How to create Association Block using scripting

[PHannah]

Using VBScript I am trying to create an association block between two exiting blocks. I can get the existing blocks and create standard relationships between them, but I can't work out how to create an Association Block. I have a reasonable amount of VBScripting experience and am using EA 12.1. Thanks.

[qwerty]

Not sure what the API offers, but the association has PDATA1 = Assoc-class.Object_ID and that assoc-class has NType=17. I would probably just set these directly using repository.execute (since I almost know what I'm doing). Too lazy now to see how these are mapped in the API.

q.

[PHannah]

Thanks qwerty, I can create a block and type it. If I am trying to create an association between BlkA and BlkB, the Association Block needs to have BlkA and BlkB as "Part Participants", And I think if I can work out how to add these as Part Participants, it might all work.

[qwerty]

You're mixing terms. Above you were talking about "Association Block" which mad me think you were talking about association classes. Now you come up with Parts. So I have no idea where you are targeting.

q.

[PHannah]

Yes, my mistake. When you create an association block through the user interface EA automatically adds each end (BlkA and BlkB) as part participants on the Association Block. I therefore thought if I could add the "part participants" , this would link the association block to the connector.
It turns out that you create a connector of type "association", and then create the association class. On the association class you then call CreateAssociationClass(con.ConnectorID) which creates the association.

   set p = Repository.GetPackageByID(22)
   set blkA = Repository.GetElementByID(2496)
   set blkB = Repository.GetElementByID(2503)
   set blkC = p.Elements.AddNew("NEW IB", "Block")
   
   set con = blkA.Connectors.AddNew("TEST", "Association")
   con.ClientID = blkA.ElementID
   con.SupplierID = blkB.ElementID
   con.Update

   blkC.CreateAssociationClass(con.ConnectorID)

Thanks for your help.

[qwerty]

By any means: are you using SysML?

q.

[PHannah]

Yes I am using SysML.

[qwerty]

Then please forget what I said above. I don't have SysML and only a rough idea about it's concept/realization in EA.

q.