Solved: Getting failure in response "Fail to forward request:null". Can somebody please help me on this.

shishuvendratiwari0509 · ‎04-25-2017

I am using XML API to make XML request to create the user by using python. but i am getting response Fail to forward request:null. what is the workaround for this issue?

XML request

<?xml version="1.0" encoding="UTF-8"?>

<serv:message xmlns:xsi="http://www.w3.org/2001/XMLSchemainstance"

xmlns:serv="http://www.webex.com/schemas/2002/06/service"

xsi:schemaLocation="http://www.webex.com/schemas/2002/06/service

http://www.webex.com/schemas/2002/06/service/service.xsd>

</securityContext>

</header>

<body>

<bodyContent

xsi:type="java:com.webex.service.binding.user.CreateUser"

xmlns:att="http://www.webex.com/schemas/2002/06/service/user"

xsi:schemaLocation="http://www.webex.com/schemas/2002/06/service/user

http://www.webex.com/schemas/2002/06/service/meeting/user.xsd">

<firstName>Shibbu</firstName>

<lastName>Tiwari</lastName>

<webExId>nseth</webExId>

</privilege>

<active>ACTIVATED</active>

</bodyContent>

</body>

</serv:message>

Response XML:

<?xml version="1.0" encoding="UTF-8" ?>

<serv:message xmlns:serv="http://www.webex.com/schemas/2002/06/service" xmlns:com="http://www.webex.com/schemas/2002/06/common">

<serv:header>

<serv:response>

<serv:result>FAILURE</serv:result>

<serv:reason>Fail to forward request:null</serv:reason>

<serv:gsbStatus>PRIMARY</serv:gsbStatus>

<serv:exceptionID>000000</serv:exceptionID></serv:response>

</serv:header>

<serv:body>

</serv:body>

</serv:message>

What am i missing?

ryanhunt · ‎04-25-2017

From your security Context, remove: <siteID> & <partnerID>, replace this with <siteName> that contains the subdomain for your site, ie, sitename.webex.com...

Also ensure your calling the correct XML API Service URL with the same siteName: https://sitename.webex.com/WBXService/XMLService

View solution in original post

ryanhunt · ‎04-25-2017

From your security Context, remove: <siteID> & <partnerID>, replace this with <siteName> that contains the subdomain for your site, ie, sitename.webex.com...

Also ensure your calling the correct XML API Service URL with the same siteName: https://sitename.webex.com/WBXService/XMLService

shishuvendratiwari0509 · ‎04-25-2017

Hello Ryan,

Thanks for picking up.. i modified as you suggested but getting same error. this time getting like this.

<serv:result>FAILURE</serv:result>

<serv:reason>Fail to forward request:null</serv:reason>

<serv:gsbStatus>BACKUP</serv:gsbStatus>

<serv:exceptionID>000000</serv:exceptionID>

Initially i was getting Primary but now is Backup...

please advise what should i try now?

shishuvendratiwari0509 · ‎04-25-2017

Script i am using....

import urllib

import httplib

import re

from xml.dom.minidom import parse, parseString

target_url = "https://cebmeetingstest.webex.com/WBXService/XMLService"

xml_request = """

<?xml version="1.0" encoding="UTF-8"?>

<serv:message xmlns:xsi="http://www.w3.org/2001/XMLSchemainstance"

xmlns:serv="http://www.webex.com/schemas/2002/06/service"

xsi:schemaLocation="http://www.webex.com/schemas/2002/06/service

http://www.webex.com/schemas/2002/06/service/service.xsd>

<siteName>cebmeetingstest.webex.com</siteName>

</securityContext>

</header>

<body>

<bodyContent

xsi:type="java:com.webex.service.binding.user.CreateUser"

xmlns:att="http://www.webex.com/schemas/2002/06/service/user"

xsi:schemaLocation="http://www.webex.com/schemas/2002/06/service/user

http://www.webex.com/schemas/2002/06/service/user/user.xsd">

</privilege>

<active>ACTIVATED</active>

</bodyContent>

</body>

</serv:message>

"""

print(xml_request)

result = urllib.urlopen( target_url, urllib.urlencode( {'request':xml_request} ) )

print parse( result ).toprettyxml()

result.close()

ryanhunt · ‎04-26-2017

sitename shold just be: cebmeetingstest, remove the .webex.com