Question

我正在尝试从名为“ inData.txt”的文本文件中加载程序的数据，并将其输出到文件“ out.txt”中，基本上模仿了保存。问题是，当我通过GUI运行此代码时，它不会加载到文本区域，但可以保存。我不知道我的GUI代码或League文件中的代码是否有问题。

我尝试检查文本文件是否更改，但是没有更改，但是代码运行时没有错误。

//League.java

private static final String READ_FILE="inData.txt";
private static final String OUT_FILE="out.txt.";

public void readFromFile() //throwsIO exception /take from file into arraylist
{
    //reading the inData.txt file and loading the arraylist
    //with each instantiated object from the data
    try
    {
        Scanner in = new Scanner(new File(READ_FILE));
    //loop through the file and read each valie and create an object
    while(in.hasNextLine())
    {

        String name = in.nextLine();
        System.out.println ("read name: " + name);

        String coach = in.nextLine();
        System.out.println ("read coach: " + coach);

        int teamID = in.nextInt();
        System.out.println ("read team ID: " + teamID);

        int wins = in.nextInt();
        System.out.println ("read wins: " + wins);

        int losses = in.nextInt();
        in.nextLine();
        System.out.println ("read losses: " + losses);
        String player1 = in.nextLine();
        System.out.println ("read player 1 name: " + player1);
        String position1 = in.nextLine();
        System.out.println ("read player 1 position: " + position1);
        Player p1 = new Player(player1, position1);
        String player2 = in.nextLine();
        System.out.println ("read player 2 name: " + player2);
        String position2 = in.nextLine();
        System.out.println ("read player 2 position: " + position2);
        Player p2 = new Player(player2, position2);
        String player3 = in.nextLine();
        System.out.println ("read player 3: " + player3);
        String position3 = in.nextLine();
        System.out.println ("read player 3 position: " + position3);
        Player p3 = new Player(player3, position3);

        //create and Item object
        Team newTeam = new Team(name, coach, teamID, wins, losses, p1, p2, p3);
        league.add(newTeam);
    }
        }
    catch(IOException e)
    {
        System.out.println ("Error with file loading");
    }
}

readFromFile（）的预期结果是从inData.txt获取信息并将其导入GUI。

Answer 1

文件内容是什么样的？

鉴于您正在逐行循环，建议您始终使用nextLine（）并根据需要解析从中获取的字符串。例如，如果您的文件内容如下所示：

line1
23:34:45:99
34

您可以这样做：

if (in.hasNextLine()){
   String firstLine = in.nextLine();
   String secondLine = in.nextLine();
   int[] nums = new int[3];
   int i = 0;
   for (String intString : lineTwo.split(":")) {
      nums[i] = Integer.valueOf(intString);
      i++;
   }
   // or simply:
   int thirdLine = Integer.valueOf(in.next());
   // if you just have one int per line anyway
}

我猜您遇到的问题涉及一些稍微高级的主题，例如代码点或换行符与回车符之间的区别，如果您想了解Scanner的复杂性，可能会对您感兴趣。 / p>

让我知道这实际上是否不是您遇到的问题，我将编辑我的答复。

将输入和输出数据加载并保存到文本文件中（不起作用）

1 个答案: