Solved

MSXML2.ServerXmlhttp.6.0 Error

Posted on 2013-11-22
6
2,165 Views
Last Modified: 2013-11-26
Set http = CreateObject("MSXML2.ServerXMLHTTP.6.0")
http.Open  "GET", "http://www.google.com", False,userid,password
http.Send
WScript.ECHO http.ResponseText

The above code creating error as below
The server name or address could not be resolved

But this is working with MSXML2.XMLHTTP
0
Comment
Question by:JayeshIyer
  • 4
  • 2
6 Comments
 
LVL 65

Accepted Solution

by:
RobSampson earned 500 total points
Comment Utility
I'm not sure that would work with Google.  Using a username and password implies you should be querying the HTTPS protocol instead of HTTP, but it may also be expecting a different URL to authenticate against.

On the other hand, if it's not resolving, are you behind a proxy that might be stopping it from reaching the URL?

If so, can you try the code from a direct link with no proxy?  Or you can try adding this:
http.setProxy 2, "myproxyserver:8080"
http.setProxyCredentials "MyUserName", "myPassword"

before the GET call.

Rob.
0
 
LVL 65

Expert Comment

by:RobSampson
Comment Utility
There's also a good article on MSDN for troubleshooting such code here:
http://blogs.msdn.com/b/jpsanders/archive/2008/06/25/troubleshooting-code-that-uses-the-http-protocol.aspx

Rob.
0
 

Author Comment

by:JayeshIyer
Comment Utility
Thanks rob for your comments.
This worked for the above code.

Set http = CreateObject("MSXML2.ServerXMLHTTP.6.0")
http.Open  "GET", "https://jazz.visteon.com:9443/ccm", False,userid,password
http.setProxy 2,  "136.17.0.7:83"
http.setProxyCredentials username, password
http.Send
WScript.ECHO http.ResponseText

But when i am using different url the below error is comming


Network Error (tcp_error)
Communication error
0
Find Ransomware Secrets With All-Source Analysis

Ransomware has become a major concern for organizations; its prevalence has grown due to past successes achieved by threat actors. While each ransomware variant is different, we’ve seen some common tactics and trends used among the authors of the malware.

 
LVL 65

Expert Comment

by:RobSampson
Comment Utility
Again, I'd make sure you can access that through the proxy, and try running the script from the same authorised user account as well.
0
 

Author Comment

by:JayeshIyer
Comment Utility
Yes rob i tried without proxy settings.then it gives the below error

A connection with the server could not be established.
0
 
LVL 65

Expert Comment

by:RobSampson
Comment Utility
Hmmm, I guess some sites may require that you send specific headers with the request. There is code here that may help:
http://zanstra.home.xs4all.nl/inTec/ServerXMLHTTP.htm

I have put it with your current code to see what happens.

Set http = CreateObject("MSXML2.ServerXMLHTTP.6.0")
http.Open  "GET", "https://jazz.visteon.com:9443/ccm", False,userid,password
http.setProxy 2,  "136.17.0.7:83"
http.setProxyCredentials username, password
http.setOption 2, 13056
http.setRequestHeader "Authorization", "Basic " & Base64Encode(userid & ":" & password)
http.Send
WScript.ECHO http.responseText

Function Base64Encode(inData)
'ripped from: 
'http://www.pstruh.cz/tips/detpg_Base64Encode.htm
  'rfc1521
  '2001 Antonin Foller, PSTRUH Software, http://pstruh.cz
  Const Base64 = _
"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
  Dim sOut, I
  
  'For each group of 3 bytes
  For I = 1 To Len(inData) Step 3
    Dim nGroup, pOut
    
    'Create one long from this 3 bytes.
    nGroup = &H10000 * Asc(Mid(inData, I, 1)) + _
      &H100 * MyASC(Mid(inData, I + 1, 1)) + _
      MyASC(Mid(inData, I + 2, 1))
    
    'Oct splits the long To 8 groups with 3 bits
    nGroup = Oct(nGroup)
    
    'Add leading zeros
    nGroup = String(8 - Len(nGroup), "0") & nGroup
    
    'Convert To base64
    pOut = Mid(Base64, CLng("&o" & Mid(nGroup, 1, 2)) + 1, 1) + _
      Mid(Base64, CLng("&o" & Mid(nGroup, 3, 2)) + 1, 1) + _
      Mid(Base64, CLng("&o" & Mid(nGroup, 5, 2)) + 1, 1) + _
      Mid(Base64, CLng("&o" & Mid(nGroup, 7, 2)) + 1, 1)
    
    'Add the part To OutPut string
    sOut = sOut + pOut
    
  Next
  Select Case Len(inData) Mod 3
    Case 1: '8 bit final
      sOut = Left(sOut, Len(sOut) - 2) + "=="
    Case 2: '16 bit final
      sOut = Left(sOut, Len(sOut) - 1) + "="
  End Select
  Base64Encode = sOut
End Function

Function MyASC(OneChar)
  If OneChar = "" Then MyASC = 0 Else MyASC = Asc(OneChar)
End Function

Open in new window


Or, if the certificate isn't quite right, you can try this to ignore certificate errors:
Set http = CreateObject("MSXML2.ServerXMLHTTP.6.0")
http.Open  "GET", "https://jazz.visteon.com:9443/ccm", False,userid,password
http.setProxy 2,  "136.17.0.7:83"
http.setProxyCredentials username, password
http.setOption 2, 13056
http.Send
WScript.ECHO http.responseText

Open in new window


Rob.
0

Featured Post

Find Ransomware Secrets With All-Source Analysis

Ransomware has become a major concern for organizations; its prevalence has grown due to past successes achieved by threat actors. While each ransomware variant is different, we’ve seen some common tactics and trends used among the authors of the malware.

Join & Write a Comment

Over the years I have built up my own little library of code snippets that I refer to when programming or writing a script.  Many of these have come from the web or adaptations from snippets I find on the Web.  Periodically I add to them when I come…
Whether you’re a college noob or a soon-to-be pro, these tips are sure to help you in your journey to becoming a programming ninja and stand out from the crowd.
The goal of the tutorial is to teach the user how to use functions in C++. The video will cover how to define functions, how to call functions and how to create functions prototypes. Microsoft Visual C++ 2010 Express will be used as a text editor an…
The viewer will learn how to user default arguments when defining functions. This method of defining functions will be contrasted with the non-default-argument of defining functions.

728 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

14 Experts available now in Live!

Get 1:1 Help Now