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(xún)，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(xiě)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?，引用人家的東東而不寫(xiě)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(xiě)
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(mǎn)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(mǎn)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(xún)，所以用了for update造成其他用戶(hù)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;