编译PL / SQL函数时出错

Question

CREATE OR REPLACE FUNCTION employer_details_func
   RETURN VARCHAR(20);
 IS 
   e_name VARCHAR(20); 
 BEGIN 
    SELECT emp_name INTO e_name
    FROM employees WHERE emp_no = '5';
    RETURN e_name;
 END employer_details_func;

编译上述程序时，我收到此错误

  

错误（2,18）：PLS-00103：遇到符号“（”，当遇到下列情况之一时：。@％;使用authid作为集群顺序使用   外部字符确定性parallel_enable流水线   聚合result_cache

更新：

CREATE OR REPLACE FUNCTION employer_details_func
   RETURN VARCHAR2(20);
 IS 
   e_name VARCHAR2(20); 
 BEGIN 
    SELECT emp_name INTO e_name FROM employees WHERE emp_no ='5';
    RETURN e_name;
 END employer_details_func;

错误：

Error(2,19): PLS-00103: Encountered the symbol "(" when expecting one of the following:     . @ % ; is authid as cluster order using external character    deterministic parallel_enable pipelined aggregate    result_cache 

我的代码出了什么问题？请告诉我。

Answer 1

使用RETURN VARCHAR代替RETURN VARCHAR(20);。

http://docs.oracle.com/cd/B19306_01/server.102/b14200/statements_5009.htm上的Oracle文档说 RETURN子句...数据类型无法指定长度，精度或比例......

---

修改

我使用此代码在SQL Fiddle上仔细检查了它，它似乎有用（返回bob）：

CREATE TABLE employees(emp_name VARCHAR2(20), emp_no VARCHAR2(20))
/
CREATE OR REPLACE FUNCTION employer_details_func
   RETURN VARCHAR
 IS 
   e_name VARCHAR(20); 
 BEGIN 
   SELECT emp_name INTO e_name FROM employees WHERE emp_no ='5';
   RETURN e_name;
 END employer_details_func;
/

insert into employees values('bob','5');
select employer_details_func() from dual;

Answer 2

这应该解决它：

CREATE OR REPLACE FUNCTION employer_details_func
   RETURN VARCHAR2
 IS 
   e_name VARCHAR2(20); 
 BEGIN 
    SELECT emp_name INTO e_name FROM employees WHERE emp_no ='5';
    RETURN e_name;
 END employer_details_func;

Answer 3

创建或替换

FUNCTION employer_details_func

RETURN VARCHAR2

IS

e_name VARCHAR2（20）;

BEGIN

SELECT last_name INTO e_name FROM employees WHERE employee_id ='100';

RETURN e_name;

END employer_details_func;

您不能提及由oracle引擎管理的返回类型的大小。

Answer 4

您也可以尝试这样

创建或替换功能employer_details_func

RETURN employees.e_name％TYPE

IS

e_name employees.e_name％TYPE;

BEGIN

SELECT emp_name INTO e_name FROM employees WHERE emp_no ='5';

RETURN e_name;

END employer_details_func;