Java字符串索引超出范围:0

Question

我有这个问题,一旦我输入我的第一个输入程序崩溃,我得到

字符串索引超出范围:0

我看了别处并试图找到我的错误,但我发现了不同的问题,而不是我的问题.有人可以告诉我哪里出错了？

感谢您的帮助,以下是代码:

import java.util.Scanner;

public class Assignment1Q2 {

    public static void main(String[] args) {

        System.out.println("Thank you for your call,\nPlease take some time to answer a few questions");
        collectData();

    }//end of main

    public static void collectData() {

        Scanner userInput = new Scanner(System.in);

        int age;
        char gender;
        char show;
        int over30MY = 0, over30FY = 0, under30MY = 0, under30FY = 0;
        int over30MN = 0, over30FN = 0, under30MN = 0, under30FN = 0;

        System.out.println("\nWhat is your age?\n");
        age = userInput.nextInt();

        System.out.println("Male or Female (Enter M or Y)");
        gender = userInput.nextLine().charAt(0);
        gender = Character.toLowerCase(gender);

        System.out.println("Do you watch the show regularly? (Enter Y or N)");
        show = userInput.nextLine().charAt(0);
        show = Character.toLowerCase(show);

        if((age > 30) && (gender == 'm') && (show == 'y')) {       
            over30MY++;             
        }
        else if((age > 30) && (gender == 'f') && (show == 'y')) {
            over30FY++;
        }
        else if((age < 30) && (gender == 'm') && (show == 'y')) {
            under30MY++;
        }
        else if((age < 30) && (gender == 'f') && (show == 'y')) {
            under30FY++;
        }
        else if((age > 30) && (gender == 'm') && (show == 'n')) {
            over30MN++;
        }
        else if((age > 30) && (gender == 'f') && (show == 'n')) {
            over30FN++;
        }
        else if((age < 30) && (gender == 'm') && (show == 'n')) {
            under30MN++;
        }
        else if((age < 30) && (gender == 'f') && (show == 'n')) {
            under30FN++;
        }//end of if else

    }//end of collectData
}// end of class

Answer 1

你的问题在于这一行:

userInput.nextLine().charAt(0);

所述nextLine()方法扫描在当前行一切,然后前进指针过去的该行.因此,当您调用charAt()方法时,您在下一行(即空格)上调用它,因此发生错误.

而是将此行更改为:

userInput.next().charAt(0)

请注意,这意味着您的代码的其他部分也需要更改.

编辑:

即将编辑我的解决方案,但@ Marc-Andre添加了他的答案,涵盖了它,所以只是把目光投向它.

Answer 2

你正在做的问题age = userInput.nextInt();是你可能输入一个数字4然后按Enter键.

因此,当您正在呼叫时,扫描仪会读取4 nextInt但新线路不会消耗.这意味着当你这样做时:userInput.nextLine().charAt(0);你正在使用新行,所以nextLine()它将返回一个空字符串.由于您正在chartAt使用空字符串,因此它会为您提供异常.

你可以这样做:

age = userInput.nextInt();
userInput.nextLine();

这将使用新行,因此流应为空.所以你不会有例外,你可以要求下一个输入.