS HOUR,
COUNT(YourTable.[order no]) AS ORDERS_PER_HOUR
FROM hours
LEFT JOIN YourTable ON DATEPART(HOUR, orderdate) = hours.h
WHERE CAST(orderdate AS DATE) = @someDate
GROUP BY hours.h
答案 1 :(得分：0)
这是一个简单的解决方案，在这里您可以了解当天的小时......
drop table if exists dbo.tOrders;
create table dbo.tOrders (
OrderNo varchar(50)
, OrderDate datetime
);
insert into dbo.tOrders (OrderNo, OrderDate)
values ('01', '2012-08-01 00:00')
, ('02', '2012-08-01 00:10')
, ('03', '2012-08-01 00:15')
, ('04', '2012-08-01 00:30')
, ('05', '2012-08-01 10:00')
, ('06', '2012-08-01 10:35')
, ('07', '2012-08-01 14:00')
, ('08', '2012-08-02 20:30')
, ('09', '2012-08-02 20:35')
, ('10', '2012-08-02 23:00')
select
case
when datepart(hour, t.OrderDate) >= 0 and datepart(hour, t.OrderDate) < 1 then '00-01'
when datepart(hour, t.OrderDate) >= 1 and datepart(hour, t.OrderDate) < 2 then '01-02'
when datepart(hour, t.OrderDate) >= 2 and datepart(hour, t.OrderDate) < 3 then '02-03'
when datepart(hour, t.OrderDate) >= 3 and datepart(hour, t.OrderDate) < 4 then '03-04'
when datepart(hour, t.OrderDate) >= 4 and datepart(hour, t.OrderDate) < 5 then '04-05'
when datepart(hour, t.OrderDate) >= 5 and datepart(hour, t.OrderDate) < 6 then '05-06'
when datepart(hour, t.OrderDate) >= 6 and datepart(hour, t.OrderDate) < 7 then '06-07'
when datepart(hour, t.OrderDate) >= 7 and datepart(hour, t.OrderDate) < 8 then '07-08'
when datepart(hour, t.OrderDate) >= 8 and datepart(hour, t.OrderDate) < 9 then '08-09'
when datepart(hour, t.OrderDate) >= 9 and datepart(hour, t.OrderDate) < 10 then '09-10'
when datepart(hour, t.OrderDate) >= 10 and datepart(hour, t.OrderDate) < 11 then '10-11'
when datepart(hour, t.OrderDate) >= 11 and datepart(hour, t.OrderDate) < 12 then '11-12'
when datepart(hour, t.OrderDate) >= 12 and datepart(hour, t.OrderDate) < 13 then '12-13'
when datepart(hour, t.OrderDate) >= 13 and datepart(hour, t.OrderDate) < 14 then '13-14'
when datepart(hour, t.OrderDate) >= 14 and datepart(hour, t.OrderDate) < 15 then '14-15'
when datepart(hour, t.OrderDate) >= 15 and datepart(hour, t.OrderDate) < 16 then '15-16'
when datepart(hour, t.OrderDate) >= 16 and datepart(hour, t.OrderDate) < 17 then '16-17'
when datepart(hour, t.OrderDate) >= 17 and datepart(hour, t.OrderDate) < 18 then '17-18'
when datepart(hour, t.OrderDate) >= 18 and datepart(hour, t.OrderDate) < 19 then '18-19'
when datepart(hour, t.OrderDate) >= 19 and datepart(hour, t.OrderDate) < 20 then '19-20'
when datepart(hour, t.OrderDate) >= 20 and datepart(hour, t.OrderDate) < 21 then '20-21'
when datepart(hour, t.OrderDate) >= 21 and datepart(hour, t.OrderDate) < 22 then '21-22'
when datepart(hour, t.OrderDate) >= 22 and datepart(hour, t.OrderDate) < 23 then '22-23'
when datepart(hour, t.OrderDate) >= 23 then '23-00'
end as TimeOfDay
, count(t.OrderNo) as OrderCount
from dbo.tOrders t
group by
case
when datepart(hour, t.OrderDate) >= 0 and datepart(hour, t.OrderDate) < 1 then '00-01'
when datepart(hour, t.OrderDate) >= 1 and datepart(hour, t.OrderDate) < 2 then '01-02'
when datepart(hour, t.OrderDate) >= 2 and datepart(hour, t.OrderDate) < 3 then '02-03'
when datepart(hour, t.OrderDate) >= 3 and datepart(hour, t.OrderDate) < 4 then '03-04'
when datepart(hour, t.OrderDate) >= 4 and datepart(hour, t.OrderDate) < 5 then '04-05'
when datepart(hour, t.OrderDate) >= 5 and datepart(hour, t.OrderDate) < 6 then '05-06'
when datepart(hour, t.OrderDate) >= 6 and datepart(hour, t.OrderDate) < 7 then '06-07'
when datepart(hour, t.OrderDate) >= 7 and datepart(hour, t.OrderDate) < 8 then '07-08'
when datepart(hour, t.OrderDate) >= 8 and datepart(hour, t.OrderDate) < 9 then '08-09'
when datepart(hour, t.OrderDate) >= 9 and datepart(hour, t.OrderDate) < 10 then '09-10'
when datepart(hour, t.OrderDate) >= 10 and datepart(hour, t.OrderDate) < 11 then '10-11'
when datepart(hour, t.OrderDate) >= 11 and datepart(hour, t.OrderDate) < 12 then '11-12'
when datepart(hour, t.OrderDate) >= 12 and datepart(hour, t.OrderDate) < 13 then '12-13'
when datepart(hour, t.OrderDate) >= 13 and datepart(hour, t.OrderDate) < 14 then '13-14'
when datepart(hour, t.OrderDate) >= 14 and datepart(hour, t.OrderDate) < 15 then '14-15'
when datepart(hour, t.OrderDate) >= 15 and datepart(hour, t.OrderDate) < 16 then '15-16'
when datepart(hour, t.OrderDate) >= 16 and datepart(hour, t.OrderDate) < 17 then '16-17'
when datepart(hour, t.OrderDate) >= 17 and datepart(hour, t.OrderDate) < 18 then '17-18'
when datepart(hour, t.OrderDate) >= 18 and datepart(hour, t.OrderDate) < 19 then '18-19'
when datepart(hour, t.OrderDate) >= 19 and datepart(hour, t.OrderDate) < 20 then '19-20'
when datepart(hour, t.OrderDate) >= 20 and datepart(hour, t.OrderDate) < 21 then '20-21'
when datepart(hour, t.OrderDate) >= 21 and datepart(hour, t.OrderDate) < 22 then '21-22'
when datepart(hour, t.OrderDate) >= 22 and datepart(hour, t.OrderDate) < 23 then '22-23'
when datepart(hour, t.OrderDate) >= 23 then '23-00'
end
答案 2 :(得分：0)
我认为您可以使用此查询。正如您所看到的，要使用的主函数是wifi（参见https://docs.microsoft.com/en-us/sql/t-sql/functions/datepart-transact-sql），其中＆＃34;返回一个整数，表示指定日期的指定日期部分＆＃34;。
然后你可以使用GROUP BY来记录每小时的记录。
最后，您可以安排＆＃34;输出格式＆＃34;。
this.http.get(...)
.retryWhen(error => error.delay(2000)) // if an error happens, wait 2 secs and try again
.timeout(6000) // no success after 3 retries (6 secs), throw a timeout-error
.map((res: Response) => res.json()) // everything went fine
.catch(yourErrorFunction()) // do something with the error
.subscribe()
或者如果您愿意：
DATE_PART(<format>, <date column>)示例数据：
SELECT CAST(X AS VARCHAR(2)) +'-'+CAST(X+1 AS VARCHAR(2)) AS HOUR_GROUP, COUNT(*) AS RC
FROM
(SELECT DATEPART(hh,ORD_DATE) X FROM ORD)A
GROUP BY X
;
输出
SELECT CAST(DATEPART(hh,ORD_DATE) AS VARCHAR(2)) +'-'+CAST(DATEPART(hh,ORD_DATE) +1 AS VARCHAR(2)) AS HOUR_GROUP, COUNT(*) AS RC
FROM ORD
GROUP BY DATEPART(hh,ORD_DATE)
;
如果您希望所有24小时组都可以使用：
INSERT INTO ORD VALUES (1,'2012-01-08 00:00');
INSERT INTO ORD VALUES (2,'2012-01-08 00:10');
INSERT INTO ORD VALUES (3,'2012-01-08 00:15');
INSERT INTO ORD VALUES (4,'2012-01-08 00:30');
INSERT INTO ORD VALUES (9,'2012-01-08 01:30');
INSERT INTO ORD VALUES (5,'2012-01-08 10:00');
INSERT INTO ORD VALUES(7,'2012-01-08 15:00');
INSERT INTO ORD VALUES (8,'2012-02-08 20:30');
输出：
HOUR_GROUP RC
0-1 4
1-2 1
10-11 1
15-16 1
20-21 1
答案 3 :(得分：0)
如果我理解正确的话，无论日期如何，您都在寻找小时计数，即您正在寻找一天中最忙碌的时间？在这种情况下，您可以使用以下内容：
declare @hours table
(
hourpart datetime
)
INSERT INTO @hours VALUES
('00:00:00'),
('01:00:00'),
('02:00:00'),
('03:00:00'),
('04:00:00'),
('05:00:00'),
('06:00:00'),
('07:00:00'),
('08:00:00'),
('09:00:00'),
('10:00:00'),
('11:00:00'),
('12:00:00'),
('13:00:00'),
('14:00:00'),
('15:00:00'),
('16:00:00'),
('17:00:00'),
('18:00:00'),
('19:00:00'),
('20:00:00'),
('21:00:00'),
('22:00:00'),
('23:00:00')
declare @orders table
(
orderNo int,
orderDate datetime)
INSERT INTO @orders VALUES
(01,'2012-08-01 00:00'),
(02,'2012-08-01 00:10'),
(03,'2012-08-01 00:15'),
(04,'2012-08-01 00:30'),
(05,'2012-08-01 10:00'),
(06,'2012-08-01 10:35'),
(07,'2012-08-01 14:00'),
(08,'2012-08-02 20:30'),
(09,'2012-08-02 20:35'),
(10,'2012-08-02 23:00'),
(11,'2012-08-01 20:25')
SELECT convert(varchar(8), h.hourpart, 108), COUNT(*) FROM @orders o
INNER JOIN @hours h
ON DATEPART(hour, o.orderdate) = DATEPART(hour, h.hourpart)
GROUP BY h.hourpart
ORDER BY h.hourpart
结果：
00:00:00 4
10:00:00 2
14:00:00 1
20:00:00 3
23:00:00 1