2011年的每一天是周几?

SELECT   TO_DATE ('20110101', 'yyyymmdd') + ROWNUM - 1 rq,
         TO_CHAR (TO_DATE ('20110101', 'yyyymmdd') + ROWNUM - 1, 'DAY')
            theday
  FROM   (SELECT       ROWNUM
                FROM   DUAL
          CONNECT BY   ROWNUM <=
                            TO_DATE ('20111231', 'yyyymmdd')
                          - TO_DATE ('20110101', 'yyyymmdd')
                          + 1)

原文地址:https://www.cnblogs.com/tracy/p/2043261.html