need help with a regular expression replace command

[Danny]

I need an asp command to strip out from a string all extra punctuation such
as apostrophe, comma, period, spaces dashes, etc etc and just leave the
letters.

Can anybody give me some ideas?

Thanks

[Roland Hall]

"Danny" wrote in message
news:pw**********************@news4.srv.hcvlny.cv. net...
:I need an asp command to strip out from a string all extra punctuation such
: as apostrophe, comma, period, spaces dashes, etc etc and just leave the
: letters.
:

Yes. It's easier to match what you want and just dump the rest. This
example returns A-Z regardless of case or in your terms, "just the letters."

<%@ Language=VBScript %>
<%
Option Explicit
Response.Buffer = True

function getAlpha(str)
dim re, match, matches, reStr
set re = New RegExp
re.Pattern = "[a-z]+"
re.IgnoreCase = True
re.Global = True
set matches = re.Execute(str)
for each match in matches
reStr = reStr & match.value
next
getAlpha = reStr
end function

Response.Write("Result: " & getAlpha("AaB5b!'C,*cD/d\Ee"))
%>

Response:
Result: AaBbCcDdEe

HTH...

--
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[Evertjan.]

Roland Hall wrote on 03 okt 2004 in
microsoft.public.inetserver.asp.general:

"Danny" wrote in message
news:pw**********************@news4.srv.hcvlny.cv. net...
:I need an asp command to strip out from a string all extra punctuation
:such
: as apostrophe, comma, period, spaces dashes, etc etc and just leave
: the letters.
:

Yes. It's easier to match what you want and just dump the rest. This
example returns A-Z regardless of case or in your terms, "just the
letters."

I don't think so,
the replace function is much quicker than matching and a loop:

function getAlpha(str)
dim re
set re = New RegExp
re.Pattern = "[^a-z]+"
re.IgnoreCase = True
re.Global = True
getAlpha = re.replace(str,"")
end function

Response.Write "Result: " & getAlpha("AaB5b!'C,*cD/d\Ee")
--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress,
but let us keep the discussions in the newsgroup)

[Dave Anderson]

Danny wrote:

I need an asp command to strip out from a string all extra
punctuation such as apostrophe, comma, period, spaces dashes, etc etc
and just leave the letters.

JScript:
return myString.replace(/[^a-z]/gi,"")

vbscript:
Use the Replace function and the RegExp object to do the same.

--
Dave Anderson

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me directly or ask me to contact you directly for assistance. If your
question is worth asking, it's worth posting.

[broby425]

I use a function for that. There might a slicker way to do it, but this is what I came up with. You can replace each puncuation mark with just an empty string "" instead of the encoding.

Function EncPunc(checkPunc)
strTmp = checkPunc
strTmp = Replace(strTmp,"'","%39")
strTmp = Replace(strTmp,"%","%25")
strTmp = Replace(strTmp,"&","%26")
strTmp = Replace(strTmp,"#","%23")
strTmp = Replace(strTmp,";","%3B")
strTmp = Replace(strTmp,"/","%2F")
strTmp = Replace(strTmp,"?","%3F")
strTmp = Replace(strTmp,":","%3A")
strTmp = Replace(strTmp,"@","%40")
strTmp = Replace(strTmp,"=","%3D")
strTmp = Replace(strTmp,"<","%3C")
strTmp = Replace(strTmp,">","%3E")
strTmp = Replace(strTmp,"""","%22")
strTmp = Replace(strTmp,"{","%7B")
strTmp = Replace(strTmp,"}","%7D")
strTmp = Replace(strTmp,"|","%7C")
strTmp = Replace(strTmp,"\\","%5C")
strTmp = Replace(strTmp,"^","%5E")
strTmp = Replace(strTmp,"~","%7E")
strTmp = Replace(strTmp,"[","%5B")
strTmp = Replace(strTmp,"]","%5D")
strTmp = Replace(strTmp,"`","%60")
EncPunc = strTmp
End Function

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[Evertjan.]

br******@hotmail.com wrote on 13 okt 2004 in
microsoft.public.inetserver.asp.general:

Function EncPunc(checkPunc)
strTmp = checkPunc
strTmp = Replace(strTmp,"'","%39")
strTmp = Replace(strTmp,"%","%25")
strTmp = Replace(strTmp,"&","%26")
strTmp = Replace(strTmp,"#","%23")

teststr = "%&#;/?:@=<>""{}|\^~[]`"

Function EncPunc(x)
EncPunc = ""
for i=1 to len(x)
t = mid(x,i,1)
if instr(teststr,t)>0 then t = "%" & asc(t)
EncPunc = EncPunc & t
next
End Function

==================================
strTmp = Replace(strTmp,"\\","%5C")

you are mistaken, this is not javascript, but vbscript

--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress,
but let us keep the discussions in the newsgroup)

[regx]

Yes, there is a much easier wa to do HTML entity, unfortunatly as far as I
can tell regexps in VB are severely impared!!!
The easy way to do this is to match the char with a regular expresion and
then get the numeric value for the char.
In VB it would look something like
dim reg1 as RegExp
set reg1= new RegExp
reg1.global = true
reg1.pattern="([^a-zA-Z0-9\s])"
text=reg1.replace(text,"&" & Asc("$1") & ";"

Unfortunatley in VB you can not pass submatches to a funtion. Submatches
only work inside the expression.
If anyone knows how to pass submatches to a function in a VB RegExp please
let me know.
For now I just call external Perl scripts from VB

by the way, the above in Perl looks like this
$text=~s/([^a-zA-Z0-9\s])/"&".ord($1).";"/ge;

Aside from being brief, it works!!!
And if we want to check to make sure entities like $ do not get encoded we
can pass to a function that handles this
$text=~s/([^a-zA-Z0-9\s])/encodeHTMLentities($1)/ge;
"Bob Roby" wrote:

I use a function for that. There might a slicker way to do it, but this is what I came up with. You can replace each puncuation mark with just an empty string "" instead of the encoding.

Function EncPunc(checkPunc)
strTmp = checkPunc
strTmp = Replace(strTmp,"'","%39")
strTmp = Replace(strTmp,"%","%25")
strTmp = Replace(strTmp,"&","%26")
strTmp = Replace(strTmp,"#","%23")
strTmp = Replace(strTmp,";","%3B")
strTmp = Replace(strTmp,"/","%2F")
strTmp = Replace(strTmp,"?","%3F")
strTmp = Replace(strTmp,":","%3A")
strTmp = Replace(strTmp,"@","%40")
strTmp = Replace(strTmp,"=","%3D")
strTmp = Replace(strTmp,"<","%3C")
strTmp = Replace(strTmp,">","%3E")
strTmp = Replace(strTmp,"""","%22")
strTmp = Replace(strTmp,"{","%7B")
strTmp = Replace(strTmp,"}","%7D")
strTmp = Replace(strTmp,"|","%7C")
strTmp = Replace(strTmp,"\\","%5C")
strTmp = Replace(strTmp,"^","%5E")
strTmp = Replace(strTmp,"~","%7E")
strTmp = Replace(strTmp,"[","%5B")
strTmp = Replace(strTmp,"]","%5D")
strTmp = Replace(strTmp,"`","%60")
EncPunc = strTmp
End Function

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"Bob Roby" wrote:
I use a function for that. There might a slicker way to do it, but this is what I came up with. You can replace each puncuation mark with just an empty string "" instead of the encoding.

Function EncPunc(checkPunc)
strTmp = checkPunc
strTmp = Replace(strTmp,"'","%39")
strTmp = Replace(strTmp,"%","%25")
strTmp = Replace(strTmp,"&","%26")
strTmp = Replace(strTmp,"#","%23")
strTmp = Replace(strTmp,";","%3B")
strTmp = Replace(strTmp,"/","%2F")
strTmp = Replace(strTmp,"?","%3F")
strTmp = Replace(strTmp,":","%3A")
strTmp = Replace(strTmp,"@","%40")
strTmp = Replace(strTmp,"=","%3D")
strTmp = Replace(strTmp,"<","%3C")
strTmp = Replace(strTmp,">","%3E")
strTmp = Replace(strTmp,"""","%22")
strTmp = Replace(strTmp,"{","%7B")
strTmp = Replace(strTmp,"}","%7D")
strTmp = Replace(strTmp,"|","%7C")
strTmp = Replace(strTmp,"\\","%5C")
strTmp = Replace(strTmp,"^","%5E")
strTmp = Replace(strTmp,"~","%7E")
strTmp = Replace(strTmp,"[","%5B")
strTmp = Replace(strTmp,"]","%5D")
strTmp = Replace(strTmp,"`","%60")
EncPunc = strTmp
End Function

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