Question

在我的程序中，我想从文本文件中读取。我文件中的数据是由空格分隔的数字。我会在程序中将它们转换为整数。如果从文件读取的数据不是数字，我想捕获异常。我想确切地知道错误发生在文件的哪一行（数据不是数字）。我知道我应该捕获NumberFormatException异常，但不知道如何找到异常行。这里有一些代码：

try 
{ 
  //...using Bufferreader ...some code here
  while ((strLine = br.readLine()) != null ) 
  {
    String[] term = strLine.split(" ");  
    //for now i just work with the first element of the array
    int num = Integer.parseInt(term[0]); 
    //....some code here
  }
  in.close();
}
catch (NumberFormatException eg)
{
  System.out.println("Error: there is a problem in line ..? " );
}

Answer 1

int line = 0;
try 
{ 
  //...using Bufferreader ...some code here
  while ((strLine = br.readLine()) != null ) 
  {
    line ++;
    String[] term = strLine.split(" ");  
    //for now i just work with the first element of the array
    int num = Integer.parseInt(term[0]); 
    //....some code here
  }
  in.close();
}
catch (NumberFormatException eg)
{
  System.out.println("Error: there is a problem in line " + line );
}

Answer 2

int num = Integer.parseInt(term[0]);

当参数不是数字时，上面的行将抛出NumberFormatException。

如何知道IOexception发生在文件的哪一行

2 个答案: