函数，存储过程

[yo_258]

表a : 表b:
Id name int_rate id name int_rate1 int_rate2 int_rate3
1 rose 37 1 rose 37 34 38
2 rose 34 2 john 35 45 55
3 rose 38
1 john 35
2 john 45
3 john 55
利用pl/sql程序块把表a变成表b？

表
Id salary name
1 4500 rose

创建存储过程，输入员工薪水范围，返回员工工号、姓名、薪水结果集。结果集按薪水升序排列
创建函数更新员工薪水，低于2000，且姓王的员工加5%，其他不变，更新成功返回0，否则返回1；

[superhsj]

select rownum,name,
sum(case id when 1 then int_rate else 0 end) int_rate1,
sum(case id when 2 then int_rate else 0 end) int_rate2,
sum(case id when 3 then int_rate else 0 end) int_rate3
from a group by name;

create or replace test return number as
begin
 update table set 薪水=薪水*(1+0.05) where 姓='王' and 薪水<2000;
 return 0;
 exception
  when others then
   return 1;
end;

[superhsj]

漏了一个

create or replace test1(val1 in number,val2 in number)
v_id number;
v_name varchar2(10);
v_salary number;
as
 cursor v_cur is select id,name,salary from table where salary<=val2 and salary>=val1 order by salary;
begin
 open v_cur;
 fetch v_cur into v_id,v_name,v_salary;
 while v_cur %found loop
  dbms_output.put_line(v_id||chr(9)||v_name||chr(9)||v_salary);
  fetch v_cur into v_id,v_name,v_salary;
 end loop;
 close v_cur;
end;
 

[superhsj]

create or replace test return number as
begin
 update table set salary=salary*(1+0.05) where name like '王%' and salary<2000;
 commit;
 return 0;
 exception
  when others then
   return 1;
end;

[woodheart]

高，直接到这边来找干活的人了

[superhsj]

函数和过程关键字都漏了，lz自己加吧

[superhsj]

我正在给lz解释，lz还是在很认真的学习的，呵呵
[Quote=引用 4 楼 woodheart 的回复:]
高，直接到这边来找干活的人了
[/Quote]

[tanzuwen]

select Id,name,
sum(case Id when 1 then int_rate else 0 end)int_rate1,
sum(case Id when 1 then int_rate else 0 end)int_rate2,
sum(case Id when 1 then int_rate else 0 end)int_rate3,
from a group by name