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Differences From Artifact [7e6b6c54ff]:

To Artifact [c461b3bc6e]:


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				    (conc (string-intersperse
					   (map (lambda (varval)
						  (string-intersperse varval "="))
						row)
					   " ")
					  "\n"))
				  items)))
	  (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)))
	       (if (tests:match test-patts hed my-item-path required: required-tests) ;; (patt-list-match my-item-path item-patts)           ;; yes, we want to process this item, NOTE: Should not need this check here!
		   (let ((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)

	  ;; (debug:print-info 0 *default-log-port* "Test " (tests:testqueue-get-testname test-record) " is itemized but has no items")

	  ;; 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
	  ;; (loop (car newtal)(cdr newtal) reg reruns)
	  (if (null? tal)







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				    (conc (string-intersperse
					   (map (lambda (varval)
						  (string-intersperse varval "="))
						row)
					   " ")
					  "\n"))
				  items)))

          (let* ((items-in-testpatt
                  (filter
                   (lambda (my-itemdat)
                     (tests:match test-patts hed (item-list->path my-itemdat) required: required-tests))
                   items) ))
            (if (null? items-in-testpatt)
                (let ((test-id   (rmt:get-test-id run-id test-name "")))
                  (debug:print-info 0 *default-log-port* "Test " (tests:testqueue-get-testname test-record) " is itemized but has no items matching test pattern -- marking status ZERO_ITEMS")
                  (if test-id
                      (mt:test-set-state-status-by-id run-id test-id "NOT_STARTED" "ZERO_ITEMS" "This test has no items which match test pattern.")))
                
                (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
	  ;; (loop (car newtal)(cdr newtal) reg reruns)
	  (if (null? tal)