Question

假设我们有员工表，我的目的是编写declare -blog语句，该员工的薪水增加10％，工作部门的工资高于平均水平，如果没有更多，则增加20％薪水，我认为代码应该是

begin
update employees e set e.salary=e.salary+e.salary*0.1;
where e.salary>(select avg(e.salary)  from employees e group by e.department_id);
update employees e set e.salary=e.salary+e.salary*0.1;
where e.salary<(select avg(e.salary)  from employees e group by e.department_id);



end;

但它返回以下错误列表

Error starting at line 4 in command:
begin
update employees e set e.salary=e.salary+e.salary*0.1;
where e.salary>(select avg(e.salary)  from employees e group by e.department_id);
update employees e set e.salary=e.salary+e.salary*0.1;
where e.salary<(select avg(e.salary)  from employees e group by e.department_id);



end;
Error report:
ORA-06550: line 3, column 1:
PLS-00103: Encountered the symbol "WHERE" when expecting one of the following:

   ( begin case declare end exception exit for goto if loop mod
   null pragma raise return select update while with
   <an identifier> <a double-quoted delimited-identifier>
   <a bind variable> << continue close current delete fetch lock
   insert open rollback savepoint set sql execute commit forall
   merge pipe purge
06550. 00000 -  "line %s, column %s:\n%s"
*Cause:    Usually a PL/SQL compilation error.
*Action:

请帮我解释为什么？，是否有必要的循环声明或？

Answer 1

删除<{p>}上的;

update employees e set e.salary=e.salary+e.salary*0.1;

;会终止该声明，但您会继续使用WHERE

Answer 2

;之前有分号（where）;分号关闭PL / SQL语句。

我还建议您以一种易于阅读和更难犯错的方式格式化您的陈述：

update employees e set 
    e.salary=e.salary+e.salary*0.1 -- no semicolon here
where 
    e.salary > (
        select avg(e.salary)  
        from employees e 
        group by e.department_id
    );

写PL / SQL博客声明

2 个答案: