LCM ???

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Thread: LCM ???

  1. #1
    Join Date
    Dec 1969
    Posts
    303

    Default LCM ???

    I am trying to get the number of years and Months between two dates (between 7/21/1993 and todays date from System). How can i get the Years and numbers? I know using LCM (Lowest Common Multiplier)i can get this, but not sure about using VBScript. help me to get XX number of Years and XX number of Months

  2. #2
    Join Date
    Dec 1969
    Posts
    11,334

    Default Check out DateDiff() <nm>

    boo

  3. #3
    Join Date
    Dec 1969
    Posts
    2,050

    Default DateDiff

    will do this for u.

  4. #4
    Join Date
    Dec 1969
    Posts
    303

    Default RE: DateDiff

    Thanks for your response.<BR>YES, i used "DateDiff"<BR><BR>YRRR = DateDiff("yyyy", Date, SvDt) <BR>I got the correct number of Years using this and i used <BR>Mo = DateDiff("m", Date, SvDt) Here i am getting the total months like 119. I need like 9 Years 11 Months...<BR>


  5. #5
    Join Date
    Dec 1969
    Posts
    11,334

    Default Check out the MOD operator <nm>

    boo

  6. #6
    Join Date
    Dec 1969
    Posts
    303

    Default RE: Check out the MOD operator <nm>

    Thank You... BUT .... :(:( still needs help...<BR><BR>Mo = DateDiff("m", Date, SvDt) &#039; SvDt = 5/17/1991<BR>YRRR = DateDiff("yyyy", Date, SvDt) <BR><BR>I need to display 12 Years and 1 Month (between 5/17/1991 and Todays Date)<BR><BR>I am displaying 145 Months and 12 Years...<BR><BR>Is there any way that i can get 12 Years and 1 Month??? <BR><BR>You told me to use "mod" Operator!!!!<BR><BR>TYM = ???? (Mo Mod 12) I am not sure... Help me<BR>

  7. #7
    Join Date
    Dec 1969
    Posts
    115

    Default RE: Check out the MOD operator <nm>

    u answered ure own question... <BR>(Mo Mod 12)<BR>145/12=12 REMAINDER 1 ;)

  8. #8
    Join Date
    Dec 1969
    Posts
    96,118

    Default Forget the DIFF of years!

    Mo = DateDiff("m", Date, SvDt)<BR>Response.Write (mo 12) & " years and " & (mo MOD 12) & " months."<BR><BR>Yes, that really is a backwards slash! (That&#039;s VBS&#039;s "integer divide" operator.)<BR><BR>You did notice, I hope, that the answer is inaccurate. There really are *NOT* 12 years and 1 month between 5/17/1991 and today. It&#039;s really only 12 years 0 months 19 days. But you are seeing how DateDiff works. If that&#039;s accurate enough for you, go for it.<BR><BR>To illustrate more clearly:<BR> DateDiff("m", #5/31/2003 11:59:59 PM#, #6/1/2003 0:00:01 AM#) <BR>is 1, even though only *two seconds* have elapsed!<BR><BR>Hokay?<BR><BR><BR><BR>

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