Rounding a Number UP to the Nearest Hundredth 2

[iamedwardkim]

Is there an easy way to round numbers up to the nearest hundredth (e.g. 7.83942 and 7.83021 would both be rounded up to 7.84)? Thanks in advance.

 

[MichaelRed]

This is probably to simplistic, but it at least can work for the specific examples:

Public Function basRndUp2(MyVal As Double, RndVal As Single) As Double

    'Michael Red    11/25/02
    'Sample Usage
    '? basRndUp2(7.83942, 0.01)
    ' 7.84

    '? basRndUp2(7.83021 , 0.01)
    ' 7.84


    If (Round(MyVal, 2) = Round(MyVal + (RndVal / 2), 2)) Then
        basRndUp2 = Round(MyVal, 2)
     Else
        basRndUp2 = Round(MyVal + (RndVal / 2), 2)
    End If

End Function

MichaelRed
m.red@att.net

Searching for employment in all the wrong places

 

[r937]

select int(yournum*100+1)/100.00


rudy

 

[iamedwardkim]

r937,

wouldn't a number like 7.83 be rounded to 7.84 in the solution you provided?

 

[r937]

edward, yes it would, thanks

let me try that again --

[tt]select int( (yournum+0.00999999)*100 )/100.00[/tt]


rudy

 

[PaulBricker]

Here's another solution.

Round([YourNumberField]+.005,2)

Paul

 

[iamedwardkim]

That's perfect PaulBricker. Thanks to all!

 

[dekcool]

hi paul
did you have an idea how to get this rounded value
sample

from 1 to 4.9 the value is equal to 5
then from 5.1 to 9.9 value is equal to 10
just adding 5

thanks in advance
DEK

 

[PaulBricker]

About the best I could do is this
IIf([NumberField]Mod 5 = 0,[NumberField],(5 - (Int([NumberField)Mod 5))+ Int([NumberField])

Paul

 

[dekcool]

thanks a lot it helps DEK

 

[PaulBricker]

DEK, I found that in the Debug window it worked without any problems, but when I ran it in a query I had some Double Precision Floating Point Error. For example the value
9.99 Mod 5
returned 0 which it should not return. It should return 4. If you have issues let me know and I'll work on it a little more.

Paul

 

[MichaelRed]

Public Function basRnd2Val(ValIn As Single, Rnd2 As Single) As Single

    'Michael Red    11/27/02    Tek-Tips thread701-414063

    '? basRnd2Val(7.49, 3.625)
    ' 10.875

    '? basRnd2Val(7.49, 3)
    ' 9

    '? basRnd2Val(7.49, 0.01)
    '7.5

    Dim ModRmdr As Single
    Dim ValPart() As String
    Dim DecPart() As String
    Dim Shift(2) As Integer
    Dim MyVal As Long
    Dim MyRnd As Long
    Dim MyRtn As Long

    ValPart = Split(Str(ValIn), ".")
    DecPart = Split(Str(Rnd2), ".")

    If (UBound(ValPart) > 0) Then
        Shift(0) = Len(ValPart(1))
     Else
        Shift(0) = 0
    End If

    If (UBound(DecPart) > 0) Then
        Shift(1) = Len(DecPart(1))
     Else
        Shift(1) = 0
    End If

    If (Shift(0) > Shift(1)) Then
        Shift(2) = Shift(0)
     Else
        Shift(2) = Shift(1)
    End If

    MyVal = Val(ValIn) * (10 ^ Shift(2))
    MyRnd = Val(Rnd2) * (10 ^ Shift(2))

    MyRtn = MyVal - (MyVal Mod MyRnd) + MyRnd
    basRnd2Val = MyRtn * 10 ^ (-1 * Shift(2))

End Function
[code]
 MichaelRed
m.red@att.net

Searching for employment in all the wrong places