Sum and DateDiff

  • Thread starter Thread starter Kelly
  • Start date Start date
K

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,
 
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,
Click to expand...

Sure:

TotalTime: Round(Sum(DateDiff("n",[StartTime],[EndTime])/60),2)
 
That worked perfect! Thank You

John W. Vinson said:
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,
Click to expand...

Sure:

TotalTime: Round(Sum(DateDiff("n",[StartTime],[EndTime])/60),2)
Click to expand...
 
Back
Top