Bug 140150

Summary: It's impossible to reuse a same use case in a use cases diagram
Product: [Applications] umbrello Reporter: Nicolas Dumoulin <nicolas>
Component: generalAssignee: Oliver Kellogg <okellogg>
Status: RESOLVED FIXED    
Severity: wishlist    
Priority: NOR    
Version First Reported In: unspecified   
Target Milestone: ---   
Platform: Ubuntu   
OS: Linux   
Latest Commit: Version Fixed/Implemented In:
Sentry Crash Report:

Description Nicolas Dumoulin 2007-01-16 11:23:16 UTC 
Version:           1.5.5 (using KDE KDE 3.5.5)
Installed from:    Ubuntu Packages
OS:                Linux

It would be nice if I could use a same use case (same name) several times in a use cases diagram.

For example, in a diagram, if I have two actors that can make action after an authentification (that I want represent), the two authentification use cases couldn't have the same caption.
Comment 1 Oliver Kellogg 2007-04-18 22:04:27 UTC 
Confirm.
UML standard says UseCase relates one-to-many to UseCaseInstance
(In Umbrello, UseCaseInstance corresponds to UseCaseWidget.)
Comment 2 Oliver Kellogg 2007-04-18 22:35:58 UTC 
SVN commit 655592 by okellogg:

contentsDragEnterEvent(), case dt_UseCase: Limit widgetOnDiagram() test to ot_Actor.
BUG:140150


 M  +1 -0      ChangeLog  
 M  +1 -1      umbrello/umlview.cpp  


--- branches/KDE/3.5/kdesdk/umbrello/ChangeLog #655591:655592
@@ -16,6 +16,7 @@
 * Sequence diagram object size incorrect after toggling "Draw as Actor" (136869)
 * Incorrect Association Properties text (139872)
 * Buttons are not displayed (139913)
+* Impossible to reuse same use case in a use case diagram (140150)
 * Java 5 generics support (140669)
 * Associations not updated during move of class on diagram (140709)
 * Crash when deleting the link between a package and a class (141602)
--- branches/KDE/3.5/kdesdk/umbrello/umbrello/umlview.cpp #655591:655592
@@ -545,7 +545,7 @@
     bool bAccept = true;
     switch (diagramType) {
         case dt_UseCase:
-            if (widgetOnDiagram(id) ||
+            if ((widgetOnDiagram(id) && ot == ot_Actor) ||
                 (ot != ot_Actor && ot != ot_UseCase))
                 bAccept = false;
             break;
Comment 3 Nicolas Dumoulin 2007-04-20 18:00:12 UTC 
Great :-)

Thank you so much !