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ASP.NET error 1309.  Very simple GET page causing the error.

Posted on 2008-10-28
Medium Priority
Last Modified: 2011-10-19
This page is causing all kinds of problems.  Throws ASP.NET error 1309 in event log, and webpage display 'Application Error'.  Any ideas on what might be wrong with this?

<%@ Page Language="vb" AutoEventWireup="false" CodeBehind="Default.aspx.vb" Inherits="WebApplication1._Default" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml" >
<head runat="server">
    <title>HttpWebRequest Example</title>
<body runat="server">

    dim url
    dim xmlhttp
    url = "https://www.sagrader.com/api/submission/"&request.querystring("id")&"/?key=xxxxxxxx"
    xmlhttp = CreateObject("MSXML2.ServerXMLHTTP.3.0")
    xmlhttp.open ("GET", url, false)
    xmlhttp.send ()
    Response.ContentType = "text/xml"
    Response.Write (xmlhttp.responseXML.xml)
    Response.write (xmlhttp.responseText)
    xmlhttp = nothing
Question by:DevenLF

Assisted Solution

bluV11t earned 60 total points
ID: 22825953
Could it have anything to do with this:

Author Comment

ID: 22826129
Thanks for the reply.  Unfortunately, my error message doesn't have anything to do with the serialization problem.  Please see attached file for full error in event log.
LVL 15

Assisted Solution

dosth earned 90 total points
ID: 22838700
can you check you have correctly getting the value from query string

 url = "https://www.sagrader.com/api/submission/"&request.querystring("id")&"/?key=xxxxxxxx"

try printing the url and check once.
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Author Comment

ID: 22842152
dosth:  Thanks, all information is being placed in the string.  (the querystring id is being printed).

Accepted Solution

Norush earned 300 total points
ID: 22893673
Can you do the following and run the page again?

In Internet Explorer goto Tools-->Internet options-->Advanced and UNcheck the entry called "Show friendly HTTP error messages"

Now run the page again and report the exact error you are getting.

Also try removing the parentheses from the .open line:
xmlhttp.open "GET", url, false

Open in new window

LVL 15

Expert Comment

ID: 22896484

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