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  • INTERVAL YEAR TO MONTH数据类型

    INTERVAL 'integer [- integer]' {YEAR | MONTH} [(precision)][TO {YEAR | MONTH}]

    该数据类型常用来表示一段时间差, 注意时间差只精确到年和月. precision为年或月的精确域, 有效范围是0到9, 默认值为2.

    eg:
    INTERVAL '123-2' YEAR(3) TO MONTH   
    表示: 123年2个月, "YEAR(3)" 表示年的精度为3, 可见"123"刚好为3为有效数值, 如果该处YEAR(n), n<3就会出错, 注意默认是2.

    INTERVAL '123' YEAR(3)
    表示: 123年0个月

    INTERVAL '300' MONTH(3)
    表示: 300个月, 注意该处MONTH的精度是3啊.

    INTERVAL '4' YEAR   
    表示: 4年, 同 INTERVAL '4-0' YEAR TO MONTH 是一样的

    INTERVAL '50' MONTH   
    表示: 50个月, 同 INTERVAL '4-2' YEAR TO MONTH 是一样

    INTERVAL '123' YEAR   
    表示: 该处表示有错误, 123精度是3了, 但系统默认是2, 所以该处应该写成 INTERVAL '123' YEAR(3) 或"3"改成大于3小于等于9的数值都可以的

    INTERVAL '5-3' YEAR TO MONTH + INTERVAL '20' MONTH =
    INTERVAL '6-11' YEAR TO MONTH
    表示: 5年3个月 + 20个月 = 6年11个月

    与该类型相关的函数:
    NUMTODSINTERVAL(n, 'interval_unit')
    将n转换成interval_unit所指定的值, interval_unit可以为: DAY, HOUR, MINUTE, SECOND
    注意该函数不可以转换成YEAR和MONTH的.

    NUMTOYMINTERVAL(n, 'interval_unit')
    interval_unit可以为: YEAR, MONTH

    eg: (Oracle Version 9204, RedHat Linux 9.0)
    SQL> select numtodsinterval(100,'DAY') from dual;

    NUMTODSINTERVAL(100,'DAY')                                                    
    ---------------------------------------------------------------------------   
    +000000100 00:00:00.000000000                                                 

    SQL> c/DAY/SECOND
    1* select numtodsinterval(100,'SECOND') from dual
    SQL> /

    NUMTODSINTERVAL(100,'SECOND')                                                 
    ---------------------------------------------------------------------------   
    +000000000 00:01:40.000000000                                                 

    SQL> c/SECOND/MINUTE
    1* select numtodsinterval(100,'MINUTE') from dual
    SQL> /

    NUMTODSINTERVAL(100,'MINUTE')                                                 
    ---------------------------------------------------------------------------   
    +000000000 01:40:00.000000000                                                 

    SQL> c/MINUTE/HOUR
    1* select numtodsinterval(100,'HOUR') from dual
    SQL> /

    NUMTODSINTERVAL(100,'HOUR')                                                   
    ---------------------------------------------------------------------------   
    +000000004 04:00:00.000000000                                                 

    SQL> c/HOUR/YEAR
    1* select numtodsinterval(100,'YEAR') from dual
    SQL> /
    select numtodsinterval(100,'YEAR') from dual
                              *
    ERROR at line 1:
    ORA-01760: illegal argument for function

    SQL> select numtoyminterval(100,'year') from dual;

    NUMTOYMINTERVAL(100,'YEAR')                                                   
    ---------------------------------------------------------------------------   
    +000000100-00                                                                 

    SQL> c/year/month
    1* select numtoyminterval(100,'month') from dual
    SQL> /

    NUMTOYMINTERVAL(100,'MONTH')                                                  
    ---------------------------------------------------------------------------   
    +000000008-04                                                                 


    时间的计算:
    SQL> select to_date('1999-12-12','yyyy-mm-dd') - to_date('1999-12-01','yyyy-mm-dd') from dual;

    TO_DATE('1999-12-12','YYYY-MM-DD')-TO_DATE('1999-12-01','YYYY-MM-DD')         
    ---------------------------------------------------------------------         
                                                                      11         
    -- 可以相减的结果为天.

    SQL> c/1999-12-12/1999-01-12
    1* select to_date('1999-01-12','yyyy-mm-dd') - to_date('1999-12-01','yyyy-mm-dd') from dual
    SQL> /

    TO_DATE('1999-01-12','YYYY-MM-DD')-TO_DATE('1999-12-01','YYYY-MM-DD')         
    ---------------------------------------------------------------------         
                                                                    -323         
    -- 也可以为负数的

    SQL> c/1999-01-12/2999-10-12
    1* select to_date('2999-10-12','yyyy-mm-dd') - to_date('1999-12-01','yyyy-mm-dd') from dual
    SQL> /

    TO_DATE('2999-10-12','YYYY-MM-DD')-TO_DATE('1999-12-01','YYYY-MM-DD')         
    ---------------------------------------------------------------------         
                                                                  365193         

    下面看看INTERVAL YEAR TO MONTH怎么用.
    SQL> create table bb(a date, b date, c interval year(9) to month);

    Table created.

    SQL> desc bb;
    Name                                     Null?   Type
    ----------------------------------------- -------- ----------------------------
                                                    DATE
                                                    DATE
                                                    INTERVAL YEAR(9) TO MONTH

    SQL> insert into bb values(to_date('1985-12-12', 'yyyy-mm-dd'), to_date('1984-12-01','yyyy-mm-dd'), null)

    1 row created.

    SQL> select * from bb;

                                                                             
    --------- ---------                                                           
                                                                                
    ---------------------------------------------------------------------------   
    12-DEC-85 01-DEC-84                                                           
                                                                                  
                                                                                   
    SQL> update bb set c = numtoyminterval(a-b, 'year');

    1 row updated.

    SQL> select * from bb;

                                                                              
    --------- ---------                                                            
                                                                                 
    ---------------------------------------------------------------------------    
    12-DEC-85 01-DEC-84                                                            
    +000000376-00                                                                  
                                                                                   
    -- 直接将相减的天变成年了, 因为我指定变成年的
    SQL> select a-b, c from bb;

          A-B                                                                     
    ----------                                                                     
                                                                                 
    ---------------------------------------------------------------------------    
          376                                                                     
    +000000376-00                                                                  
                                                                                   

    SQL> insert into bb values(null,null,numtoyminterval(376,'month'));

    1 row created.

    SQL> select * from bb;

                                                                            
    --------- ---------   --------------------------------------------    
    12-DEC-85 01-DEC-84   +000000376-00                                                                  
                            +000000031-04                                         

    SQL> insert into bb values ( null,null, numtoyminterval(999999999,'year'));

    1 row created.

    SQL> select * from bb;

                                                      
    ---------  ---------    ---------------------------------------------------------------------    
    12-DEC-85  01-DEC-84  +000000376-00                                                                  
                             +000000031-04
                             +999999999-00         

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  • 原文地址:https://www.cnblogs.com/einyboy/p/2624886.html
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