Oracle查询优化日期运算实例详解

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');