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Odd Numbers |

Jan 13 2007, 02:57 PM Post#1 | |

Posts: 16 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 |

Jan 13 2007, 03:15 PM Post#2 | |

Retired Moderator Posts: 13,562 Joined: 23-June 02 From: Texas (Is there anywhere else?) | Welcome to Utter Access.
Use the Mod operator. IIf(8 Mod 2,"Odd", "Even" ) |

Jan 13 2007, 03:29 PM Post#3 | |

Posts: 16 Joined: 10-January 07 | 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 |

Jan 13 2007, 04:01 PM Post#4 | |

UtterAccess VIP Posts: 13,031 Joined: 2-March 04 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... |

Jan 13 2007, 04:20 PM Post#5 | |

Retired Moderator Posts: 13,562 Joined: 23-June 02 From: Texas (Is there anywhere else?) | 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< |

Jan 13 2007, 11:05 PM Post#6 | |

Posts: 16 Joined: 10-January 07 | Thanks everyone, That helps a lot. Works like a charm |

May 1 2014, 06:43 PM Post#7 | |

Posts: 1 Joined: 1-May 14 | 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! |

May 1 2014, 08:13 PM Post#8 | |

UtterAccess VIP Posts: 6,670 Joined: 6-December 03 From: Telegraph Hill | Hi and welcome to UA! 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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