select to_date('2002-09-01','yyyy-mm-dd')-to_date('2002-08-19','yyyy-mm-dd') from dual;

解决方案 »

  1.   

    用Months_Between函数
       select months_between(to_date('2002.08.19','yyyy.mm.dd'),
          to_date('2002.09.01','yyyy.mm.dd')) from dual;
    实现得到2002-8-19到2002-9-1之间相差的天数.
      

  2.   

    15:01:04 jlanzpa817>select trunc(to_date('2002-9-1','yyyy-mm-dd') - to_date('2002-08-19',
    15:02:16   2  'yyyy-mm-dd')) from dual;TRUNC(TO_DATE('2002-9-1','YYYY-MM-DD')-TO_DATE('2002-08-19','YYYY-MM-DD'))
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                                                                            13已用时间:  00: 00: 00.10
      

  3.   

    Jugular_Phoenix(凤) : months_between返回是单位是月.