Question

我有这样的表

Id  Student     Test_Start          Test_End
1   A       2013-01-08 15:00:17.100     2013-01-08 16:05:55.100 
2   B       2013-02-08 17:01:11.800     2013-02-08 19:05:15.880
3   C       2013-02-08 14:41:10.880     2013-02-08 16:49:22.880

我要做的是获取Test_End - Test_Start时间并将其更改为小时并按升序显示。我试过了：

Select convert(date, Test_End) - convert(date,Test_Start) as 'Time_Taken' from students

或尝试使用DATEDIFF但似乎无法使其正常工作。是否有最好的方法来实现这一目标：

Time_Taken
1
2
2

Answer 1

select 
    id,
    student,
    datediff(hour, test_start, test_end) as time_taken
from students
order by time_taken

如果你想要四舍五入，确定分钟数，加上四舍五入到下一个小时所需的分钟数，再除以60得到舍入小时......

select 
    id,
    student,
    (datediff(minute, test_start, test_end) + 30) / 60 as rounded_time_taken
from students
order by rounded_time_taken

Answer 2

你真的想要整数小时吗？在大多数情况下，我宁愿将结果视为小时数：

select id, student, datediff(s, test_start, test_end)/(60.0*60.0) as hours_taken
from students
order by test_start, test_end)/(60.0*60.0)

Sql查询以小时为单位减去日期和顺序

2 个答案: