Edit String number with StringFormat

[FireFox]

Hi,

I want to convert the result of TimerDiff function in this format :

"3006.91739329968" to "3.006917" (example), I have found the solution for this (no need to use StringFormat)

But if I have 3006.91000000000 the result will be 3006.91 (with my method)

I just want to add 1,2,3,4,5 or 6 zeros if needed in order to get 6 decimal numbers >_<

Im sure Its very easy but right now I have not the good solution...

Cheers, FireFox.

[99ojo]

Hi,

I want to convert the result of TimerDiff function in this format :

"3006.91739329968" to "3.006917" (example), I have found the solution for this (no need to use StringFormat)

But if I have 3006.91000000000 the result will be 3006.91 (with my method)

I just want to add 1,2,3,4,5 or 6 zeros if needed in order to get 6 decimal numbers >_<

Im sure Its very easy but right now I have not the good solution...

Cheers, FireFox.

Hi,

maybe i missed something, but:

$string1 = "3006.91739329968"
$string2 = "3006.91000000000"

$string1new = StringLeft ($string1, 8)
$string2new = StringLeft ($string2, 8)

MsgBox (0,"", _preparestring ($string1new))
MsgBox (0,"", _preparestring ($string2new))

Func _preparestring ($string)
    local $stringnew = ""
    For $i = 1 to StringLen ($string)
        If $i = 2 Then
            $stringnew = $stringnew & "."
            ContinueLoop
        EndIf
        If StringMid ($string, $i, 1) <> "." Then
            $stringnew = $stringnew & StringMid ($string, $i, 1)
        EndIf
    Next
    Return $stringnew
EndFunc

;-))

Stefan

[Bowmore]

Hi,

I want to convert the result of TimerDiff function in this format :

"3006.91739329968" to "3.006917" (example), I have found the solution for this (no need to use StringFormat)

But if I have 3006.91000000000 the result will be 3006.91 (with my method)

I just want to add 1,2,3,4,5 or 6 zeros if needed in order to get 6 decimal numbers

Im sure Its very easy but right now I have not the good solution...

Cheers, FireFox.

Try these

Using StringFormat()

$T = TimerInit()
Sleep(3000)
$sRounded = StrinFormat("%0.6f",(TimerDiff($T)/1000))

Using Round()

$T = TimerInit()
Sleep(3000)
$sRounded = round(TimerDiff($T)/1000,6)

Edited August 25, 2009 by Bowmore

[Melba23]

FireFox,

Why not use Stringformat? It is really easy to get the right result then:

$nNumber = 3006.91000000000

ConsoleWrite(StringFormat("%.6f", $nNumber / 1000) & @CRLF)

$nNumber = 3006.91739329968

ConsoleWrite(StringFormat("%.6f", $nNumber / 1000) & @CRLF)

$nNumber = 3006.00000000000

ConsoleWrite(StringFormat("%.6f", $nNumber / 1000) & @CRLF)

M23

Edited August 25, 2009 by Melba23

[FireFox]

@Everybody

Thank you for your help !

@Melba23

Yeah, I knew that it will work with string format, I was just looking for how to >_<

Cheers, FireFox.