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oracle update數(shù)據(jù)更新的實(shí)現(xiàn)語(yǔ)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));

插入數(shù)據(jù)

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為1的數(shù)據(jù)

--方式1

update test set name='zhurhyme1',

code='007' where i_d='1';

commit;

這樣可以成功

--方式2

update test set (name,code)=(

'zhurhyme2','007')

where i_d='1';

注意,這樣是不行，update set 必須為子查詢，所以需要將其改為 :

--方式3

update test set (name,code)=(

select 'zhurhyme3','007' from dual)

where i_d='1';

commit;

2.update 說(shuō)完了，下面寫一下關(guān)于for update，for update of

下面的資料是從網(wǎng)上找到的，可是具體的網(wǎng)址現(xiàn)在找不到了，請(qǐng)?jiān)徯〉艿拇中模萌思业臇|東而不寫出處.

for update 經(jīng)常用，而for updade of 卻不常用，現(xiàn)在將這兩個(gè)作一個(gè)區(qū)分

a. select * from test for update 鎖定表的所有行，只能讀不能寫

b. select * from test where i_d = 1 for update 只鎖定i_d=1的行,對(duì)于其他的表的其他行卻不鎖定

下面再創(chuàng)建一個(gè)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; 這樣則會(huì)鎖定兩張表的所有數(shù)據(jù)

d. select * from test a join t on a.i_d=t.dept_id where a.i_d=1 for update; 這樣則會(huì)鎖定滿足條件的數(shù)據(jù)

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; 注意區(qū)分 d與e，e只分鎖定表test中滿足條件的數(shù)據(jù)行，而不會(huì)鎖定表t中的數(shù)據(jù)，因?yàn)橹霸趐rocedure中作一個(gè)update，而需要update的數(shù)據(jù)需要關(guān)聯(lián)查詢，所以用了for update造成其他用戶更新造成阻塞，所以才查到這段資料.

for update of 是一個(gè)行級(jí)鎖，這個(gè)行級(jí)鎖，開(kāi)始于一個(gè)cursor 打開(kāi)時(shí)，而終止于事務(wù)的commit或rollback,而并非cursor的close.

如果有兩個(gè)cursor對(duì)于表的同一行記錄同時(shí)進(jìn)行update，實(shí)際上只有一個(gè)cursor在執(zhí)行，而另外一個(gè)一直在等待，直至另一個(gè)完成，它自己再執(zhí)行.如果第一個(gè)cursor不能被很好的處理，第二個(gè)cursor也不主動(dòng)釋放資源，則死鎖就此產(chǎn)生.

執(zhí)行如下代碼就會(huì)死鎖(在兩個(gè)command window中執(zhí)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;

/

注意兩個(gè)pl/sql塊中沒(méi)有commit;

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