# Can't use math operations within print results

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By auk3, Junior Member on 19th March 2014, 11:25 PM
Hello!
When I write, for example:
Quote:

\$#C1D#=1? (#C1SRd#+1)\$

Where C1D = 1 & C1SRd = 3

How could I make to print '4', unless '3+1'?

Thanks a lot!

20th March 2014, 07:22 AM |#2
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Quote:
Originally Posted by auk3

Hello!
When I write, for example:

Where C1D = 1 & C1SRd = 3

How could I make to print '4', unless '3+1'?

Thanks a lot!

It works if you remove the space you have between your ? and the first ( . If you need a space in front your number, you'll have to put it in a separate conditional.

Hope that helps you out
20th March 2014, 08:51 AM |#3
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Quote:
Originally Posted by kwerdenker

It works if you remove the space you have between your ? and the first ( . If you need a space in front your number, you'll have to put it in a separate conditional.

Hope that helps you out

Thank you! but it doesn't work to me. I've proved so many different combinations, and still doesn't work. Keep searching...

EDIT: It finally works! In my case (not in the example) after '(#C1SRd#+1)' I put a 'd' -> '(#C1SRd#+1)d'. That's why it didn't work, because of the 'd'. Thanks a lot again!!
20th March 2014, 08:59 AM |#4
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Quote:
Originally Posted by auk3

Thank you! but it doesn't work to me. I've proved so many different combinations, and still doesn't work. Keep searching...

That's odd. As you can see from the attached screenshot it works as expected for me:

Send from my secret moonbase via space carrier pigeons
20th March 2014, 09:10 AM |#5
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Quote:
Originally Posted by kwerdenker

That's odd. As you can see from the attached screenshot it works as expected for me:
Send from my secret moonbase via space carrier pigeons

It finally works! In my case (not in the example) after '(#C1SRd#+1)' I put a 'd' -> '(#C1SRd#+1)d'. That's why it didn't work, because of the 'd'. Thanks a lot again!!

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