"nivi" <Ni****************@gmail.comwrote in message
news:11**********************@m73g2000cwd.googlegr oups.com...
Hi Gurus,
I have a problem with the way my output gets displayed in my asp page
I have the UI in ASP which asks for input from a user on the asp page
and then passes that data to a perl file which further processes it and
passes it back to the asp in an xml format
Finally the asp page has to open the output in html in the same page.
The created xml file has the details to the xsl stylesheet .
but presently the o/p gets diaplyed in an xml format in the asp and not
in a html format
<?xml version="1.0"?>
<?xml-stylesheet .......
<component>.....</component>
I do know that when this output file is opened in a browser it is a
neatly (xsl) transformed page
The transform is here being performed by the browser. The server has simply
sent the XML.
how do i make the output also appear the same way in my asp page
can i force the asp to understand it as a html?
I am presently doing this in my asp
Response.Write "receive: "&objXMLHTTP.status&"
"&objXMLHTTP.statusText&"<xmp>"&objXMLHTTP.Respons eText&"</xmp>"
Again the ResponseText will simply contain XML text.
>
Please advise
Thanks in advance
Nivi
If you want to perform the transform on your ASP server you will need to
execute it yourself.
I'll assume the remote server has specified the content type it is send is
"text/xml" therefore the ResponseXML property should return an XML DOM.
Dim oNode
Dim sTransformResult
Dim sTransformHref
Set xml = objXMLHTTP.ResponseXML
sTransformHRef = "<url goes here" 'Replace with the URL to the XSL
Set xsl = Server.CreateObject("MSXML2.DOMDocument.3.0")
xsl.async = false
xsl.setProperty "ServerHTTPRequest", True
xsl.load sTransformHref
sTransformResult = oDOM.transformNode(xsl)
Of course this presupposes that you know what the value of sTransformHref
will be. If not then it needs to be
harvested from the xml-stylesheet processing instruction in the DOM and then
resolved to a full URL. Thats a lot more work. Add these functions to your
page:-
Function GetStyleSheetHRef(roDOM)
Dim roDOM
For Each oNode in roDOM.childNodes
If oNode.nodeTypeString = "processinginstruction" Then
If oNode.nodeName = "xml-stylesheet" Then
GetStyleSheetHRef = ExtractHREF(oNode.nodeValue)
Exit For
End If
End If
Next
End Function
Function ExtractHREF(rsValue)
Dim rgx
Dim oMatches
Set rgx = New RegExp
rgx.pattern = "href=""(.*?)"""
rgx.IgnoreCase = True
Set oMatches = rgx.Execute(rsValue)
If oMatches.Count 0 Then
ExtractHREF = oMatches.Item(0).SubMatches(0)
End If
End Function
Function ResolvePath(rsOriginalPath, rsNewPath)
Dim rgx
Dim oOrigMatch
Dim oNewMatch
Set rgx = New RegExp
rgx.pattern = "(https?\://.+?(?=/))?((?:.*/)*)(.*)$"
rgx.IgnoreCase = True
oNewMatch = rgx.Execute(rsNewPath)(0)
oOrigMatch = rgx.Execute(rsOriginalPath)(0)
If oNewMatch.SubMatches(0) <"" Then
ResolvePath = rsNewPath
ElseIf oNewMatch.SubMatches(1) = "" Then
ResolvePath = oOrigMatch.SubMatches(0) & oOrigMatch.SubMatches(1) &
rsNewPath
ElseIf Left(oNewMatch.SubMatches(1),1) <"/" Then
ResolvePath = oOrigMatch.SubMatches(0) & oOrigMatch.SubMatches(1) &
rsNewPath
Else
ResolvePath = oOrigMatch.SubMatches(0) & rsNewPath
End If
End Function
Obviously you will have the URL to the original XML you've retrieved and
given that you have that in a variable called sOriginalHRef you can use this
line of code to resolve sTransformHRef:-
sTransformHRef = ResolvePath(sOriginalHRef, GetStyleSheetHRef(xml))
Note this is untested code and doesn't handle exceptionals well.
HTH,
Anthony.
