我正在开发一个C#项目,我尝试使用以下SQL对Oracle数据库插入记录:
String Qry = INSERT INTO browsers (browsers.browser,browsers.engine,browsers.platform,browsers.version,browsers.grade) VALUES ('Alans browser','Gecko','every','1.0','U') RETURNING id INTO :ID
我在以下示例代码:http://www.sqlines.com/oracle-to-sql-server-cs-conversion/insert-returning-clause
我的代码如下:
OracleConnection conn = new OracleConnection(...);
OracleCommand cmd = null;
...
cmd = new OracleCommand(Qry, conn);
OracleParameter prm = new OracleParameter();
prm = new OracleParameter(":ID", OracleDbType.Int32, ParameterDirection.ReturnValue);
cmd.Parameters.Add(prm);
cmd.ExecuteNonQuery(); //this line throws error
query.setNewRecordID(cmd.Parameters[":ID"].Value.ToString());
...
当然,浏览器表的id列设置为Sequence和Trigger,以便在任何插入时自动递增。当我没有尝试设置id列时,我不明白为什么我会收到错误。我只是想检索给新记录的id值。
Uddate:以下是用于创建序列和触发器的SQL命令:
序列:
CREATE SEQUENCE "SYSTEM"."BROWSERS_SEQ" MINVALUE 1 MAXVALUE 9999999999999999999999999999 INCREMENT BY 1 START WITH 21 CACHE 20 NOORDER NOCYCLE NOPARTITION ;
触发:
create or replace trigger BROWSERS_TRG
before insert on "SYSTEM"."BROWSERS"
for each row
begin
if inserting then
if :NEW."ID" is null then
select BROWSERS_SEQ.nextval into :NEW."ID" from dual;
end if;
end if;
end;
答案 0 :(得分:1)
我认为您的序列存在问题(我无法按原样编译)。我从中取出了NOPARTITION,它起作用了。
我在SCOTT(示例)模式中创建了类似的场景,如下所示:
SCOTT@dev> CREATE TABLE "SCOTT"."EMP2"
2 (
3 "EMPNO" NUMBER(4,0),
4 "ENAME" VARCHAR2(10 BYTE),
5 "JOB" VARCHAR2(9 BYTE),
6 "MGR" NUMBER(4,0),
7 "HIREDATE" DATE,
8 "SAL" NUMBER(7,2),
9 "COMM" NUMBER(7,2),
10 "DEPTNO" NUMBER(2,0)
11 )
12 TABLESPACE "SYSTEM" ;
Table created.
SCOTT@dev> CREATE UNIQUE INDEX "SCOTT"."EMP2_EMPNO" ON "SCOTT"."EMP2"
2 (
3 "EMPNO"
4 )
5 TABLESPACE "SYSTEM" ;
Index created.
SCOTT@dev> set define off;
SCOTT@dev> ALTER TABLE "SCOTT"."EMP2" ADD PRIMARY KEY ("EMPNO");
Table altered.
SCOTT@dev> CREATE SEQUENCE EMP2_SEQ MINVALUE 1 MAXVALUE 9999999999999999999999999999 INCREMENT BY 1 START WITH 21 CACHE 20 NOORDER NOCYCLE ;
Sequence created.
SCOTT@dev> CREATE OR REPLACE TRIGGER EMP2_TRG before
2 INSERT ON SCOTT.EMP2 FOR EACH row BEGIN IF inserting THEN IF :NEW.EMPNO IS NULL THEN
3 SELECT EMP2_SEQ.NEXTVAL INTO :NEW.EMPNO FROM dual;
4 END IF;
5 END IF;
6 END;
7 /
Trigger created.
SCOTT@dev> commit;
Commit complete.
SCOTT@dev> INSERT INTO EMP2
2 (ENAME)
3 VALUES
4 ('FRED')
5 /
1 row created.
SCOTT@dev> commit;
Commit complete.
因此,序列似乎是罪魁祸首。
答案 1 :(得分:0)
您确定顺序AND触发器设置正确吗?通常我不会使用触发器,而只是将sequencer.nextval直接用于查询。如果您需要在代码的其他部分使用生成的ID,您可以执行从双重'中选择sequence.nextval。并缓存结果