DIIOP Access and debugging Sam N Waymouth 29.Nov.02 04:11 AM a Web browser Applications Development All ReleasesWindows 2000
Thanks for you information on the DIIOP topic, It's saved me a lot of time and I'm sure a lot of other people feel the same!
However! I have a different problem, using the following senario...
A PC with WebSphere App Studio 4.0.3, a single HTML form, and single servlet, a separate Domino R6 box (yes a real 2 teir development environment! gosh), WIN2K throughout.
The idea is that I can build a single WebSphere app that will portal four different notesdb's together, provide the super worflow and all the other things WebSphere can do.
After getting all the initial security stuff fixed - IIOP internet site doc, web internet site doc, db ACL etc etc. I can test my application in WASDS but there trick as you can see from my server log...
29/11/2002 08:54:44 DIIOP Server: 192.168.1.10 connected
29/11/2002 08:54:44 DIIOP Server: Executing createSession
29/11/2002 08:54:44 DIIOP Session SN00030F506: Opened for Sam Waymouth/DSL (v1.9)
29/11/2002 08:54:47 DIIOP Server: 18.104.22.168 connected
29/11/2002 08:54:47 DIIOP Server: Rejected access by Sam Waymouth/DSL (192.168.1.10) through second connection
29/11/2002 08:54:48 DIIOP Server: Rejected access by Sam Waymouth/DSL (192.168.1.10) through second connection
I terms of code what I have is...
Session s = NotesFactory.createSession(host, user, pwd);
String p = s.getPlatform();
out.println("<h2>" + p + "</h2>");
Database db = s.getDatabase(null, dbname, false);
String title = db.getTitle();
out.println("<h2>" + title + "</h2>");
As you can see from the log, the create session runs fine and i get back the session platform, what bombs out is instantiating the database ojbect which throws...
What is getting to me is that I have the session open and it has permissions because I can call its methods, and the database is anonymous access with editor in the maximum internet access section. So the question is why would it start a second connection when the getdatabase method is called and why would the credentials not be valid?