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oracle时间相减(国外英语资料).doc

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oracle时间相减(国外英语资料)

oracle时间相减(国外英语资料) The problem of subtracting time from SQL in Oracle 2011-01-07 09:50 --MONTHS_BETWEEN (date2, date1) Give the month of date2-date1 SQL, select, months_between (19-12, -1999,19-3, -1999) mon_between, from, dual; MON_BETWEEN At Nine SQLselect, months_between (to_date (2000.05.20,yyyy.mm.dd), to_date (2005.05.20,yyyy.dd)) mon_betw, from, dual; MON_BETW - -60 Oracle computing time difference expression - get the difference between seconds of two seconds Select, ceil ((To_date (2008-05-02, 00:00:00,yyyy-mm-dd, hh24-mi-ss) - To_date (2008-04-30, 23:59:59,yyyy-mm-dd, hh24-mi-ss)) * 24 * 60 * 60 * 1000) difference in the number of seconds FROM DUAL; * Difference between seconds Dear sirsre Eighty-six million four hundred and one thousand 1 row selected * / - get the difference between seconds in two seconds Select, ceil ((To_date (2008-05-02, 00:00:00,yyyy-mm-dd, hh24-mi-ss) - To_date (2008-04-30, 23:59:59,yyyy-mm-dd, hh24-mi-ss)) * 24 * 60 * 60) difference in seconds FROM DUAL; * Difference seconds Dear sirsre Eighty-six thousand four hundred and one 1 row selected * / - get the difference between minutes of two minutes Select, ceil ((To_date (2008-05-02, 00:00:00,yyyy-mm-dd, hh24-mi-ss) - To_date (2008-04-30, 23:59:59,yyyy-mm-dd, hh24-mi-ss)) * 24 * 60) phase difference minutes FROM DUAL; * Minutes of difference Dear sirsre One thousand four hundred and forty-one 1 row selected * / - get the difference between hours of two hours Select, ceil ((To_date (2008-05-02, 00:00:00,yyyy-mm-dd, hh24-mi-ss) - To_date (2008-04-30, 23:59:59,yyyy-mm-dd, hh24-mi-ss)) * 24) difference hours FROM DUAL; * Hours of difference Dear sirsre Twenty-five 1 row selected * / - get the difference between two hours Select, ceil ((To_date (2008-05-02, 00:00:00,yyyy-mm-dd, hh24-mi-ss) - To_date (2008-04-30, 23:59:59,yyyy-mm-dd, hh24-mi-ss)) the difference is FROM DUAL; * return int Dear sirsre Two 1 row selected * / ---------------------------------------- Note: the number of days ca