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MS Office Forum / General MS InfoPath Questions / August 2005Rounding to the nearest .5
Does anyone know how to do this? I've tried adjusting the decimal place to 1, but that didn't auto round or anything. I have about 9 textboxes that need to do this. Any Ideas? You can use rules to achieve this goal. Play with the "floor", "ceiling" and "mod" XPath functions, they should help. > Does anyone know how to do this? I've tried adjusting the decimal place > to > 1, but that didn't auto round or anything. I have about 9 textboxes that > need to do this. > Any Ideas? Well, i see the "round" function, but I need it to go by the nearst .5 Also, I did not see a mod function. Is this still possible? > You can use rules to achieve this goal. Play with the "floor", "ceiling" and > "mod" XPath functions, they should help. [quoted text clipped - 4 lines] > > need to do this. > > Any Ideas? Do you mean to the "nearest" .5 or to the "next" .5? For example, if you have 123.6, does that mean you want 123.5 or 124.0? The above is more for curiosity than anything. My recommendation would be that you don't use rules, but instead use code. I'm sure if there's a good mathematician here, they could give you some kind of a formula to do this, but it seems to me that writing code is just as easy. Then you have fine-grain control over such neuances as I mentioned above.  SignatureGreg Collins [InfoPath MVP] Please visit: http://www.InfoPathDev.com Well, i see the "round" function, but I need it to go by the nearst .5 Also, I did not see a mod function. Is this still possible? "Alex [MSFT]" wrote: > You can use rules to achieve this goal. Play with the "floor", "ceiling" and > "mod" XPath functions, they should help. [quoted text clipped - 4 lines] > > need to do this. > > Any Ideas? I guess I would want it to go to the nearest .5 That would probably work the best for me. > Do you mean to the "nearest" .5 or to the "next" .5? For example, if you have 123.6, does that mean you want 123.5 or 124.0? > [quoted text clipped - 16 lines] > > > need to do this. > > > Any Ideas? Dim oldno as Decimal = 23 Dim newno = int(oldno.Round((oldno /10 * 2),0) /2 * 10) Response.write(newno) > *Does anyone know how to do this? I've tried adjusting the decimal > place to > 1, but that didn't auto round or anything. I have about 9 textboxes > that > need to do this. > Any Ideas? * -- Databasic Oops sorry I misread. This rounds to the nearest 5 not .5 > *Dim oldno as Decimal = 23 > Dim newno = int(oldno.Round((oldno /10 * 2),0) /2 * 10) > Response.write(newno) * -- Databasic How tough would it be to get it to round to the .5? > Oops sorry I misread. This rounds to the nearest 5 not .5 > [quoted text clipped - 9 lines] > View this thread: http://www.mcse.ms/message1738704.html > |
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