原文鏈接:http://blog.csdn.net/ld422586546/article/details/9626921/
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獲取一段時間內所有的小時
SELECT to_date('2013-07-01 12', 'yyyy-mm-dd hh24') + (ROWNUM - 1) / 24 sdate FROM dual
CONNECT BY ROWNUM <= (to_date('2013-07-02 22', 'yyyy-mm-dd hh24') - to_date('2013-07-01 12', 'yyyy-mm-dd hh24')) * 24 + 1;
獲取一段時間內所有的天
SELECT to_date('2013-07-01', 'yyyy-mm-dd') + ROWNUM - 1 sdate FROM dual?
CONNECT BY ROWNUM <= to_date('2013-07-05', 'yyyy-mm-dd') - to_date('2013-07-01', 'yyyy-mm-dd') + 1;
獲取一段時間內所有的月
SELECT to_char(add_months(to_date('2013-01', 'yyyy-mm'), ROWNUM - 1), 'yyyy-mm') day_id FROM dual
CONNECT BY ROWNUM <= months_between(to_date('2013-07', 'yyyy-mm'), to_date('2013-01', 'yyyy-mm')) + 1;
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