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mtfenn
post Jan 13 2007, 02:57 PM
Post#1


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Joined: 10-January 07



I'm relatively new to access and have what I hope is a basic question.
Can I get access to return numbers based on whether they are odd or even.
What I want is to make a query that basically looks like this
IIf ([Field]=Odd, yes, no)
For
Iif (Odd([field]), yes, no) --(thats how I do it in excel)
Any thoughts?
Thanks a bunch
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truittb
post Jan 13 2007, 03:15 PM
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Welcome to Utter Access.

Use the Mod operator.

IIf(8 Mod 2,"Odd", "Even" )
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mtfenn
post Jan 13 2007, 03:29 PM
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sorry I'm not familiar with this.
There do I select the field to determine if the selected field is odd or even?
what is the 8 and 2?
thanks
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dannyseager
post Jan 13 2007, 04:01 PM
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From: Leicester, UK


Using truittb's method would be
If([FieldName] Mod 2,"Odd", "Even" )
Alternativly you could use
If(Int([FieldName]/2)<>([FieldName]/2),"Odd","Even")
You can use this expression in many places... form conrols, as a new column in a query, controls in reports...
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truittb
post Jan 13 2007, 04:20 PM
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Thanks Danny
o explain further, Mod is used to divide two numbers and return only the remainder. So 8 Mod 2 is dividing 8 by 2 and there is no remainer so the result is 0. An odd number will always have a remainder 7 Mod 2 will be 1
From Access VBA Help<
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mtfenn
post Jan 13 2007, 11:05 PM
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Thanks everyone, That helps a lot. Works like a charm
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AllyChels
post May 1 2014, 06:43 PM
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Hi Everybody,
This is my first post so I hope all is well! Just want to ask why did the IIF statement return a false value on this IIf(8 Mod 2,"Odd", "Even" ) when there is no logic operator? I'm sorry for the question but I really don't get it and by the way I'm a rookie..... hehehehe!
Thanks in advance!
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cheekybuddha
post May 1 2014, 08:13 PM
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From: Telegraph Hill


Hi and welcome to UA!
welcome2UA.gif
This takes advantage of the convention in programming that 0 = FALSE and [any number other than 0, though usually 1 or -1] = TRUE.
Odd numbers will have a remainder of 1 - so this is evaluated as TRUE
So, the expression essentially says: IIf(TRUE,"Odd", "Even" )
Even numbers will have a remainder of 0 - so this is evaluated as FALSE
So, the expression essentially says: IIf(FALSE,"Odd", "Even" )
hth,
d
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