oracle update数据更新的实现语句

SQL> -- create demo table
SQL> create table Employee(
2    ID                 VARCHAR2(4 BYTE)         NOT NULL,
3    First_Name         VARCHAR2(10 BYTE),
4    Last_Name          VARCHAR2(10 BYTE),
5    Start_Date         DATE,
6    End_Date           DATE,
7    Salary             Number(8,2),
8    City               VARCHAR2(10 BYTE),
9    Description        VARCHAR2(15 BYTE)
10 )
11 /

Table created.

SQL>
SQL> -- prepare data
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values ('01','Jason', 'Martin', to_date('19960725','YYYYMMDD'), to_date('20060725','YYYYMMDD'), 1234.56, 'Toronto', 'Programmer')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('02','Alison', 'Mathews', to_date('19760321','YYYYMMDD'), to_date('19860221','YYYYMMDD'), 6661.78, 'Vancouver','Tester')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('03','James', 'Smith', to_date('19781212','YYYYMMDD'), to_date('19900315','YYYYMMDD'), 6544.78, 'Vancouver','Tester')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('04','Celia', 'Rice', to_date('19821024','YYYYMMDD'), to_date('19990421','YYYYMMDD'), 2344.78, 'Vancouver','Manager')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('05','Robert', 'Black', to_date('19840115','YYYYMMDD'), to_date('19980808','YYYYMMDD'), 2334.78, 'Vancouver','Tester')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('06','Linda', 'Green', to_date('19870730','YYYYMMDD'), to_date('19960104','YYYYMMDD'), 4322.78,'New York', 'Tester')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('07','David', 'Larry', to_date('19901231','YYYYMMDD'), to_date('19980212','YYYYMMDD'), 7897.78,'New York', 'Manager')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('08','James', 'Cat', to_date('19960917','YYYYMMDD'), to_date('20020415','YYYYMMDD'), 1232.78,'Vancouver', 'Tester')
3 /

1 row created.

SQL>
SQL>
SQL>
SQL> -- display data in the table
SQL> select * from Employee
2 /

ID   FIRST_NAME LAST_NAME START_DAT END_DATE      SALARY CITY       DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01   Jason      Martin     25-JUL-96 25-JUL-06    1234.56 Toronto    Programmer
02   Alison     Mathews    21-MAR-76 21-FEB-86    6661.78 Vancouver Tester
03   James      Smith      12-DEC-78 15-MAR-90    6544.78 Vancouver Tester
04   Celia      Rice       24-OCT-82 21-APR-99    2344.78 Vancouver Manager
05   Robert     Black      15-JAN-84 08-AUG-98    2334.78 Vancouver Tester
06   Linda      Green      30-JUL-87 04-JAN-96    4322.78 New York   Tester
07   David      Larry      31-DEC-90 12-FEB-98    7897.78 New York   Manager
08   James      Cat        17-SEP-96 15-APR-02    1232.78 Vancouver Tester

8 rows selected.

SQL>
SQL>
SQL>
SQL>
SQL>
SQL>
SQL>
SQL> --Modify multiple rows with a single UPDATE statement;
SQL>
SQL>
SQL> UPDATE Employee
2 SET City ='L.A.'
3 WHERE City = 'New York';

2 rows updated.

SQL>
SQL> select * from employee;

ID   FIRST_NAME LAST_NAME START_DAT END_DATE      SALARY CITY       DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01   Jason      Martin     25-JUL-96 25-JUL-06    1234.56 Toronto    Programmer
02   Alison     Mathews    21-MAR-76 21-FEB-86    6661.78 Vancouver Tester
03   James      Smith      12-DEC-78 15-MAR-90    6544.78 Vancouver Tester
04   Celia      Rice       24-OCT-82 21-APR-99    2344.78 Vancouver Manager
05   Robert     Black      15-JAN-84 08-AUG-98    2334.78 Vancouver Tester
06   Linda      Green      30-JUL-87 04-JAN-96    4322.78 L.A.       Tester
07   David      Larry      31-DEC-90 12-FEB-98    7897.78 L.A.       Manager
08   James      Cat        17-SEP-96 15-APR-02    1232.78 Vancouver Tester

以下所列sql都是基于下表

create table test (name varchar2(30),code varchar2(10),i_d varchar2(10));

插入数据

insert into test(name,code,i_d) values('zhu1','001','1');
insert into test(name,code,i_d) values('zhu2','002','2');
insert into test(name,code,i_d) values('zhu3','003','3');
commit;
select * from test s;

1. update 更新i_d为1的数据

--方式1
update test set name='zhurhyme1',
code='007' where i_d='1';
commit;

这样可以成功

--方式2

update test set (name,code)=(
'zhurhyme2','007')
where i_d='1';

注意,这样是不行，update set 必须为子查询，所以需要将其改为 :

--方式3

update test set (name,code)=(
select 'zhurhyme3','007' from dual)
where i_d='1';

commit;

2.update 说完了，下面写一下关于for update，for update of

下面的资料是从网上找到的，可是具体的网址现在找不到了，请原谅小弟的粗心，引用人家的东东而不写出处.

for update 经常用，而for updade of 却不常用，现在将这两个作一个区分

a. select * from test for update 锁定表的所有行，只能读不能写

b. select * from test where i_d = 1 for update 只锁定i_d=1的行,对于其他的表的其他行却不锁定

下面再创建一个表

create table t (dept_id varchar(10),dept_name varchar2(50));

c. select * from test a join t on a.i_d=t.dept_id for update; 这样则会锁定两张表的所有数据

d. select * from test a join t on a.i_d=t.dept_id where a.i_d=1 for update; 这样则会锁定满足条件的数据

e. select * from test a join t on a.i_d=t.dept_id where a.i_d=1 for update of a.i_d; 注意区分 d与e，e只分锁定表test中满足条件的数据行，而不会锁定表t中的数据，因为之前在procedure中作一个update，而需要update的数据需要关联查询，所以用了for update造成其他用户更新造成阻塞，所以才查到这段资料.

for update of 是一个行级锁，这个行级锁，开始于一个cursor 打开时，而终止于事务的commit或rollback,而并非cursor的close.

如果有两个cursor对于表的同一行记录同时进行update，实际上只有一个cursor在执行，而另外一个一直在等待，直至另一个完成，它自己再执行.如果第一个cursor不能被很好的处理，第二个cursor也不主动释放资源，则死锁就此产生.

执行如下代码就会死锁(在两个command window中执行）

declare
cursor cur_test
is
select name,code from test where i_d=1 for update of name;
begin

for rec in cur_test loop
update test set name='TTTT1' where current of cur_test;
end loop;
end;
/

declare
cursor cur_test
is
select name,code from test where i_d=1 for update of name;
begin

for rec in cur_test loop
update test set name='TTTT2' where current of cur_test;
end loop;
end;
/

注意两个pl/sql块中没有commit;

补充：数据库,Oracle教程