Question

DECLARE
var_employee_number VARCHAR2(30);
var_employee_name VARCHAR2(30);
var_salary VARCHAR2(30);
BEGIN
SELECT employee_number,employee_name,
CASE var_salary
    WHEN salary <= 10000 THEN salary:= salary* .02 /*I am getting an error over here*/
    WHEN salary BETWEEN 10000 AND 15000 THEN salary := salary * .02
    WHEN salary BETWEEN 15000 AND 20000 THEN salary := salary * .025
    END AS salary INTO var_employee_number,var_employee_name,var_salary
FROM EMPLOYEES
WHERE employee_number := var_employee_number;
dbms_output.put_line('The salary is '||var_salary);
END;

我试图写一个案例陈述，如果你的工资在一个特定的范围内，工资应该乘以20％或25％，具体取决于工资范围。

Answer 1

您的CASE声明有几个问题：

CASE var_salary -- you don't need var_salary here
    WHEN salary <= 10000 THEN salary:= salary* .02 /*I am getting an error over here*/ -- you don't need "salary:=" here
    WHEN salary BETWEEN 10000 AND 15000 THEN salary := salary * .02
    WHEN salary BETWEEN 15000 AND 20000 THEN salary := salary * .025
END AS salary

所以纠正这些问题应该只读：

CASE WHEN salary <= 10000 THEN salary * 0.2
  WHEN salary BETWEEN 10000 AND 15000 THEN salary * 0.2
  WHEN SALARY BETWEEN 15000 AND 20000 THEN salary * 0.25
END AS salary

您可能还会注意到<= 10000和BETWEEN 10000 AND 15000的案例相同，因此可以合并。

另请注意，BETWEEN ... AND包含在内，因此值15000将返回15000 * 0.2（3000），而不是* 0.25。

此外，请注意我更正了您的数字... 20％是0.2，而不是.02。

＆＃34; ORA-00905：缺少关键字＆＃34;用于案例陈述

1 个答案: