Oracle查询优化日期运算实例详解_oracle_

 select a.hiredate 雇用日期, a.hiredate + 5 加5天, a.hiredate - 5 减5天, add_months(hiredate, 5) 加5个月, add_months(hiredate, -5) 减5个月, add_months(hiredate, 5 * 12) 加5年, add_months(hiredate, -5 * 12) 减5年 from emp a where rownum <=1; 雇用日期 加5天 减5天 加5个月 减5个月 加5年 减5年 ----------- ----------- ----------- ----------- ----------- ----------- ----------- 1980-12-17 1980-12-22 1980-12-12 1981-05-17 1980-07-17 1985-12-17 1975-12-17 

 select a.hiredate, a.hiredate - 5 / 24 / 60 / 60 减5秒, a.hiredate + 5 / 24 / 60 / 60 加5秒, a.hiredate - 5 / 24 / 60 减5分钟, a.hiredate + 5 / 24 / 60 加5分钟, a.hiredate - 5 / 24 减5小时, a.hiredate + 5 / 24 加5小时 from emp a where rownum <= 1; 

 SQL> select duration, duration * 24 间隔小时, duration * 24 * 60 间隔分钟, duration * 24 * 60 * 60 间隔秒 from (select (max(a.hiredate) - min(a.hiredate)) as duration from emp a where a.ename in ('ALLEN', 'WARD')) x; DURATION 间隔小时 间隔分钟 间隔秒 ---------- ---------- ---------- ---------- 2 48 2880 172800 

 select max_hd - min_hd 间隔天, months_between(max_hd, min_hd) 间隔月, months_between(max_hd, min_hd) /12 间隔年 from (select min(hiredate) min_hd, max(hiredate) max_hd from emp) x; 间隔天 间隔月 间隔年 ---------- ---------- ---------- 2348 77.1935483 6.43279569 

 with x as (select level lvl from dual connect by level <= (add_months(trunc(sysdate, 'y'), 12) - trunc(sysdate, 'y'))) select to_char(trunc(sysdate, 'y') + lvl - 1, 'DAY'), count(*) from x group by to_char(trunc(sysdate, 'y') + lvl - 1, 'DAY');