The full error I am getting is:

   An error occurred while creating a new DataLayer for the business object.

   The DataSourceKey [XXX] is not defined.



This error is generated on the call to:

SFSPreferencesBO.RetrieveSecurityPreferences()



Everything works fine* when I don't add a named datasourcekey but leave the default "" :

SecurityBasics.SecurityDataSourceKey = "XXX"



I tried adding this but it didnt help:

DataLayer.DataSources.Add(New SqlDataSourceItem("XXX", "Data Source=IPADDRESS\SQLNAME;Initial Catalog=DBNAME;Integrated Security=True;Persist Security Info=False;Asynchronous Processing=True"))



I have a security project with a security key that matches.



I have my security tables in the same database as my application files, added via the DDT and deployed to the SQL Server.



I am sure I am doing something dumb, as this is the first chance i have had to implement the security module.



Keith



* when I left the SecurityDataSourceKey = "", I could only see the users within my app, when I called up the SecurityDialog from within VS, I saw no users.




Keith Chisarik

The above should say "permissions" not "roles" as far as the design time questions.



<< still hoping for an edit capability on ones own posts>>

Keith Chisarik

Then you do not need to set the SecurityDataSourceKey as the security key will reference your application database or default connection.

The first question is that did you deploy your users to your application database?  By default this will be empty.  Also, make sure you call the SecurityBasics.SetSecurityKeyAndVectorForUserAuthentication("yourPresharedKey") in the InitApplication otherwise you will not see any user.  In fact, you will get an error.

No.  You cannot create permissions as run-time.  This would not make sense since the end-users should not be able to create permissions.  If the end-user created the permissions they would never be implemented within the code (or tested on by you the developer).  They would just be "hollow" permissions that did not do anything.