SQL解惑1——interval关键字用法 .

时间:2021-12-02 11:36:00

做个例子描述吧,也许更易于理解。

准备:

1.建表

create table INTERVAL_TEST
(
  t_id     INTEGER not null,  --纯id
  t_date   VARCHAR2(40) not null,--时间,如果此处类型为date最好,不过现在很多应用都喜欢把时间建成varchar2
  t_flagCHAR(40)  --没什么业务含义,只是个标记
)

2.造数据

001        2011-01-01 12:00:00        1                                      
001        2011-01-02 12:00:00        1                                      
001        2011-01-03 12:00:00        1                                      
002        2011-01-01 12:00:00        1                                      
002        2011-01-03 12:00:00        1                                      
002        2011-01-05 12:00:00        1

3.写SQL:同一ID,如果时间连续两天flag为1,则把第二天...第N天的flag修改为0。

update interval_test
   set interval_test.flag= '0'
 where exists
 (select *
          from interval_test i
         where interval_test.t_id = i.t_id
           and to_date(interval_test.t_date, 'yyyy-mm-dd hh24:mi:ss') =
               to_date(i.t_date, 'yyyy-mm-dd hh24:mi:ss') + interval '1' day
)

--因为建表时时间字段为varchar2,所以这里需要to_date一下。

 

commit后,查询结果:

1 2011-01-01 12:00:00 1                                      
1 2011-01-02 12:00:00 0                                      
1 2011-01-03 12:00:00 0                                      
2 2011-01-01 12:00:00 1                                      
2 2011-01-03 12:00:00 1                                      
2 2011-01-05 12:00:00 1                                      

结论,interval为取间隔的含义,在这个SQL中,子查询的where条件中红色部分含义说白了就是:当A表时间 = B表时间 + 1天

也可以扩展使用为间隔多少小时:A.T_DATE = B.T_DATE+ interval '1' hour——即 A表时间 = B表时间 + 1小时,也可以为分钟、秒。

 

Oracle里有INTERVAL DAY TO SECOND类型,在建表时可以定义,只提供下表作为释义。

具体类型的使用方式参考:http://book.51cto.com/art/200812/103848.htm


时间间隔字面量

说明

INTERVAL ‘3’ DAY

时间间隔为3天

INTERVAL ‘2’ HOUR

时间间隔为2小时

INTERVAL ‘25’ MINUTE

时间间隔为25分钟

INTERVAL ‘45’ SECOND

时间间隔为45秒

INTERVAL ‘3 2’ DAY TO HOUR

时间间隔为3天零2小时

INTERVAL ‘3 2:25’ DAY TO MINUTE

时间间隔为3天零2小时25分

INTERVAL ‘3 2:25:45’ DAY TO SECOND

时间间隔为3天零2小时25分45秒

INTERVAL ‘123 2:25:45.12’ DAY(3)

TO SECOND(2)

时间间隔为123天零2小时25分45.12秒; 天的精度是3位数字,秒的小数部分的精度是2位数字

INTERVAL ‘3 2:00:45’ DAY TO SECOND

时间间隔为3天2小时0分45秒

INTERVAL ‘-3 2:25:45’ DAY TO SECOND

时间间隔为负数,值为3天零2小时25分45秒

INTERVAL ‘1234 2:25:45’ DAY(3)

TO SECOND

时间间隔无效,因为天的位数超过了指定的精度3

INTERVAL ‘123 2:25:45.123’ DAY

TO SECOND(2)

时间间隔无效,因为秒的小数部分的位数超过了指定的精度2

 

 

INTERVAL YEAR TO MONTH数据类型

Oracle语法:
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
----------------------------------------- -------- ----------------------------
A                                                  DATE
B                                                  DATE
C                                                  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;

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

1 row updated.

SQL> select * from bb;

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

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

1 row created.

SQL> select * from bb;

A         B             C                                                       
--------- ---------    --------------------------------------------    
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;

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

========================
今天来添加点新的东西![2008-07-26] 这部分东东来源:http://www.oraclefans.cn/forum/showblog.jsp?rootid=139

INTERVAL YEAR TO MONTH类型2个TIMESTAMP类型的时间差别。内部类型是182,长度是5。其中4个字节存储年份差异,存储的时候在差异上加了一个0X80000000的偏移量。一个字节存储月份的差异,这个差异加了60的偏移量。

SQL> ALTER TABLE TestTimeStamp ADD E INTERVAL YEAR TO MONTH;
SQL> update testTimeStamp set e=(select interval '5' year + interval '10' month year  from dual);

已更新3行。

SQL> commit;
提交完成。

SQL> select dump(e,16) from testTimeStamp;

DUMP(E,16)
---------------------------------------------
Typ=182 Len=5: 80,0,0,5,46
Typ=182 Len=5: 80,0,0,5,46
Typ=182 Len=5: 80,0,0,5,46

年:0X80000005-0X80000000=5
月:0x46-60=10

INTERVAL DAY TO SECOND数据类型

Oracle语法:
INTERVAL '{ integer | integer time_expr | time_expr }'
{ { DAY | HOUR | MINUTE } [ ( leading_precision ) ]
| SECOND [ ( leading_precision [, fractional_seconds_precision ] ) ] }
[ TO { DAY | HOUR | MINUTE | SECOND [ (fractional_seconds_precision) ] } ]

leading_precision值的范围是0到9, 默认是2. time_expr的格式为:HH[:MI[:SS[.n]]] or MI[:SS[.n]] or SS[.n], n表示微秒.
该类型与INTERVAL YEAR TO MONTH有很多相似的地方,建议先看INTERVAL YEAR TO MONTH再看该文.

范围值:
HOUR:    0 to 23
MINUTE: 0 to 59
SECOND: 0 to 59.999999999

eg:
INTERVAL '4 5:12:10.222' DAY TO SECOND(3)
表示: 4天5小时12分10.222秒

INTERVAL '4 5:12' DAY TO MINUTE
表示: 4天5小时12分

INTERVAL '400 5' DAY(3) TO HOUR
表示: 400天5小时, 400为3为精度,所以"DAY(3)", 注意默认值为2.

INTERVAL '400' DAY(3)
表示: 400天

INTERVAL '11:12:10.2222222' HOUR TO SECOND(7)
表示: 11小时12分10.2222222秒

INTERVAL '11:20' HOUR TO MINUTE
表示: 11小时20分

INTERVAL '10' HOUR
表示: 10小时

INTERVAL '10:22' MINUTE TO SECOND
表示: 10分22秒

INTERVAL '10' MINUTE
表示: 10分

INTERVAL '4' DAY
表示: 4天

INTERVAL '25' HOUR
表示: 25小时

INTERVAL '40' MINUTE
表示: 40分

INTERVAL '120' HOUR(3)
表示: 120小时

INTERVAL '30.12345' SECOND(2,4)    
表示: 30.1235秒, 因为该地方秒的后面精度设置为4, 要进行四舍五入.

INTERVAL '20' DAY - INTERVAL '240' HOUR = INTERVAL '10-0' DAY TO SECOND
表示: 20天 - 240小时 = 10天0秒

==================
该部分来源:http://www.oraclefans.cn/forum/showblog.jsp?rootid=140
INTERVAL DAY TO SECOND类型存储两个TIMESTAMP之间的时间差异,用日期、小时、分钟、秒钟形式表示。该数据类型的内部代码是183,长度位11字节:

l         4个字节表示天数(增加0X80000000偏移量)
l         小时、分钟、秒钟各用一个字节表示(增加60偏移量)
l         4个字节表示秒钟的小时差异(增加0X80000000偏移量)

以下是一个例子:

SQL> alter table testTimeStamp add f interval day to second ;

表已更改。

SQL> update testTimeStamp set f=(select interval '5' day + interval '10' second from dual);

已更新3行。

SQL> commit;

提交完成。

SQL> select dump(f,16) from testTimeStamp;

DUMP(F,16)

--------------------------------------------------------------------------------

Typ=183 Len=11: 80,0,0,5,3c,3c,46,80,0,0,0
Typ=183 Len=11: 80,0,0,5,3c,3c,46,80,0,0,0
Typ=183 Len=11: 80,0,0,5,3c,3c,46,80,0,0,0

日期:0X80000005-0X80000000=5

小时:60-60=0
分钟:60-60=0
秒钟:70-60=10
秒钟小数部分:0X80000000-0X80000000=0