6 Replies Latest reply: Aug 19, 2009 3:38 AM by kyaw zorr

Hello,

In Pivot table , Expression i used the num( ) function. The function is show the decimal place is correct but when the formula was manipulated they used the original value. That mean

i used the function like that,

num('0.32409178531','.##') * num('40648.216','.##') => 13173.752893

I checked each function's output is correct , num('0.32409178531','.##') output is => 0.32

num('40648.216','.##') output is => 40648.22

but combine these functions together

num('0.32409178531','.##') * num('40648.216','.##') output=> 13173.752893

Can anybody help me? Thanks a lot for advance.

• ###### Help for about decimal.

I think they taken 0.32 instead of 0.32409178531

floor(num('0.32','.##') * num('40648.216','.##')) . This will give you the exact value 13007

-Raghu

• ###### Help for about decimal.

Hi Raghu,

But , i don't want to change the the value '0.32409178531' to '0.32'.

Because in the report the value may not be the fix.

Do you have any other way to do that.

Thanks a lot.

• ###### Help for about decimal.

Take first two decimal places using functions and multiply with the number.

I think the only reason the value get changed because of '0.32409178531'. If you take only the first two decimal places then the answer will be correct.

-Raghu.

• ###### Help for about decimal.

Hello,

Can you try that :

=round(num('0.32409178531','.##'),0.01) * round(num('40648.216','.##'),0.01)

• ###### Help for about decimal.

we can use the floor function to get the numeric part.

=Floor(=round(num('0.32409178531','.##'),0.01) * round(num('40648.216','.##'),0.01))

-Raghu

• ###### Help for about decimal.

Hi Raghu and Philippe,

Thanks you very much.

Your function is work. Thanks a lot.