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Oracle:如何创建一个身份列?[复制]

sql

直到11g版本(包括11g)为止,Oracle中似乎都没有AUTO_INCREMENT的概念。

如何在Oracle 11g中创建行为类似于自动增量的列?


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2021-03-10

共1个答案

小编典典

从Oracle 11g开始,Oracle中 没有诸如“ auto_increment”或“
identity”列之类的东西。但是,您可以使用序列和触发器轻松对其进行建模:

表定义:

CREATE TABLE departments (
  ID           NUMBER(10)    NOT NULL,
  DESCRIPTION  VARCHAR2(50)  NOT NULL);

ALTER TABLE departments ADD (
  CONSTRAINT dept_pk PRIMARY KEY (ID));

CREATE SEQUENCE dept_seq START WITH 1;

触发定义:

CREATE OR REPLACE TRIGGER dept_bir 
BEFORE INSERT ON departments 
FOR EACH ROW

BEGIN
  SELECT dept_seq.NEXTVAL
  INTO   :new.id
  FROM   dual;
END;
/

更新:

IDENTITY
列现在在Oracle 12c上可用:

create table t1 (
    c1 NUMBER GENERATED by default on null as IDENTITY,
    c2 VARCHAR2(10)
    );

或指定起始值和增量值,还防止任何插入到标识列(GENERATED ALWAYS)中(同样,仅适用于Oracle 12c +)

create table t1 (
    c1 NUMBER GENERATED ALWAYS as IDENTITY(START with 1 INCREMENT by 1),
    c2 VARCHAR2(10)
    );

另外,Oracle 12还允许使用序列作为默认值:

CREATE SEQUENCE dept_seq START WITH 1;

CREATE TABLE departments (
  ID           NUMBER(10)    DEFAULT dept_seq.nextval NOT NULL,
  DESCRIPTION  VARCHAR2(50)  NOT NULL);

ALTER TABLE departments ADD (
  CONSTRAINT dept_pk PRIMARY KEY (ID));
2021-03-10