Question

我目前正在用Java编写一个项目，它将获取一个输入文件并将其读入几个并行数组。有几个限制 - 我们不能使用数组列表，必须使用Scanner读取文件。在将其读入数组之后，还需要编写其他一些步骤，但我已经遇到了挂断。

    public static void main(String[] args) throws FileNotFoundException {

    final int ARRAY_SIZE = 10;
    int choice;
    int i, variableNumber;
    String[] customerName = new String[ARRAY_SIZE];
    int[] customerID = new int[ARRAY_SIZE];
    String[] os = new String[ARRAY_SIZE];
    String[] typeOfProblem = new String[ARRAY_SIZE];
    int[] turnAroundTime = new int[ARRAY_SIZE];

    readFile(customerName, customerID, os, typeOfProblem, turnAroundTime);

}

public static void readFile(String[] customerName, int[] customerID, String[] os, String[] typeOfProblem, int[] turnAroundTime) throws FileNotFoundException
{
    File hotlist = new File("hotlist.txt");
    int i = 0;

    if (!hotlist.exists())
    {
        System.out.println("The input file was not found.");
        System.exit(0);
    }
    Scanner inputFile = new Scanner(hotlist);
    while (inputFile.hasNext())
    {
        customerName[i] = inputFile.nextLine();
        System.out.println(customerName[i]);
        customerID[i] = inputFile.nextInt();
        os[i] = inputFile.nextLine();
        typeOfProblem[i] = inputFile.nextLine();
        turnAroundTime[i] = inputFile.nextInt();
        i++;
    }
    System.out.println("This is only a test." + customerName[1] + "\n" + customerID[1] + "\n"
                        + os[1] + "\n" + typeOfProblem[1] + "\n" + turnAroundTime[1]);
}

当我尝试运行上面的代码时，它失败并出现以下错误：

run:
Mike Rowe
Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Scanner.java:864)
    at java.util.Scanner.next(Scanner.java:1485)
    at java.util.Scanner.nextInt(Scanner.java:2117)
    at java.util.Scanner.nextInt(Scanner.java:2076)
    at mckelvey_project3.McKelvey_Project3.readFile(McKelvey_Project3.java:70)
    at mckelvey_project3.McKelvey_Project3.main(McKelvey_Project3.java:33)
Java Result: 1
BUILD SUCCESSFUL (total time: 0 seconds)

hotlist.txt文件的内容如下：

Mike Rowe
1
Windows DOS
Too Much ASCII Porn
3
Some Guy
2
Windows 10
Too Much Windows
200

非常感谢任何帮助！顺便说一句，所有的System.out语句都是测试语句，因为我试图调试我的代码。我已将错误隔离到了

customerID[i] = inputFile.nextInt();

和类似的

turnAroundTime[i] = inputFile.nextInt();

但无法弄清楚这些陈述为什么不起作用。

Answer 1

当你致电Scanner.nextInt()时，它只会消耗int，并且会在那里留下任何尾随空格或换行符。相反，你可能会使用像

这样的东西

Scanner inputFile = new Scanner(hotlist);
while (inputFile.hasNext()) {
    customerName[i] = inputFile.nextLine();
    System.out.println(customerName[i]);
    String custId = inputFile.nextLine();
    customerID[i] = Integer.parseInt(custId);
    os[i] = inputFile.nextLine();
    typeOfProblem[i] = inputFile.nextLine();
    String turnAround = inputFile.nextLine();
    turnAroundTime[i] = Integer.parseInt(turnAround);
    i++;
}

我得到（使用你的代码/文件），

Mike Rowe
Some Guy
This is only a test.Some Guy
2
Windows 10
Too Much Windows
200

Answer 2

您的主要问题是您没有设置正确的分隔符。初始化扫描程序后，执行inputFile.useDelimiter("\n")将分隔符设置为换行符 - 默认为空格。

然后，您可以使用inputFile.next()读取字符串，使用inputFile.nextInt()读取int，不会有任何问题。

从输入文件读取到整数数组

2 个答案: