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stubarnes
Contributor II
Contributor II

Format Class ranges as currency

Hi I have the following but would like the formatting of numbers to be displayed as currency

=Dual(Replace(Class(TP,500),'<= x <','-'),Class(TP,500))

I tried inserting Money('£#,##0') before replace but did not work..

Any ideas?

1 Solution

Accepted Solutions
Gysbert_Wassenaar

Perhaps something like this:

Dual( Money( Floor(TP,500) , '£#,##0') & '-' & Money( Ceil(TP,500) , '£#,##0') , Floor(TP,500))


talk is cheap, supply exceeds demand

View solution in original post

8 Replies
Gysbert_Wassenaar

Perhaps something like this:

Dual( Money( Floor(TP,500) , '£#,##0') & '-' & Money( Ceil(TP,500) , '£#,##0') , Floor(TP,500))


talk is cheap, supply exceeds demand
MarcoWedel

Hi,

some solutions might be:

=Dual('£'&Replace(Class(TP,500),' <= x < ',' - £'),Class(TP,500))

=Dual(Money(Floor(TP,500),'£#,##0')&'-'&Money(Floor(TP,500)+500,'£#,##0'),Floor(TP,500))

hope this helps

regards

Marco

balabhaskarqlik

May be this:

=Dual(Replace(Class(TP,500),'<= x <','to'), Num(Class(TP,500)))

balabhaskarqlik

stubarnes
Contributor II
Contributor II
Author

Thanks for your help on this, works perfectly!

If I could ask an additional question, is it possible to format the numbers as a percentage #,##0.0%

Gysbert_Wassenaar

Yes, with the num function instead of the money function.


talk is cheap, supply exceeds demand
MarcoWedel

Hello gwassenaar,

I noticed a difference in the results of the Class() vs. Floor()/Ceil() approach due to Floor and Ceil both letting integer values unchanged. Can you check?

thanks

regards

Marco

Gysbert_Wassenaar

Yeah, if a value falls exactly on a border value then an incorrect result is returned. Your expression always returns the correct result.


talk is cheap, supply exceeds demand