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Is it possible to create a report in JRS to list all test cases that have being reused in more than one test plan ?


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Joao Carusi (26313) | asked Dec 22 '16, 3:41 p.m.
Is it possible to create a report in JRS to list all test cases that have being reused in more than one test plan ?

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hani zaidi (7421) | answered Dec 23 '16, 10:48 a.m.
 Hi Joao,

Yes this report is possible with JRS. Select Test Plan from Quality Management artifacts, then create a trace ability link to related test cases, in the format section add a calculated values columns for test cases and limit 

You'll have a report in this format

Test Cases | Count of test linked test plans.

If you want to add a condition and not view the linked test plans count column, to view only reused test cases play with this count by editing the query manually. Try my query pasted below: (I am using JRS 6.0.2 with SQL as DB and DW as datasource)
//-- I've added a having clause for test cases having more than one linked test plan --//

SELECT T1.REFERENCE_ID,
       T1.NAME AS URL1_title,
       T1.URL AS URL1
FROM RIDW.VW_TESTCASE T1
INNER JOIN RIDW.VW_TESTPLAN_TESTCASE_LOOKUP LT1
ON (T1.TESTCASE_ID = LT1.TESTCASE_ID)
  INNER JOIN RIDW.VW_TESTPLAN T2
  ON (T2.TESTPLAN_ID = LT1.TESTPLAN_ID)
  
WHERE (T1.PROJECT_ID = 1 OR T1.PROJECT_ID = 0 OR T1.PROJECT_ID = 8 OR T1.PROJECT_ID = 7 OR T1.PROJECT_ID = 6 OR T1.PROJECT_ID = 5 OR T1.PROJECT_ID = 4 OR T1.PROJECT_ID = 9 OR T1.PROJECT_ID = 3 OR T1.PROJECT_ID = 2) AND
  (T2.PROJECT_ID = 1 OR T2.PROJECT_ID = 0 OR T2.PROJECT_ID = 8 OR T2.PROJECT_ID = 7 OR T2.PROJECT_ID = 6 OR T2.PROJECT_ID = 5 OR T2.PROJECT_ID = 4 OR T2.PROJECT_ID = 9 OR T2.PROJECT_ID = 3 OR T2.PROJECT_ID = 2) 
 AND
(T1.ISSOFTDELETED = 0 AND T2.ISSOFTDELETED = 0) AND
(T1.TESTCASE_ID <> -1 AND T1.TESTCASE_ID IS NOT NULL) AND
(T2.TESTPLAN_ID <> -1 AND T2.TESTPLAN_ID IS NOT NULL)GROUP BY T1.PROJECT_NAME,
         T1.REFERENCE_ID,
         T1.NAME,
         T1.URL having COUNT( DISTINCT T2.REFERENCE_ID) > 1


Hope it helps,
Regards,
Hani 
Joao Carusi selected this answer as the correct answer

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