1

我目前正在用 Java 编写一个项目，该项目将获取一个输入文件并将其读入多个并行数组。有几个限制——我们不能使用数组列表，必须使用 Scanner 读取文件。将其读入数组后，我还需要编写一些其他步骤，但我已经挂断了。

    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]);
}

当我尝试运行上述代码时，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文件内容如下：

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

任何帮助是极大的赞赏！顺便说一句，当我试图调试我的代码时，所有 System.out 语句都是测试语句。我已将错误专门隔离到

customerID[i] = inputFile.nextInt();

同样地

turnAroundTime[i] = inputFile.nextInt();

但无法弄清楚为什么这些陈述不起作用。

4

2 回答 2

0

当您调用Scanner.nextInt()它时，只使用int，它会在那里留下任何尾随空格或换行符。相反，您可能会使用类似的东西，

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（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
于 2014-11-23T22:15:08.033 回答
0

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

然后，您可以毫无问题地读取 String usinginputFile.next()和 int using 。inputFile.nextInt()

于 2014-11-23T22:20:57.337 回答