Java程序中的方法不会接收用户输入

Question

所以我在这里有一些代码：

import java.util.*;

public class tester 
{
    public static void main(String args[])
    {
        Scanner kb = new Scanner(System.in);
        String b;
        System.out.println("Choose a number 1-10.");
        int a = kb.nextInt();
        a = a * 2;
        a = a + 5;
        a = a * 50;
        System.out.println("Enter the year you were born in.");
        int c = kb.nextInt();
        System.out.println("Did you already have your birthday this year?");
        b = kb.nextLine();
        if (b.equals("yes"))
        {
            a = a + 1764;
        }
        else
        {
            a = a + 1763;
        }
        a = a - c;
        System.out.println(a);
        kb.close();
    }
}

我在这里得到输出：

Choose a number 1-10.
5
Enter the year you were born in.
2014
Did you already have your birthday this year?
499

在我看来，（String）b被完全忽略了。 谁能解释我在做什么错？

Answer 1

这是因为scanner.nextInt()不使用用户键入的输入的换行符。 你应该叫kb.nextLine()后scanner.nextInt()仅仅是消耗了新行字符留下。

int c = kb.nextInt();
System.out.println("Did you already have your birthday this year?");
kb.nextLine(); //consumes new line character left by the last scanner.nextInt() call
b = kb.nextLine();

或替换所有的kb.nextInt(); 对于Integer.parseInt(kb.nextLine());

Scanner kb = new Scanner(System.in);
String b;
System.out.println("Choose a number 1-10.");
int a = 0;
try {
    a = Integer.parseInt(kb.nextLine());
} catch (NumberFormatException numberFormatException) {
    a=0;
}
a = a * 2;
a = a + 5;
a = a * 50;
System.out.println("Enter the year you were born in.");
int c = 0;
try {
    c = Integer.parseInt(kb.nextLine()); //consumes the new line character
} catch (NumberFormatException numberFormatException) {
    c=0;
}
System.out.println("Did you already have your birthday this year?");
b = kb.nextLine();
if (b.equals("yes")) {
    a = a + 1764;
} else {
    a = a + 1763;
}
a = a - c;
System.out.println(a);
kb.close();