请问，编写一个PL/SQL程序时，是不是要将Exception语句块放到程序的最后？
RT。
谢谢！！
问题点数：10、回复次数：4Top
1 楼cenlmmx（学海无涯苦作舟）回复于 2006-03-08 21:01:22 得分 2
不一定,如
declare
......
begin
begin
insert into test(c1,c2) value(dd,ff);
exception
when others then
...........
end;
update .................
exception
when others then
................
end ;Top
2 楼RUBY_uu（）回复于 2006-03-08 21:24:13 得分 0
可是，为什么我下面的这个程序有错误：
create or replace procedure NO_new_stu(stu_name in student.stuname%type,
stu_id out student.stuid%type)
as
num char(4);
s_id char(4);
begin
select max(stuid) into num from student;
select stuid into s_id from student where stuname=stu_name;
if s_id is null then
if substr(num,3,2) between '01' and '09' then
stu_id:='1000'+substr(num,4,1)+1;
else
stu_id:='1000'+substr(num,3,2)+1;
end if;
else
RAISE_APPLICATION_ERROR(-20001,'The student is existed!');
end if;
exception
when others then dbms_output.put_line(SQLERRM);
end;
update student set stuid=stu_id where stu_name=stuname;
end;
/
警告: 创建的过程带有编译错误。
SQL> show err
PROCEDURE NO_NEW_STU 出现错误:
LINE/COL ERROR
-------- --------------------------------
20/1 PLS-00103: 出现符号 "UPDATE"
而且，如果我把update语句放到when others then dbms_output.put_line(SQLERRM);下面的话：
SQL> create or replace procedure NO_new_stu(stu_name in student.stuname%type,
2 stu_id out student.stuid%type)
3 as
4 num char(4);
5 s_id char(4);
6 begin
7 select max(stuid) into num from student;
8 select stuid into s_id from student where stuname=stu_name;
9 if s_id is null then
10 if substr(num,3,2) between '01' and '09' then
11 stu_id:='1000'+substr(num,4,1)+1;
12 else
13 stu_id:='1000'+substr(num,3,2)+1;
14 end if;
15 else
16 RAISE_APPLICATION_ERROR(-20001,'The student is existed!');
17 end if;
18 exception
19 when others then dbms_output.put_line(SQLERRM);
20 update student set stuid=stu_id where stu_name=stuname;
21 end;
22 /
过程已创建。
update这句干脆就不执行了，表student没有任何变化。
只有把update这句放到Exception的上方才对表进行了更改：
create or replace procedure NO_new_stu(stu_name in student.stuname%type,
stu_id out student.stuid%type)
as
num char(4);
s_id char(4);
begin
select max(stuid) into num from student;
select stuid into s_id from student where stuname=stu_name;
if s_id is null then
if substr(num,3,2) between '01' and '09' then
stu_id:='1000'+substr(num,4,1)+1;
else
stu_id:='1000'+substr(num,3,2)+1;
end if;
else
RAISE_APPLICATION_ERROR(-20001,'The student is existed!');
end if;
update student set stuid=stu_id where stu_name=stuname;
exception
when others then dbms_output.put_line(SQLERRM);
end;
/
这是怎么回事呢？？Top
3 楼sbaz（万神渡劫）回复于 2006-03-09 09:35:44 得分 4
create or replace procedure NO_new_stu(stu_name in student.stuname%type,
stu_id out student.stuid%type)
as
num char(4);
s_id char(4);
begin
begin
select max(stuid) into num from student;
select stuid into s_id from student where stuname=stu_name;
if s_id is null then
if substr(num,3,2) between '01' and '09' then
stu_id:='1000'+substr(num,4,1)+1;
else
stu_id:='1000'+substr(num,3,2)+1;
end if;
else
RAISE_APPLICATION_ERROR(-20001,'The student is existed!');
end if;
exception
when others then dbms_output.put_line(SQLERRM);
end;
update student set stuid=stu_id where stu_name=stuname;
end;
/
少了个BEGIN.
有BEGIN就有END.
Top
4 楼mosquitoxh（(IT,中间件,架构)http://it.paiming.org/bbs）回复于 2006-03-09 09:36:03 得分 4
begin
................
begin
...............
exception
end;
..................
begin
...............
exception
end;
exception
begin
......
exception
end
end;
就是这样一个原则
一个begin和end 之间可以有一个exceptionTop
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