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oracle update数据更新的实现语句
来源:互联网   发布日期:2016-01-28 13:03:18   浏览:1248次  

导读:/p><p> 3 First_Name VARCHAR2(10 BYTE),</p><p> 4 Last_Name VARCHAR2(10 BYTE),</p><p> 5 Start_Date DATE,</p><p> 6 End_Date DATE,</p><p> 7 Salary Number(8,2),</p><p> 8 City VARCHAR2(10 ...

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_dvarchar2(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 updateof 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forupdate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;

注:更多精彩文章请关注编程教程栏目。

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