如何在SQL Server中比较时间？(How can I compare time in SQL Server?)

如何在SQL Server中比较时间？(How can I compare time in SQL Server?)

我试图在SQL查询的datetime字段中比较时间，但我不知道它是对的。 我不想比较日期部分，只是时间部分。

我在做这个：

SELECT timeEvent FROM tbEvents WHERE convert(datetime, startHour, 8) >= convert(datetime, @startHour, 8)

这是对的吗？

我问这个是因为我需要知道'08：00：00'是否小于或大于'07：30：00'，我不想比较日期，只是时间部分。

谢谢！

I'm trying to compare time in a datetime field in a SQL query, but I don't know if it's right. I don't want to compare the date part, just the time part.

I'm doing this:

SELECT timeEvent FROM tbEvents WHERE convert(datetime, startHour, 8) >= convert(datetime, @startHour, 8)

Is it correct?

I'm asking this because I need to know if 08:00:00 is less or greater than 07:30:00 and I don't want to compare the date, just the time part.

Thanks!

最满意答案

您的比较将会起作用，但是它会很慢，因为日期会转换为每行的字符串。 要有效地比较两个时间部分，请尝试：

declare @first datetime set @first = '2009-04-30 19:47:16.123' declare @second datetime set @second = '2009-04-10 19:47:16.123' select (cast(@first as float) - floor(cast(@first as float))) - (cast(@second as float) - floor(cast(@second as float))) as Difference

长说明：SQL Server中的日期存储为浮点数。 小数点前的数字代表日期。 小数点后的数字代表时间。

所以这里是一个例子：

declare @mydate datetime set @mydate = '2009-04-30 19:47:16.123'

我们把它转换成一个float：

declare @myfloat float set @myfloat = cast(@mydate as float) select @myfloat -- Shows 39931,8244921682

现在把数字后面的部分，即时间：

set @myfloat = @myfloat - floor(@myfloat) select @myfloat -- Shows 0,824492168212601

将其转换回datetime：

declare @mytime datetime set @mytime = convert(datetime,@myfloat) select @mytime -- Shows 1900-01-01 19:47:16.123

1900-01-01只是“零”的日期; 您可以使用转换显示时间部分，例如格式为108，这只是时间：

select convert(varchar(32),@mytime,108) -- Shows 19:47:16

datetime和float之间的转换相当快，因为​​它们基本上以相同的方式存储。

Your compare will work, but it will be slow because the dates are converted to a string for each row. To efficiently compare two time parts, try:

declare @first datetime set @first = '2009-04-30 19:47:16.123' declare @second datetime set @second = '2009-04-10 19:47:16.123' select (cast(@first as float) - floor(cast(@first as float))) - (cast(@second as float) - floor(cast(@second as float))) as Difference

Long explanation: a date in SQL server is stored as a floating point number. The digits before the decimal point represent the date. The digits after the decimal point represent the time.

So here's an example date:

declare @mydate datetime set @mydate = '2009-04-30 19:47:16.123'

Let's convert it to a float:

declare @myfloat float set @myfloat = cast(@mydate as float) select @myfloat -- Shows 39931,8244921682

Now take the part after the digit, i.e. the time:

set @myfloat = @myfloat - floor(@myfloat) select @myfloat -- Shows 0,824492168212601

Convert it back to a datetime:

declare @mytime datetime set @mytime = convert(datetime,@myfloat) select @mytime -- Shows 1900-01-01 19:47:16.123

The 1900-01-01 is just the "zero" date; you can display the time part with convert, specifying for example format 108, which is just the time:

select convert(varchar(32),@mytime,108) -- Shows 19:47:16

Conversions between datetime and float are pretty fast, because they're basically stored in the same way.