Sum and DateDiff

[Kelly]

Hi,
I doing a calculation in a query of the difference in times. I'm wondering
if I can control the decimal precision to be just (2) places. The number I
end up with is ie; 5.6666666666, can I have this to round off to something
like 5.67?

Here is the query expression I'm using for this;

TotalTime: Sum(DateDiff("n",[StartTime],[EndTime])/60)

Thanks you,

 

[John W. Vinson]

Hi,
I doing a calculation in a query of the difference in times. I'm wondering
if I can control the decimal precision to be just (2) places. The number I
end up with is ie; 5.6666666666, can I have this to round off to something
like 5.67?

Here is the query expression I'm using for this;

TotalTime: Sum(DateDiff("n",[StartTime],[EndTime])/60)

Thanks you,

Sure:

TotalTime: Round(Sum(DateDiff("n",[StartTime],[EndTime])/60),2)

 

[Kelly]

That worked perfect! Thank You

Hi,
I doing a calculation in a query of the difference in times. I'm wondering
if I can control the decimal precision to be just (2) places. The number I
end up with is ie; 5.6666666666, can I have this to round off to something
like 5.67?

Here is the query expression I'm using for this;

TotalTime: Sum(DateDiff("n",[StartTime],[EndTime])/60)

Thanks you,

Sure:

TotalTime: Round(Sum(DateDiff("n",[StartTime],[EndTime])/60),2)