NTERVAL YEAR TO MONTH數據類型 oracle語法: INTERVAL integer [- integer] {YEAR | MONTH} [(precision)][TO {YEAR | MONTH}] 該數據類型常用來表示一段時間差, 注意時間差只精確到年和月. precision為年或月的精確域, 有效范圍是0到9, 默認值為2. eg: INTER
NTERVAL YEAR TO MONTH數據類型
Oracle語法:
INTERVAL ‘integer [- integer]’ {YEAR | MONTH} [(precision)][TO {YEAR | MONTH}]
該數據類型常用來表示一段時間差, 注意時間差只精確到年和月. precision為年或月的精確域, 有效范圍是0到9, 默認值為2.
eg:
INTERVAL ‘123-2’ YEAR(3) TO MONTH????
表示: 123年2個月, “YEAR(3)” 表示年的精度為3, 可見”123″剛好為3為有效數值, 如果該處YEAR(n), n
INTERVAL ‘123’ YEAR(3)
表示: 123年0個月
INTERVAL ‘300’ MONTH(3)
表示: 300個月, 注意該處MONTH的精度是3啊.
INTERVAL ‘4’ YEAR????
表示: 4年, 同 INTERVAL ‘4-0’ YEAR TO MONTH 是一樣的
INTERVAL ’50’ MONTH????
表示: 50個月, 同 INTERVAL ‘4-2’ YEAR TO MONTH 是一樣
INTERVAL ‘123’ YEAR????
表示: 該處表示有錯誤, 123精度是3了, 但系統默認是2, 所以該處應該寫成 INTERVAL ‘123’ YEAR(3) 或”3″改成大于3小于等于9的數值都可以的
INTERVAL ‘5-3′ YEAR TO MONTH + INTERVAL ’20’ MONTH =
INTERVAL ‘6-11’ YEAR TO MONTH
表示: 5年3個月 + 20個月 = 6年11個月
與該類型相關的函數:
NUMTODSINTERVAL(n, ‘interval_unit’)
將n轉換成interval_unit所指定的值, interval_unit可以為: DAY, HOUR, MINUTE, SECOND
注意該函數不可以轉換成YEAR和MONTH的.
NUMTOYMINTERVAL(n, ‘interval_unit’)
interval_unit可以為: 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?????????????????????????????????????????????????????????????????
時間的計算:
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??????????
— 可以相減的結果為天.
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??????????
— 也可以為負數的
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??????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????
— 直接將相減的天變成年了, 因為我指定變成年的
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???????????????????????????????????????????????????????????????
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INTERVAL YEAR TO MONTH類型2個TIMESTAMP類型的時間差別。內部類型是182，長度是5。其中4個字節存儲年份差異，存儲的時候在差異上加了一個0X80000000的偏移量。一個字節存儲月份的差異，這個差異加了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數據類型
Oracle語法:
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值的范圍是0到9, 默認是2. time_expr的格式為:HH[:MI[:SS[.n]]] or MI[:SS[.n]] or SS[.n], n表示微秒.
該類型與INTERVAL YEAR TO MONTH有很多相似的地方,建議先看INTERVAL YEAR TO MONTH再看該文.
范圍值:
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小時12分10.222秒
INTERVAL ‘4 5:12’ DAY TO MINUTE
表示: 4天5小時12分
INTERVAL ‘400 5’ DAY(3) TO HOUR
表示: 400天5小時, 400為3為精度,所以”DAY(3)”, 注意默認值為2.
INTERVAL ‘400’ DAY(3)
表示: 400天
INTERVAL ’11:12:10.2222222′ HOUR TO SECOND(7)
表示: 11小時12分10.2222222秒
INTERVAL ’11:20′ HOUR TO MINUTE
表示: 11小時20分
INTERVAL ’10’ HOUR
表示: 10小時
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小時
INTERVAL ’40’ MINUTE
表示: 40分
INTERVAL ‘120’ HOUR(3)
表示: 120小時
INTERVAL ‘30.12345’ SECOND(2,4)????
表示: 30.1235秒, 因為該地方秒的后面精度設置為4, 要進行四舍五入.
INTERVAL ’20’ DAY – INTERVAL ‘240’ HOUR = INTERVAL ’10-0′ DAY TO SECOND
表示: 20天 – 240小時 = 10天0秒
==================
該部分來源：http://www.Oraclefans.cn/forum/showblog.jsp?rootid=140
INTERVAL DAY TO SECOND類型存儲兩個TIMESTAMP之間的時間差異，用日期、小時、分鐘、秒鐘形式表示。該數據類型的內部代碼是183，長度位11字節：
l???????? 4個字節表示天數（增加0X80000000偏移量）
l???????? 小時、分鐘、秒鐘各用一個字節表示（增加60偏移量）
l???????? 4個字節表示秒鐘的小時差異（增加0X80000000偏移量）
以下是一個例子：
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
小時：60－60＝0
分鐘：60－60＝0
秒鐘：70－60＝10
秒鐘小數部分：0X80000000-0X80000000=0
