Question

当我去添加员工时，此代码首先会提示输入新的员工ID。然后它检查数组以确保该ID尚未使用。我遇到的问题是，它让我输入ID两次，然后再转到其他的场地。其他字段只需输入一次。我无法弄清楚造成它的原因。你们能帮助我指出正确的方向吗？

示例输出：

Please enter the maximum number of employees for your store

2

Welcome to the employee management system

a) Add a new Employee

b) Delete an Employee

c) Print the Employee List

d) Print a Specific Employee

e) Exit the Employee Management System

a

Please enter the employee ID.

1234  //should only have to type it in once but as you can see in the example i have to type it a second time 
             //before it moves on

1234

Please enter the employee name.

我认为问题出在下面的方法中。

public void addEmployee() // choice a
{
    int id = 0;
    int x = 0;
    int y = 0;      
    while (x < maxEmployees && empNumber[x] != 0) // allows user to add employee as long as x is less than maxEmployees and empNumber in next array is 0
        x++;

    if(x == maxEmployees)// tells user that they can not add more employees if they have too many before kicking them back to the menu
    {
        System.out.println("Unable to add more employees. Max number of employees reached.");
    }
    else
    {
        System.out.println("Please enter the employee ID.");
        id = scan.nextInt();
        scan.nextLine();

        while (y < maxEmployees && empNumber[y] != id) // checks to see if employee id already exsist
            y++;
        if(y == maxEmployees) // if no matching id is found continues the adding process
        {
            empNumber[x] = scan.nextInt();
            scan.nextLine();
            System.out.println("Please enter the employee name.");
            empName[x] = scan.nextLine();
            System.out.println("Please enter the employee address.");
            empAddress[x] = scan.nextLine();
            System.out.println("Please enter the employee salary.");
            empSalary[x] = scan.nextInt();
            scan.nextLine();
        }

        else
            System.out.println("This ID is already associated with an employee");
    }

Answer 1

看起来您要求用户输入两次ID。

第一次来了：

id = scan.nextInt();
nothing = scan.nextLine();

第二次来了：

empNumber[x] = scan.nextInt();
scan.nextLine();

尝试

empNumber[x] = id;

取代

empNumber[x] = scan.nextInt();
scan.nextLine();

为什么我必须重复输入才能让我继续前进？

1 个答案: