format string

**Yarn** · Jul 19 '05, 07:26 AM

Re: format string

You could try this, although it's a little funky-

counter = 2
myString = "12345678"
WHILE counter <= 8
response.write right(left("123 45678",counter) ,2)&" "
counter=counter +2
WEND

"Christophe r Brandsdal" <christopherwb@ c2i.net> wrote in message
news:u4muyvnfDH A.3700@TK2MSFTN GP11.phx.gbl...[color=blue]
> Just a little question...
>
> How do I split 12345678 into 12 34 56 78???????
>
>
> Christopher Brandsdal
>
>[/color]

**Christopher Brandsdal** · Jul 19 '05, 07:26 AM

Re: format string

Thanks!
But is there not a more easy way to do this?

:)

"Yarn" <nope@noaddress .com> skrev i melding
news:e4cUN8nfDH A.392@TK2MSFTNG P12.phx.gbl...[color=blue]
> You could try this, although it's a little funky-
>
>
> counter = 2
> myString = "12345678"
> WHILE counter <= 8
> response.write right(left("123 45678",counter) ,2)&" "
> counter=counter +2
> WEND
>
>
>
> "Christophe r Brandsdal" <christopherwb@ c2i.net> wrote in message
> news:u4muyvnfDH A.3700@TK2MSFTN GP11.phx.gbl...[color=green]
> > Just a little question...
> >
> > How do I split 12345678 into 12 34 56 78???????
> >
> >
> > Christopher Brandsdal
> >
> >[/color]
>
>[/color]

**Yarn** · Jul 19 '05, 07:26 AM

Re: format string

Probably. This really all there is to it though-

response.write right(left("123 45678",2),2)&" "
response.write right(left("123 45678",4),2)&" "
response.write right(left("123 45678",6),2)&" "
response.write right(left("123 45678",8),2)&" "

First we grab the two over from the left, than two over from the right.
Than we grab the fourth over from the left, than two over from the right.
and so on..

"Christophe r Brandsdal" <christopherwb@ c2i.net> wrote in message
news:uDhrvGofDH A.3200@tk2msftn gp13.phx.gbl...[color=blue]
> Thanks!
> But is there not a more easy way to do this?
>
> :)
>
>
> "Yarn" <nope@noaddress .com> skrev i melding
> news:e4cUN8nfDH A.392@TK2MSFTNG P12.phx.gbl...[color=green]
> > You could try this, although it's a little funky-
> >
> >
> > counter = 2
> > myString = "12345678"
> > WHILE counter <= 8
> > response.write right(left("123 45678",counter) ,2)&" "
> > counter=counter +2
> > WEND
> >
> >
> >
> > "Christophe r Brandsdal" <christopherwb@ c2i.net> wrote in message
> > news:u4muyvnfDH A.3700@TK2MSFTN GP11.phx.gbl...[color=darkred]
> > > Just a little question...
> > >
> > > How do I split 12345678 into 12 34 56 78???????
> > >
> > >
> > > Christopher Brandsdal
> > >
> > >[/color]
> >
> >[/color]
>
>[/color]

**Evertjan.** · Jul 19 '05, 07:26 AM

Re: format string

Christopher Brandsdal wrote on 19 sep 2003 in[color=blue][color=green]
>> "Christophe r Brandsdal" <christopherwb@ c2i.net> wrote in message[color=darkred]
>> > How do I split 12345678 into 12 34 56 78???????
>> >[/color][/color]
> "Yarn" <nope@noaddress .com> skrev i melding[color=green]
>> You could try this, although it's a little funky-
>>
>> counter = 2
>> myString = "12345678"
>> WHILE counter <= 8
>> response.write right(left("123 45678",counter) ,2)&" "
>> counter=counter +2
>> WEND[/color]
>
> But is there not a more easy way to do this?[/color]

<%
myString = "12345678"
Set regEx = New RegExp
regEx.Pattern = "(..)(?=.)"
regEx.Global = True
newString = regEx.Replace(m yString, "$1 ")

response.write newString & " "
%>

=============== =============== ==============

Works for any length any char.

The (..) selects to char's,
that are replaced "$1 " by the same plus a space.

The (?=.) looks ahead for another char,
so that an end space is suppressed.

--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress)

**Tom B** · Jul 19 '05, 07:26 AM

Re: format string

myString="12345 678"
Dim iLoop
for iLoop = 1 to len(myString) Step 2
Response.write mid(myString,iL oop,2)
next

'Untested....bu t it should work

"Yarn" <nope@noaddress .com> wrote in message
news:e4cUN8nfDH A.392@TK2MSFTNG P12.phx.gbl...[color=blue]
> You could try this, although it's a little funky-
>
>
> counter = 2
> myString = "12345678"
> WHILE counter <= 8
> response.write right(left("123 45678",counter) ,2)&" "
> counter=counter +2
> WEND
>
>
>
> "Christophe r Brandsdal" <christopherwb@ c2i.net> wrote in message
> news:u4muyvnfDH A.3700@TK2MSFTN GP11.phx.gbl...[color=green]
> > Just a little question...
> >
> > How do I split 12345678 into 12 34 56 78???????
> >
> >
> > Christopher Brandsdal
> >
> >[/color]
>
>[/color]

**Phill. W** · Jul 19 '05, 07:28 AM

Re: format string

"Yarn" <nope@noaddress .com> wrote in message
news:eGqfnPofDH A.460@TK2MSFTNG P12.phx.gbl...
.. . .[color=blue]
> response.write right(left("123 45678",2),2)&" "[/color]
.. . .

(Yes, it's /definitely/ VBScript, so...)

What, pray tell, is wrong with using Mid() ???

sData = "12345678"
Response.Write Mid( sData, 1, 2 ) _
& " " & Mid( sData, 3, 2 ) _
& " " & Mid( sData, 5, 2 ) _
& " " & Mid( sData, 7, 2 )

Regards,
Phill W.

**dlbjr** · Jul 19 '05, 07:28 AM

Re: format string

Try this test to make your decision on what to do.

<%
strString = "12345678"
start = Timer()
For i = 1 To 1000
RexEx(strString )
Next
response.write "RegularExpersi on = " & GetTime(start) & " "
start = Timer()
For i = 1 To 1000
MidLoop(strStri ng)
Next
response.write "MidLoop = " & GetTime(start) & " "
start = Timer()
For i = 1 To 1000
MidHardCode(str String)
Next
response.write "MidHardCod e = " & GetTime(start) & " "
start = Timer()
For i = 1 To 1000
ArrayBreak(strS tring)
Next
response.write "ArrayBreak = " & GetTime(start) & " "

Function GetTime(start)
GetTime = FormatNumber(Ti mer() - start,4)
End Function

Function MidLoop(strData )
for iLoop = 1 to len(strData) Step 2
MidLoop = mid(strData,iLo op,2) & " "
next
End Function

Function MidHardCode(str Data)
MidHardCode = Mid( strData, 1, 2 ) _
& " " & Mid( strData, 3, 2 ) _
& " " & Mid( strData, 5, 2 ) _
& " " & Mid( strData, 7, 2 )
End Function

Function RexEx(strData)
Set regEx = New RegExp
regEx.Pattern = "(..)(?=.)"
regEx.Global = True
RexEx = regEx.Replace(s trData, "$1 ")
End Function

Function ArrayBreak(strD ata)
Dim aryData()
intLen = Len(strData)
If intLen > 0 Then
ReDim aryData((intLen \ 2) + 1)
intCount = 0
for iLoop = 1 to len(strData) Step 2
strItem = Trim(mid(strDat a,iLoop,2))
If Len(strItem) > 0 Then
aryData(intCoun t) = strItem
intCount = intCount + 1
End If
next
ArrayBreak = Trim(Join(aryDa ta," "))
End If
End Function
%>

--
-dlbjr

Discerning resolutions for the alms

**Yarn** · Jul 19 '05, 07:28 AM

Re: format string

I liked my first one better.
Keeps my clients from trying to hack my code themselves.
Obsfucating is the key to long term clients!

(what a weasel, I know..)

counter = 2
myString = "12345678"
WHILE counter <= 8
response.write right(left("123 45678",counter) ,2)&" "
counter=counter +2
WEND

"Phill. W" <P.A.Ward@open. ac.uk> wrote in message
news:bkes24$l44 $1@yarrow.open. ac.uk...[color=blue]
> "Yarn" <nope@noaddress .com> wrote in message
> news:eGqfnPofDH A.460@TK2MSFTNG P12.phx.gbl...
> . . .[color=green]
> > response.write right(left("123 45678",2),2)&" "[/color]
> . . .
>
> (Yes, it's /definitely/ VBScript, so...)
>
> What, pray tell, is wrong with using Mid() ???
>
> sData = "12345678"
> Response.Write Mid( sData, 1, 2 ) _
> & " " & Mid( sData, 3, 2 ) _
> & " " & Mid( sData, 5, 2 ) _
> & " " & Mid( sData, 7, 2 )
>
> Regards,
> Phill W.
>
>[/color]

**dlbjr** · Jul 19 '05, 07:28 AM

Re: format string

'Here is a timing test

<%
strString = "12345678"
start = Timer()
For i = 1 To 1000
RexEx(strString )
Next
response.write "RegularExpersi on = " & GetTime(start) & " "
start = Timer()
For i = 1 To 1000
MidLoop(strStri ng)
Next
response.write "MidLoop = " & GetTime(start) & " "
start = Timer()
For i = 1 To 1000
MidHardCode(str String)
Next
response.write "MidHardCod e = " & GetTime(start) & " "
start = Timer()
For i = 1 To 1000
ArrayBreak(strS tring)
Next
response.write "ArrayBreak = " & GetTime(start) & " "

Function GetTime(start)
GetTime = FormatNumber(Ti mer() - start,4)
End Function

Function MidLoop(strData )
for iLoop = 1 to len(strData) Step 2
MidLoop = mid(strData,iLo op,2) & " "
next
End Function

Function MidHardCode(str Data)
MidHardCode = Mid( strData, 1, 2 ) _
& " " & Mid( strData, 3, 2 ) _
& " " & Mid( strData, 5, 2 ) _
& " " & Mid( strData, 7, 2 )
End Function

Function RexEx(strData)
Set regEx = New RegExp
regEx.Pattern = "(..)(?=.)"
regEx.Global = True
RexEx = regEx.Replace(s trData, "$1 ")
End Function

Function ArrayBreak(strD ata)
Dim aryData()
intLen = Len(strData)
If intLen > 0 Then
ReDim aryData((intLen \ 2) + 1)
intCount = 0
for iLoop = 1 to len(strData) Step 2
strItem = Trim(mid(strDat a,iLoop,2))
If Len(strItem) > 0 Then
aryData(intCoun t) = strItem
intCount = intCount + 1
End If
next
ArrayBreak = Trim(Join(aryDa ta," "))
End If
End Function
%>

--
-dlbjr

Discerning resolutions for the alms

format string

format string

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