DB2中产生唯一值三种方式

一.Genearate_unique函数

[db2inst1@xifenfei ~]$ db2 "create table xifenfei.t_gu(custno char(13) for bit data,
> custname varchar(16))"
DB20000I  The SQL command completed successfully.
[db2inst1@xifenfei ~]$  db2 "insert into xifenfei.t_gu values
> (generate_unique(),'www.xifenfei.com'),(generate_unique(),'xifenfei')"
DB20000I  The SQL command completed successfully.
[db2inst1@xifenfei ~]$ db2 "select * from xifenfei.t_gu"

CUSTNO                        CUSTNAME        
----------------------------- ----------------
x'20120403054630527862000000' www.xifenfei.com
x'20120403054630527940000000' xifenfei        

  2 record(s) selected.

generate_unique是按照国际标准时间(UTC)生成的当前时间戳加上当前数据库分区号,包含13个字节的字符串。如果调整了系统时间可能会出现重复

二.序列(Sequence)

[db2inst1@xifenfei ~]$ db2 "create sequence xifenfei.xff_seq
> as bigint
> start with 1
> increment by 1
> no maxvalue
> cycle
> cache 10"
DB20000I  The SQL command completed successfully.
[db2inst1@xifenfei ~]$ db2 "create table xifenfei.t_seq(xff_id bigint,
> custname varchar(16))"
DB20000I  The SQL command completed successfully.
[db2inst1@xifenfei ~]$ db2 "insert into xifenfei.t_seq values(nextval 
> for xifenfei.xff_seq,'www.xifenfei.com'), (nextval for xifenfei.xff_seq,'xifenfei')"
DB20000I  The SQL command completed successfully.
[db2inst1@xifenfei ~]$ db2 "select * from xifenfei.t_seq"

XFF_ID               CUSTNAME        
-------------------- ----------------
                   1 www.xifenfei.com
                   2 xifenfei        

  2 record(s) selected.

和ORACLE的sequence基本相同,只是在oracle中是sequence.nextval这里改为了nextnvl for seqence

三.自增字段

--1.generated always as identity方式(不能人工干预插入数值)
[db2inst1@xifenfei ~]$ db2 "create table xifenfei.t_ind_a(xff_id bigint not null generated always 
> as identity(start with 10,increment by 1,minvalue 10,maxvalue 1000000,no cycle,nocache,no order),
> custname varchar(16))"
DB20000I  The SQL command completed successfully.
[db2inst1@xifenfei ~]$ db2 "insert into xifenfei.t_ind_a(custname) values('www.xifenfei.com')"
DB20000I  The SQL command completed successfully.
--指定值插入失败
[db2inst1@xifenfei ~]$ db2 "insert into xifenfei.t_ind_a(xff_id,custname) values(11,'XIFENFEI')"
DB21034E  The command was processed as an SQL statement because it was not a 
valid Command Line Processor command.  During SQL processing it returned:
SQL0798N  A value cannot be specified for column "XFF_ID" which is defined as 
GENERATED ALWAYS.  SQLSTATE=428C9
[db2inst1@xifenfei ~]$ db2 "select * from xifenfei.t_ind_a"

XFF_ID               CUSTNAME        
-------------------- ----------------
                  10 www.xifenfei.com

  1 record(s) selected.

--1.generated by default as identity方式(可以人工干预插入数值)
[db2inst1@xifenfei ~]$ db2 "create table xifenfei.t_ind_d(xff_id bigint not null generated by default 
> as identity (start with 10,increment by 1,minvalue 10,maxvalue 1000000,no cycle,nocache,no order),
> custname varchar(16))"
DB20000I  The SQL command completed successfully.
[db2inst1@xifenfei ~]$ db2 "insert into xifenfei.t_ind_d(custname) values('www.xifenfei.com')"
DB20000I  The SQL command completed successfully.
--指定值插入成功
[db2inst1@xifenfei ~]$ db2 "insert into xifenfei.t_ind_d(xff_id,custname) values(11,'XIFENFEI')"
DB20000I  The SQL command completed successfully.
[db2inst1@xifenfei ~]$ db2 "select * from xifenfei.t_ind_d"

XFF_ID               CUSTNAME        
-------------------- ----------------
                  10 www.xifenfei.com
                  11 XIFENFEI        

  2 record(s) selected.

这个功能和sql server/mysql的自增长列很相似,给出了两种方式选择,使得比它们更加灵活

发表在 DB2 | 评论关闭

ORACLE 十进制与二进制互转函数

十进制转换二进制

CREATE OR REPLACE FUNCTION NUMBER_TO_BIT(V_NUM NUMBER) 
RETURN VARCHAR IS V_RTN VARCHAR(8);--注意返回列长度
  V_N1  NUMBER;
  V_N2  NUMBER;
BEGIN
V_N1 := V_NUM;
    LOOP
      V_N2  := MOD(V_N1, 2);
      V_N1  := ABS(TRUNC(V_N1 / 2));
      V_RTN := TO_CHAR(V_N2) || V_RTN;
      EXIT WHEN V_N1 = 0;
    END LOOP;
--返回二进制长度
 SELECT lpad(V_RTN,8,0) 
    INTO   V_RTN
    FROM dual;
return V_RTN;
end;

SQL> select NUMBER_TO_BIT(208) from dual;

NUMBER_TO_BIT(208)
-----------------------------
11010000

二进制转换十进制

CREATE OR REPLACE FUNCTION BIT_TO_NUMBER(P_BIN IN VARCHAR2) RETURN NUMBER AS
  V_SQL    VARCHAR2(30000) := 'SELECT BIN_TO_NUM(';
  V_RETURN NUMBER;
BEGIN
  IF LENGTH(P_BIN) >= 256 THEN
    RAISE_APPLICATION_ERROR(-20001, 'INPUT BIN TOO LONG!');
  END IF;
  IF LTRIM(P_BIN, '01') IS NOT NULL THEN
    RAISE_APPLICATION_ERROR(-20002, 'INPUT STR IS NOT VALID BIN VALUE!');
  END IF;
  FOR I IN 1 .. LENGTH(P_BIN) LOOP
    V_SQL := V_SQL || SUBSTR(P_BIN, I, 1) || ',';
  END LOOP;
  V_SQL := RTRIM(V_SQL, ',') || ') FROM DUAL';
  EXECUTE IMMEDIATE V_SQL
    INTO V_RETURN;
  RETURN V_RETURN;
END;

SQL> SELECT BIT_TO_NUMBER('11010000') FROM DUAL;

BIT_TO_NUMBER('11010000')
-------------------------
                      208
发表在 Oracle 开发 | 评论关闭

ORACLE ROWID 分析

ROWID知识补充
Rowid改为:OOOOOOFFFBBBBBBRRR,占用10个字节(32bit data_object_id +10bit rfile# +22bit block +16bit)。其中,O是对象ID,F是文件ID,B是块ID,R是行ID。

Rowid对应值对应10十进制值

A-Z <==> 0 - 25 (26)
a-z <==> 26 - 51 (26)
0-9 <==> 52 - 61 (10)
+/ <==> 62 - 63 (2)

模拟环境

SQL> select * from v$version;

BANNER
----------------------------------------------------------------
Oracle Database 10g Enterprise Edition Release 10.2.0.1.0 - 64bi
PL/SQL Release 10.2.0.1.0 - Production
CORE    10.2.0.1.0      Production
TNS for Linux: Version 10.2.0.1.0 - Production
NLSRTL Version 10.2.0.1.0 - Production

SQL> create table t_xff (id number,name varchar2(100));

Table created.

SQL> insert into t_xff values(1,'www.xifenfei.com');

1 row created.

SQL> commit;

Commit complete.

SQL>  alter table t_xff move;

Table altered.

SQL> select rowid,a.* from t_xff a;

ROWID                      ID NAME
------------------ ---------- ------------------------------
AADye6AAEAAAtCcAAA          1 www.xifenfei.com

相关值计算

Data Object number = AADye6
File               = AAE
Block              = AAAtCc
ROW                = AAA

DATA_OBJECT_ID
AADye6
58*64^0+30*64^1+50*64^2+3*64^3+0*64^4+0*64^5
58+1920+204800+786432+0+0=993210

RFILE#
AAE
4*64^0+0*64^1+0*64^2
4+0+0=4

BLOCK
AAAtCc
28*64^0+2*64^1+45*64^2+0*64^3+0*64^4+0*64^5
28+128+184320+0+0+0=184476

验证结果

SQL> select object_id,data_object_id from dba_objects where object_name='T_XFF';

 OBJECT_ID DATA_OBJECT_ID
---------- --------------
    993209         993210

SQL> select dbms_rowid.rowid_object('AADye6AAEAAAtCcAAA') data_object_id#,
  2         dbms_rowid.rowid_relative_fno('AADye6AAEAAAtCcAAA') rfile#,
  3         dbms_rowid.rowid_block_number('AADye6AAEAAAtCcAAA') block#,
  4     dbms_rowid.rowid_row_number('AADye6AAEAAAtCcAAA') row# from dual;  

DATA_OBJECT_ID#     RFILE#     BLOCK#       ROW#
--------------- ---------- ---------- ----------
         993210          4     184476          0

dump方式分析

SQL> select rowid,dump(rowid) from t_xff;

ROWID              DUMP(ROWID)
------------------ ------------------------------------------
AADye6AAEAAAtCcAAA Typ=69 Len=10: 0,15,39,186,1,2,208,156,0,0

DATA_OBJECT_ID
0,15,39,186
15*256*256+39*256+186=993210

RFILE#
1,2(取前10位)
000000001 00 =4

BLOCK
2,208,156(取后22位)
000010 11010000 10011100
2^17+208*256+156=184476
发表在 Oracle | 评论关闭