为什么我的程序会循环跳过提示？

Question

我的代码应该不断循环，直到输入“stop”作为员工姓名。我遇到的一个问题是，一旦计算出第一个员工的工作时间和费率，它将再次跳过输入员工姓名的提示。为什么？ （顺便说一句，代码位于 2 个单独的类中）请帮忙。这是我的代码：

package payroll_program_3;
import java.util.Scanner;

public class payroll_program_3
{
    public static void main(String[] args)
    {

        Scanner input = new Scanner( System.in );

        employee_info theEmployee = new employee_info();

        String eName = "";
        double Hours = 0.0;
        double Rate = 0.0;
        while(true)
        {
            System.out.print("\nEnter Employee's Name: ");
            eName = input.nextLine();
            theEmployee.setName(eName);
            if (eName.equalsIgnoreCase("stop"))
            {
                return;
            }

            System.out.print("\nEnter Employee's Hours Worked: ");
            Hours = input.nextDouble();
            theEmployee.setHours(Hours);
            while (Hours <0)    //By using this statement, the program will not
            {                   //allow negative numbers.
                System.out.printf("Hours cannot be negative\n");
                System.out.printf("Please enter hours worked\n");
                Hours = input.nextDouble();
                theEmployee.setHours(Hours);
            }

            System.out.print("\nEnter Employee's Rate of Pay: ");
            Rate = input.nextDouble();
            theEmployee.setRate(Rate);
            while (Rate <0)    //By using this statement, the program will not
            {                  //allow negative numbers.
                System.out.printf("Pay rate cannot be negative\n");
                System.out.printf("Please enter hourly rate\n");
                Rate = input.nextDouble();
                theEmployee.setRate(Rate);
            }

            System.out.print("\n Employee Name:     " + theEmployee.getName());
            System.out.print("\n Employee Hours Worked:     " + theEmployee.getHours());
            System.out.print("\n Employee Rate of Pay:     " + theEmployee.getRate() + "\n\n");
            System.out.printf("\n %s's Gross Pay: $%.2f\n\n\n", theEmployee.getName(),
                theEmployee.calculatePay());
        }
    }
}

并且

package payroll_program_3;

        public class employee_info
{
            String employeeName;
            double employeeRate;
            double employeeHours;

public employee_info()
    {
    employeeName = "";
    employeeRate = 0;
    employeeHours = 0;
    }

public void setName(String name)
    {
    employeeName = name;
    }

public void setRate(double rate)
    {
    employeeRate = rate;
    }

public void setHours(double hours)
    {
    employeeHours = hours;
    }

public String getName()
    {
    return employeeName;
    }

public double getRate()
    {
    return employeeRate;
    }

public double getHours()
    {
    return employeeHours;
    }

public double calculatePay()
    {
    return (employeeRate * employeeHours);
    }
}

Answer 1

问题在于：

eName = input.nextLine();

将其更改为：

eName = input.next();

它将起作用。

默认情况下，一旦您按下 enter 键，java.util.Scanner 类就会开始解析文本。因此，在循环内，您应该请求 input.next(); 因为一行已经在处理中。

请参阅 javadoc 了解这两种方法。

Answer 2

我假设它正在等待 input.nextLine() 输入，但提示没有出现。

我的猜测是某处存在缓冲问题 - 正在打印提示但输出没有到达屏幕。尝试在调用 input.nextLine() 之前调用 System.out.flush()。

Answer 3

我一生中从未使用过 Java，但现在是这样...

Rate = input.nextDouble();

为了响应这一行，您可以输入“15.00”，然后按 {enter}，这将在输入流中放置一个“新行”，使输入流15.00\n。提取 Double 时，它​​会将“新行”保留在输入流中。

当您尝试提示用户输入另一个员工的姓名时，它会读取输入流中仍然有一个“新行”，并且它将立即返回之前的内容，即一个空字符串。

为了清除输入流，请运行虚拟 .nextLine。