SQL时间查询从晚上6点到早上6点

Question

如何统计晚上6点和早上6点之后的记录数？

这将持续到下午 12 点。

SELECT distinct count(barcode) c 
FROM table1 
where DAY(timestamp) = DAY(GETDATE())
  AND MONTH(timestamp) = MONTH(GETDATE())
  AND YEAR(timestamp) = YEAR(GETDATE())
  AND datepart(hh,timestamp) >= 18 
  AND datepart(hh,timestamp) >= 6;

Answer 1

我想这可能就是您想要的。它从 GETDATE() 调用中删除任何秒数，并为其添加适当的小时数。

SELECT COUNT(barcode)
FROM table1
WHERE timestamp >= DATEADD(HOUR,18,CONVERT(VARCHAR(10), GETDATE(),110))
  AND timestamp <= DATEADD(HOUR,6,CONVERT(VARCHAR(10), GETDATE()+1,110))

SQL Fiddle Demo

Answer 2

这个怎么样？

其中 datefunction(timestamp)>=getdate()-1 和 (datepart(hh,timestamp)>=18 或 datepart(hh,timestamp)

datefunction 将日期时间转换为日期

Answer 3

如果您想按天计算事物，但要让一天从下午 6 点开始。而不是午夜，只需在时间上添加一个偏移量：

select cast(timestamp + 0.25 as date) as theday, count(barcode)
from table1
group by cast(timestamp + 0.25 as date)
order by theday desc;

如果您想在下午 6 点进行计数。 - 早上 6 点多天：

select cast(timestamp + 0.25 as date) as theday, count(barcode)
from table1
where datepart(hh, timestamp) in (18, 19, 20, 21, 22, 23, 0, 1, 2, 3, 4, 5)
group by cast(timestamp + 0.25 as date)
order by theday desc;

最近一天，您可以：

select top 1 cast(timestamp + 0.25 as date) as theday, count(barcode)
from table1
where datepart(hh, timestamp) in (18, 19, 20, 21, 22, 23, 0, 1, 2, 3, 4, 5)
group by cast(timestamp + 0.25 as date)
order by theday desc;