Question

我是编程（学习Java）的初学者。我正在尝试编写一个程序，其中列出了四个不同的选项供用户选择。

以下是其中的一部分：

import java.util.*;
    public class fight {

            public static int upgrade1 = 0;
            public static int upgrade2 = 0;
            public static int upgrade3 = 0;
            public static int upgrade4 = 0;

    public static void main(String[] args) {

        Scanner scan = new Scanner(System.in);

        System.out.println("Please enter your name:");

            String player = scan.next();    

System.out.println("You have earned 2 upgrade points. Which of the following traits would you like to boost by 2 points?\n"
    + " 1. upgrade1\n 2. upgrade2\n 3. upgrade3\n"
    + " 4. upgrade4");

                    if (scan.nextInt() == 1) {
                        upgrade1 = upgrade1 + 2;
                            System.out.println("Your upgrade1 level is now: " + upgrade1);
                    }
                    else if (scan.nextInt() == 2) {
                        upgrade2 = upgrade2 + 2;
                            System.out.println("Your upgrade2 level is now: " + upgrade2);
                    }
                    else if (scan.nextInt() == 3) {
                        upgrade3 = upgrade3 + 2;
                            System.out.println("Your upgrade3 level is now: " + upgrade3);
                    }
                    else if (scan.nextInt() == 4) {
                        upgrade4 = upgrade4 + 2;
                            System.out.println("Your upgrade4 level is now: " + upgrade4);
                    }                                                       
        }
}

问题是：当用户输入他们想要选择的选项时，他们必须输入数字（x是他们选择的数字）x次。例如，用户想要选择选项3.他们必须在控制台中输入数字3三次才能理解并完成下一行。

这是运行程序后的控制台：

请输入您的姓名：干草堆你好，里克。您已获得2个升级积分。您希望将以下哪些特征提升2点？升级1 升级2 3.升级3 4.升级4 3 3 3 您的升级3级现在是：2

我希望这是有道理的，并且非常感谢任何帮助（我假设我只是犯了一个愚蠢的菜鸟错误）。此外，如果您对其结构方式有任何建设性的批评，请不要犹豫。谢谢！

Answer 1

您无法重复调用scan.nextInt（）。当然，除非您希望读取多个不同的整数。

相反：

Scanner scan = new Scanner(System.in);
System.out.println("Please enter your name:");
String player = scan.next();
int ichoice = scan.nextInt();
switch (ichoice) {
  case 1:
  ...

Answer 2

每次在if语句中调用scan.nextInt时，它会读取另一个int。改为：

int userChoice = scan.nextInt();
if (userChoice == 1)
{
    ...
}
else if (userChoice == 2)

...

至于建设性的批评，选择你喜欢的风格并使用它。你的缩进遍布各地;这使得代码更难以阅读。如果它是一种常用的风格并不重要，并且无关紧要其他人的想法，只要确保你喜欢它并坚持下去。

Eclipse可以为您自动格式化代码，这种行为是可自定义的（您可以使用它来调整它以匹配您的样式）。

Answer 3

每次调用nextInt()时，都会从输入中读取另一个int。因此，您只想打电话一次！

int choice = scan.nextInt();

if (choice == 1) ...
if (choice == 2) ...

Answer 4

这是因为您在每个if / else if语句中调用scan.nextInt（）。你想要做的是存储他们输入的值，然后检查if / else if语句中的值，否则你基本上会提示用户输入多次。

int input = scan.nextInt();

if (input == 1) {
    upgrade1 = upgrade1 + 2;
        System.out.println("Your upgrade1 level is now: " + upgrade1);
}
else if (input == 2) {
    upgrade2 = upgrade2 + 2;
        System.out.println("Your upgrade2 level is now: " + upgrade2);
}
else if (input == 3) {
    upgrade3 = upgrade3 + 2;
        System.out.println("Your upgrade3 level is now: " + upgrade3);
}
else if (input == 4) {
    upgrade4 = upgrade4 + 2;
        System.out.println("Your upgrade4 level is now: " + upgrade4);
}

Java扫描程序：nextInt

4 个答案: