顺序读取文件时出现NumberFormatException？答案

【问题标题】：NumberFormatException when reading from file sequentially?顺序读取文件时出现NumberFormatException？
【发布时间】：2019-04-17 13:28:17
【问题描述】：

该程序的主要目的是将学生信息写入文件并从同一文件中读取。该程序包括一个 if 语句，用于判断学生是否处于良好的信誉或学术试用期，每个文件都有各自的文件（goodstanding.txt 和 probation.txt）。如果我注释掉 goodstanding.txt 的读取逻辑，程序工作文件，但对我来说，除了文件路径之外，它们似乎几乎相同，显然。我查看了其他问题，但其中大多数似乎与错误的类型转换有关。它抛出的具体错误是NumberFormatException For input string: ""在线ID = Integer.parseInt(array[0]);

下面是写逻辑：

if(GPA >= 2.0) //If GPA >= 2.0, write to good standing file. This is 
               identical to probation writer, but causes an issue somewhere
{
     String result = ID + "," + fname + " " + lname + "," + GPA;
     OutputStream out = new BufferedOutputStream(Files.newOutputStream(good, 
     StandardOpenOption.CREATE));
     BufferedWriter wt = new BufferedWriter(new OutputStreamWriter(out));
     wt.write(result, 0, result.length()); //Write to file from position 0 to 
                                        length
     wt.newLine();
     System.out.println("Please enter next WIN or 999 to quit: ");
     ID = input.nextInt();
     input.nextLine();
     wt.close();
}
     if(GPA < 2.0) //If GPA < 2.0, write to probation file {
     String result = ID + "," + fname + " " + lname + "," + GPA;
     OutputStream out = new 
                   BufferedOutputStream(Files.newOutputStream(probation, 
                   StandardOpenOption.CREATE));
     BufferedWriter wt = new BufferedWriter(new OutputStreamWriter(out));
     wt.write(result, 0, result.length());
     wt.newLine();
     System.out.println("Please enter next WIN or 999 to quit: ");
     ID = input.nextInt();
     input.nextLine();
     wt.close();
}

以及读取逻辑：

try
{
    while(line != null)
{
    array = line.split(",");
    ID = Integer.parseInt(array[0]);
    name = array[1];
    GPA = Double.parseDouble(array[2]);
    double ahead = GPA - 2;
    System.out.println("WIN: " + ID + " | " + " Name: " + name + " |" + " GPA                 
                        = " + GPA + " | Ahead by " + ahead);
    line = reader.readLine(); 
}
}
catch(NullPointerException e)
    {System.out.println("NullPointerException occurred");}
reader.close();


InputStream input2 = new
                     BufferedInputStream(Files.newInputStream(probation));
BufferedReader reader2 = new BufferedReader(new InputStreamReader(input2));
String line2 = reader2.readLine();
System.out.println();
System.out.println("---------------------------");
System.out.println("--Academic Probation List--");
System.out.println("---------------------------");
try{
while(line2 != null)
{
    array = line2.split(",");
    ID2 = Integer.parseInt(array[0]);
    name2 = array[1];
    GPA2 = Double.parseDouble(array[2]);
    double fallShort = 2 - GPA2;
    System.out.println("WIN: " + ID2 + " | " + " Name: " + name2 + " |" + " 
                       GPA = " + GPA2 + " | Behind by " + fallShort);
    line2 = reader2.readLine();
}
}
catch(NullPointerException e)
{e.printStackTrace();}
reader.close();

我也尝试在 ID 上使用 trim() 方法，但异常仍然存在。是否有一个我需要阅读更多的概念来解释这一点？

【问题讨论】：

您的错误意味着您无法将空字符串解析为数字格式。我会添加一个 NumberFormatException 流行语，以允许程序继续并记录导致它的特定字符串值。然后你就会知道如何在处理输入之前修复它。
如果您希望 array[0] 具有数值，请在尝试将其作为整数传递之前进行验证。
现在也通读您的代码，您应该大大简化它。当您真的想打开一次缓冲写入器时，您有多个创建，完成所有写入，然后在完成后关闭它。在您了解流的目的（即指向文件并保持写入位置等）之前，您的问题将无法解决

标签： java file-io bufferedinputstream

【解决方案1】：

错误消息很清楚，我认为 NumberFormatException For input string: "" 因为您无法将空字符串解析为 int。

您的问题至少有两个原因，或者您的文件中有一些空白行，或者您的其中一行以逗号开头。通过执行以下操作来检查或忽略这些行：

....
    while (line != null) {
        if(!line.startsWith(",") && !line.isEmpty()){
            array = line.split(",");
            ID = Integer.parseInt(array[0]);
            name = array[1];
            GPA = Double.parseDouble(array[2]);
            double ahead = GPA - 2;
            System.out.println("WIN: " + ID + " | " + " Name: " + name + " |" + " GPA = " + GPA + " | Ahead by " + ahead);
        }
        line = reader.readLine();
    }

【讨论】：