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Registering UPS On-Line tools web service
Message
De
25/03/2008 14:23:07
 
 
À
25/03/2008 11:27:52
Information générale
Forum:
Visual FoxPro
Catégorie:
Web Services
Versions des environnements
Visual FoxPro:
VFP 9 SP1
OS:
Windows XP SP2
Divers
Thread ID:
01305076
Message ID:
01305349
Vues:
8
>Viv, thanks for responding. Two questions:
>
>1. Trying to register the web service: To try your suggestion, pardon me for pressing you for more detail.
>The UPS test web site is https://wwwcie.ups.com/ups.app/xml/rate. I assume I should be referencing something like "https://wwwcie.ups.com/ups.app/xml/rate.wsdl" in the init statement. If I have a developer access code xxx, user name yyy and password zzz, where would you put them?
>
>2. The UPS documentation says the UPS system wants to receive two concatenated XML documents. The first is an access request with the access code-user-pass info. The second is the actual rate request document. If I build the XML documents myself, is there any way I can initialize the service and send the file with the documents? The UPS documentation suggested this Visual Studio (Visual Basic) formulation:
>
>
>' create the object that manages the communication
>Dim oXMLHttp As XMLHTTP
>Set oXMLHttp = New XMLHTTP
>' prepare the HTTP POST request
>oXMLHttp.open "POST", "https://www.server.com/path", False
>oXMLHttp.setRequestHeader "Content-Type", _
>"application/x-www-form-urlencoded"
>' send the request
>oXMLHttp.send requestString
>' server's response will be available in oXMLHttp.responseXML
>
>
>Anything analogous in VFP? Thanks for your trouble.

Hi,
Looks like the application is handling its own authentication so my suggestion is not relevant. Sergey gave you the info you should need but, JIC, here's a full translation:
' create the object that manages the communication
oXMLHttp = CREATEOBJECT("Microsoft.xmlhttp")
' prepare the HTTP POST request
oXMLHttp.open ("POST", "https://www.server.com/path", .F.)
oXMLHttp.setRequestHeader ("Content-Type","application/x-www-form-urlencoded")
' send the request
oXMLHttp.send (requestString)
' server's response will be available in oXMLHttp.responseXML
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