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두 날짜의 차이를 년월일 시분초로 표현하기

· 2008-04-22 (화) 02:13:30 · 5143
-- 년월일
DECLARE
        d1 DATE;
        d2 DATE;
        years NUMBER;
        months  NUMBER;
        days    NUMBER;
BEGIN
        d1 := to_date( '09-JUL-2007', 'DD-MON-YYYY' );
        d2 := to_date( '24-JUL-1995', 'DD-MON-YYYY' );

        years := trunc( months_between( d1, d2 ) / 12 );
        months := mod( trunc( months_between( d1, d2 ) ), 12 );
        days := d1 - add_months(d2,trunc( months_between( d1, d2 ) ));

        dbms_output.put_line( 'years = ' || years);
        dbms_output.put_line( 'months = ' || months);
        dbms_output.put_line( 'days = ' || days);
END;
/


<결과>
years = 11
months = 11
days = 15




-- 년월일시분초
DECLARE
        d1 DATE;
        d2 DATE;
        years NUMBER;
        months  NUMBER;
        days    NUMBER;
        hours NUMBER;
        minutes NUMBER;
        seconds NUMBER;
       
BEGIN
        d1 := to_date( '09-JUL-2007 23:10:43', 'DD-MON-YYYY HH24:MI:SS' );
        d2 := to_date( '24-JUL-1995 10:09:40', 'DD-MON-YYYY HH24:MI:SS' );

        years := trunc( months_between( d1, d2 ) / 12 );
        months := mod( trunc( months_between( d1, d2 ) ), 12 );
        days := trunc(d1 - add_months(d2,trunc( months_between( d1, d2 ) )));
        hours := trunc( mod( (d1-d2)*24, 24 ) );
        minutes := trunc( mod( (d1-d2)*24*60, 60 ) );
        seconds := trunc( mod( (d1-d2)*24*60*60, 60 ) );

        dbms_output.put_line( 'years = ' || years);
        dbms_output.put_line( 'months = ' || months);
        dbms_output.put_line( 'days = ' || days);
        dbms_output.put_line( 'hours = ' || hours);
        dbms_output.put_line( 'minutes = ' || minutes);
        dbms_output.put_line( 'seconds = ' || seconds);
       
END;
/


<결과>
years = 11
months = 11
days = 15
hours = 13
minutes = 1
seconds = 3
[이 게시물은 관리자님에 의해 2011-10-31 16:47:36 Oracle에서 이동 됨]
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