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Re: loadXML() parse error 1072896760

From: Scott@-----------.---------.---
To: NULL
Date: 10/4/2004 4:53:00 PM
Thanks Martin, it was an encoding problem with the responseText field. I will 
also take you advice on the not setting the Content-Type.

Thanks again,

Scott

"Martin Honnen" wrote:

> 
> 
> Scott wrote:
> 
> >
> > Dim sourceUrl, xmlDoc, xmlHttp
> > sourceUrl = 
> > "http://www.somesource.com/search?client=someclient&output=xml_no_dtd&q=computers&num=0&ad=w15&adtest=on&adpage=3&adsafe=high&ip=10.1.4.80&useragent=Mozilla/4.0%20(compatible;%20MSIE%206.0;%20Windows%20NT%205.2;%20.NET%20CLR%201.1.4322)"
> > 
> > Set xmlDoc	= Server.CreateObject("MSXML2.DOMDocument.4.0")
> > Set xmlHttp = Server.CreateObject("MSXML2.XMLHTTP.4.0")
> > 
> > xmlHttp.open "GET", sourceUrl, False
> > xmlHttp.setRequestHeader "Content-Type", "text/xml"
> 
> If you do a HTTP GET request you are not sending any request body thus 
> you should not send the request header for Content-Type, there is no 
> content.
> 
> > xmlHttp.send()
> 
> Setting the Content-Type above would only make sense if here you would 
> send some XML e.g.
>    xmlHttp.send("<root><child>Kibo</child></root>")
> thus as already said above you shouldn't set the request header for 
> Content-Type.
> 
> > Response.Write xmlHttp.responseText & "<br><br><br>" & chr(13) & chr(10)
> > 
> > xmlDoc.validateOnParse = false
> > 
> > xmlDoc.loadXML(xmlHttp.responseText)
> 
> There is no need to parse responseXML, you can simply use
>    xmlHttp.responseXML
> as that is automatically created. And if you parse responseText you 
> might run into encoding related problems which do not occur if you 
> simply use responseXML, a Msxml2.DOMDocument object.
> 
> 
> 
> -- 
> 
> 	Martin Honnen
> 	http://JavaScript.FAQTs.com/
> 


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