[20191219]oracle timestamp数据类型的存储.txt
--//在优化一个项目,里面的日期类型基本选择都是timestamp类型.自己在以前工作中很少遇到这种类型.
--//倒是遇到几例定义date类型,而绑定变量是timestamp的情况,导致出现隐式转换.
--//顺便说一下,我个人一直任何使用date类型足以,不知道在什么情况下需要这么高精度的时间类型.
--//我仅仅想到一种可能,假设1秒内存在上百上千的业务,里面的日期精度这个时候也许很重要.
1.环境:
SCOTT@book> @ ver1
PORT_STRING VERSION BANNER
------------------------------ -------------- --------------------------------------------------------------------------------
x86_64/Linux 2.4.xx 11.2.0.4.0 Oracle Database 11g Enterprise Edition Release 11.2.0.4.0 - 64bit Production
2.测试:
SCOTT@book> create table t(id number,cdate timestamp);
Table created.
SCOTT@book> insert into t values (1,sysdate);
1 row created.
SCOTT@book> insert into t values (2,systimestamp);
1 row created.
SCOTT@book> commit ;
Commit complete.
SCOTT@book> select id,dump(cdate,16) c50,dump(cdate,10) c60 ,cdate from t;
ID C50 C60 CDATE
-- -------------------------------------------------- ----------------------------------------------------- ---------------------------
1 Typ=180 Len=7: 78,77,c,13,10,2b,8 Typ=180 Len=7: 120,119,12,19,16,43,8 2019-12-19 15:42:07.000000
2 Typ=180 Len=11: 78,77,c,13,10,2b,f,1e,e6,db,b0 Typ=180 Len=11: 120,119,12,19,16,43,15,30,230,219,176 2019-12-19 15:42:14.518446
--//可以发现timestamp类型如果没有秒后面的精度,仅仅占用7个字节.对于格式很容易猜测出来.
--//前1,2位各减去100,对应就是2019年. 月日不可能出现0的情况,也就是相互对应.而时分秒有0的出现,在原来的时间基础上+1.
--//比如时分秒=15:42:07,对应的编码(10进制)就是16 43 8.
--//剩下的难点就是秒后面的数值.比如1e,e6,db,b0如何表示.518446.
SCOTT@book> @ 16to10 1ee6dbb0
16 to 10 DEC
------------
518446000
--//可以猜测相当于0.518446*power(10,9)转换为16进制就是对应编码.
--//0x10000000 = 268435456,插入秒后这样的时间看看?
SCOTT@book> insert into t values (3,to_timestamp('2019-12-20 0:0:0.268435456', 'syyyy-mm-dd hh24:mi:ss.ff9'));
1 row created.
SCOTT@book> commit ;
Commit complete.
SCOTT@book> select id,dump(cdate,16) c50,dump(cdate,10) c60 ,cdate from t;
ID C50 C60 CDATE
--- ---------------------------------------------- ----------------------------------------------------- --------------------------
1 Typ=180 Len=7: 78,77,c,13,10,2b,8 Typ=180 Len=7: 120,119,12,19,16,43,8 2019-12-19 15:42:07.000000
2 Typ=180 Len=11: 78,77,c,13,10,2b,f,1e,e6,db,b0 Typ=180 Len=11: 120,119,12,19,16,43,15,30,230,219,176 2019-12-19 15:42:14.518446
3 Typ=180 Len=11: 78,77,c,14,1,1,1,f,ff,fe,38 Typ=180 Len=11: 120,119,12,20,1,1,1,15,255,254,56 2019-12-20 00:00:00.268435
--//噢明白了缺省timestamp类型是保留6位,除非明确精度9.
SCOTT@book> @ desc t
Name Null? Type
------ -------- -------------
1 ID NUMBER
2 CDATE TIMESTAMP(6)
SCOTT@book> alter table t add ( ccdate timestamp(9));
Table altered.
SCOTT@book> insert into t (id ,ccdate) values (4,to_timestamp('2019-12-20 0:0:0.268435456', 'syyyy-mm-dd hh24:mi:ss.ff9'));
1 row created.
SCOTT@book> insert into t (id ,ccdate) values (5,to_timestamp('2019-12-20 0:0:0.999999999', 'syyyy-mm-dd hh24:mi:ss.ff9'));
1 row created.
SCOTT@book> commit ;
Commit complete.
SCOTT@book> select id,dump(ccdate,16) c50,dump(ccdate,10) c60 ,ccdate from t where id>=4;
ID C50 C60 CCDATE
--- -------------------------------------------- -------------------------------------------------- -----------------------------
4 Typ=180 Len=11: 78,77,c,14,1,1,1,10,0,0,0 Typ=180 Len=11: 120,119,12,20,1,1,1,16,0,0,0 2019-12-20 00:00:00.268435456
5 Typ=180 Len=11: 78,77,c,14,1,1,1,3b,9a,c9,ff Typ=180 Len=11: 120,119,12,20,1,1,1,59,154,201,255 2019-12-20 00:00:00.999999999
--//999999999 = 0x3b9ac9ff.
--//主要看看后面的0是否没有,可以发现还是存在的,也就是要么占用7字节,要么占用11字节.
--//补充一点实际上精度自己可以控制,最大是9.
SCOTT@book> alter table t add ( cccdate timestamp(8));
Table altered.
SCOTT@book> alter table t add ( xdate timestamp(10));
alter table t add ( xdate timestamp(10))
*
ERROR at line 1:
ORA-30088: datetime/interval precision is out of range
SCOTT@book> @ desc t
Name Null? Type
--------- -------- -------------
1 ID NUMBER
2 CDATE TIMESTAMP(6)
3 CCDATE TIMESTAMP(9)
4 CCCDATE TIMESTAMP(8)
来自 “ ITPUB博客 ” ,链接:https://blog.itpub.net/267265/viewspace-2669500/ ,如需转载,请注明出处,否则将追究法律责任。