"ORA-00905: missing keyword" 用于案例陈述

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;

我想写一个案例陈述，如果你的薪水在一个特定的范围内，薪水应该乘以 20% 或 25%，具体取决于薪水范围。

Answer 1

您的 CASE 陈述有几个问题：

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

所以更正这些问题应该只是：

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

您可能还注意到 <= 10000 和 BETWEEN 10000 AND 15000 的情况相同，因此可以合并。

还要记住 BETWEEN ... AND 是包含值，因此值 15000 将 return 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。

"ORA-00905: missing keyword" 用于案例陈述

"ORA-00905: missing keyword" for case statement

oracle

plsql

case