Question

在寻找解决方案的同时，我发现了使用sc.next（）而不是sc.nextLine（）的建议，但我不能使用它，所以我需要额外的帮助。

我正在开发一个在线多人文字游戏，需要过滤玩家名称，但我不断收到错误＆＃34;字符串索引超出范围：0＆＃34;我已经尝试了几个小时来解决这个问题并且无法找到解决方案。导致错误的函数是：

public static Player charCreate(Player player) {

    int points = 60;
    int i = 0;
    boolean cont = true;
    int str = 0;
    int dex = 0;
    int end = 0;
    int INT = 0;
    int lck = 0;
    int cha = 0;

    Output.slowDiscription("You stand there and think about all"
            + " that has transpired, who are you realy?");

    System.out.print("What is your Name? ");

    boolean bool = false;
    String temp;
    String name = null; //name is forced to be declared by while loop

    do {

        name = sc.nextLine();
        System.out.println(name);
        if (Utills.profan(name) && (name != null)
                && ((!name.equals("")) || (!name.equals(" ")))) {

            bool = true;
            player.setName(Utills.filter(name)); //Error is here

        }
        else {

            bool = false;

        }

    } while (bool == false);

    player.Height = getUsrHeight();

    System.out.print("Please select your stats. ");

    do {

        System.out.println("Points Remaining: " + points);

        switch (i) {

        case 0:

            System.out.println();
            System.out.println("Please Enter the number of points to alot to "
                    + "Strength.\n Min:1 Max:18");

            str = sc.nextInt();
            if ((str >= 1) && (str <= 18) && (str <= points) &&
                    ((points - str) >= 5)) {

                points -= str;
                i++;

            }
            break;

        case 1:

            System.out.println();
            System.out.println("Please Enter the number of points to alot to "
                    + "Dexterity.\n Min:1 Max:18");

            dex = sc.nextInt();
            if ((dex >= 1) && (dex <= 18) && (dex <= points) &&
                    ((points - dex) >= 4)) {

                points -= dex;
                i++;

            }
            break;

        case 2:

            System.out.println();
            System.out.println("Please Enter the number of points to alot to "
                    + "Endurance.\n Min:1 Max:18");

            end = sc.nextInt();
            if ((end >= 1) && (end <= 18) && (end <= points) &&
                    ((points - end) >= 3)) {

                points -= end;
                i++;

            }
            break;

        case 3:

            System.out.println();
            System.out.println("Please Enter the number of points to alot to "
                    + "Inteligence.\n Min:1 Max:18");

            INT = sc.nextInt();
            if ((INT >= 1) && (INT <= 18) && (INT <= points) &&
                    ((points - INT) >= 2)) {

                points -= INT;
                i++;

            }
            break;

        case 4:

            System.out.println();
            System.out.println("Please Enter the number of points to alot to "
                    + "Luck.\n Min:1 Max:18");

            lck = sc.nextInt();
            if ((lck >= 1) && (lck <= 18) && (lck <= points) &&
                    ((points - lck) >= 1)) {

                points -= lck;
                i++;

            }
            break;

        case 5:

            System.out.println();
            System.out.println("Please Enter the number of points to alot to "
                    + "Charisma.\n Min:1 Max:18");

            cha = sc.nextInt();
            if ((cha >= 1) && (cha <= 18) && (cha <= points)) {

                points -= cha;
                i++;

            }
            break;

        case 6:

            int[] stats = {str, dex, end, INT, lck, cha};

            player.setStats(stats);
            cont = false;
            break;

        }

    }while (cont);


    return player;

}

这里的错误来自Utills.filter（名称）：

public static String filter(String name) {

    //Variables
    String[] Name = name.toLowerCase().split(" ");

    StringBuilder sb = new StringBuilder();
    StringBuilder endStr = new StringBuilder();

    char temp;

    int i = 0;

    //Sorting
    for(String w: Name) {

        sb.append(w);

        temp = Character.toUpperCase(sb.charAt(0)); //And Error is here
        sb.setCharAt(0, temp);

        if(i >= 1) {

            endStr.append(" " + sb.toString());

        }
        else {

            endStr.append(sb);

        }

        i++;
        empty(sb);

    }


    return endStr.toString();

}

我会很乐意提供任何帮助

Answer 1

我认为你的问题是：

if (Utills.profan(name) && (name != null)
            && ((!name.equals("")) || (!name.equals(" "))))

应该是

if (Utills.profan(name) && (name != null)
            && !name.equals("") && !name.equals(" "))

甚至

if (Utills.profan(name) && (name != null)
            && !name.trim().isEmpty())

如果您仍然没有看到我的意思，检查((!name.equals("")) || (!name.equals(" ")))始终为true，因为name始终不是""或" " 1}}。

字符串索引超出范围：0

1 个答案: