我想更新表格，但每次尝试时都会出现错误代码“PLS-00103”答案

【问题标题】：I would like to update the table but the error code 'PLS-00103' occurs every time I try this我想更新表格，但每次尝试时都会出现错误代码“PLS-00103”
【发布时间】：2016-12-12 15:42:06
【问题描述】：

DECLARE 
    Emp_Number NUMBER(5);
    Salary NUMBER(5);
    Salary_increment NUMBER(5);
BEGIN
     SELECT Employee_Salary 
     INTO Salary 
     FROM EMPLOY_TABLE;
END;
BEGIN
     SELECT Employee_ID 
     INTO Emp_Number 
     FROM EMPLOY_TABLE;
END;
BEGIN
    IF Salary <= 20000 THEN
       Salary_increment := (Salary/100) * 12.5;
    ELSIF Salary > 20000 AND employee_Salary <= 40000 THEN
       Salary_increment := (Salary/100) * 11.5;
    ELSE
        Salary_increment := (Salary/100) * 10;
    END IF;

    UPDATE EMPLOY_TABLE 
    SET Employee_Salary = Salary + Salary_increment  
    WHERE Employee_ID = Emp_Number;
END;

【问题讨论】：

错字提醒：这是薪水 - 不是“薪水” ....
但我知道这不会导致错误 marc_s

标签： stored-procedures plsql oracle11g

【解决方案1】：

您有太多的 BEGIN 和 END。变量的范围仅限于它们被声明的块（以及其中的任何嵌套块）。例如，变量salary 在第三块中是不可访问的。试试：

DECLARE 
    Emp_Number NUMBER(5);
    Salary NUMBER(5);
    Salary_increment NUMBER(5);
BEGIN
     SELECT Employee_Salary 
     INTO Salary 
     FROM EMPLOY_TABLE;

     SELECT Employee_ID 
     INTO Emp_Number 
     FROM EMPLOY_TABLE;

    IF Salary <= 20000 THEN
       Salary_increment := (Salary/100) * 12.5;
    ELSIF Salary > 20000 AND employee_Salary <= 40000 THEN
       Salary_increment := (Salary/100) * 11.5;
    ELSE
        Salary_increment := (Salary/100) * 10;
    END IF;

    UPDATE EMPLOY_TABLE 
    SET Employee_Salary = Salary + Salary_increment  
    WHERE Employee_ID = Emp_Number;
END;

如果您需要嵌套块，例如用于异常处理，您可以这样做：

DECLARE 
    Emp_Number NUMBER(5);
    Salary NUMBER(5);
    Salary_increment NUMBER(5);
BEGIN
     BEGIN
       SELECT Employee_Salary 
       INTO Salary 
       FROM EMPLOY_TABLE;
     END;
     BEGIN
       SELECT Employee_ID 
       INTO Emp_Number 
       FROM EMPLOY_TABLE;
     END;
     BEGIN
       IF Salary <= 20000 THEN
           Salary_increment := (Salary/100) * 12.5;
       ELSIF Salary > 20000 AND employee_Salary <= 40000 THEN
           Salary_increment := (Salary/100) * 11.5;
       ELSE
           Salary_increment := (Salary/100) * 10;
       END IF;

       UPDATE EMPLOY_TABLE 
       SET Employee_Salary = Salary + Salary_increment  
       WHERE Employee_ID = Emp_Number;
    END;
END;

但这在这里是多余的。

【讨论】：