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Calling Web Service that calls other Web Service with Windows Authentication

Hi all

I would like to now the answer to the following problem, if someone can
help.

Given a windows application client that's calling a web service (using
default credentials) the
web service gets invoked fine with the users credentials. However they
need
to call a second web service from the first one. When it hits the
second web
service, the web service (even though configured correctly) always
reports (401) Unauthorized.

Is there any possibility to do this or is impossible ?

Can someone give me a sample of Kerberos Implementation for this case ?

Is possible to implement any workaround for this problem ?

Thanks in advance

Manuel Brandão

Jun 16 '06 #1
1 1286
It would really depend on the second web service. Some web services use
"integrated " credentials and some force you to manually pass credentials. In
the second case, you'll have to store the credentials that the web service
needs and pass them however the web service requires. ESRI ArcWebServices,
for example, requires that you ask it to create an authorization token (with
another WebService invoke) that is uses as a "parameter" to other web service
invocations. MapPoint requires you populate the Credentials property with
the login and password (as it doesn't support Windows authentication) on
every invocation.

--
http://www.peterRitchie.com/
"Manuel Brandao" wrote:
Hi all

I would like to now the answer to the following problem, if someone can
help.

Given a windows application client that's calling a web service (using
default credentials) the
web service gets invoked fine with the users credentials. However they
need
to call a second web service from the first one. When it hits the
second web
service, the web service (even though configured correctly) always
reports (401) Unauthorized.

Is there any possibility to do this or is impossible ?

Can someone give me a sample of Kerberos Implementation for this case ?

Is possible to implement any workaround for this problem ?

Thanks in advance

Manuel Brandão

Jun 16 '06 #2

This thread has been closed and replies have been disabled. Please start a new discussion.

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