@@ -1501,20 +1501,20 @@
                 (for-each (lambda (my-itemdat)
                             (let* ((new-test-record (let ((newrec (make-tests:testqueue)))
                                                       (vector-copy! test-record newrec)
                                                       newrec))
                                    (my-item-path (item-list->path my-itemdat))
+
                                    (newtestname (db:test-make-full-name hed my-item-path)))    ;; test names are unique on testname/item-path
                               (tests:testqueue-set-items!     new-test-record #f)
                             (tests:testqueue-set-itemdat!   new-test-record my-itemdat)
                             (tests:testqueue-set-item_path! new-test-record my-item-path)
                             (hash-table-set! test-records newtestname new-test-record)
                             (set! tal (append tal (list newtestname)))))  ;; since these are itemized create new test names testname/itempath
                           items-in-testpatt)))
                           
             
-
 
 	  ;; At this point we have possibly added items to tal but all must be handed off to 
 	  ;; INNER COND logic. I think loop without rotating the queue 
 	  ;; (loop hed tal reg reruns))
 	  ;; (let ((newtal (append tal (list hed))))  ;; We should discard hed as it has been expanded into it's items? Yes, but only if this *is* an itemized test