Question

我正在编写一个读入数字列表的程序。如：

我只是希望我的程序跳过包含其中任何字母的行。任何帮助将不胜感激。谢谢！

如果它有所不同，这是我的代码：

import java.io.*;

public class ScoreReader {

    public static void main(String[] args) {
        BufferedReader reader = null;

        try {
            String currentLine;

            reader = new BufferedReader(new FileReader("QuizScores.txt"));
            while ((currentLine = reader.readLine()) != null) {

                int sum = 0;

                String[] nums = currentLine.split("\\s+");
                for (int i = 0; i < nums.length; i++) {
                    int num = Integer.parseInt(nums[i]);
                    if (num != -1) {
                        sum += num;
                    }
                }

                System.out.println(sum / nums.length);
            }
        } catch (IOException err) {
            err.printStackTrace();
        } 
        catch (NumberFormatException err) {

        }

        finally {
            try {
                if (reader != null)
                    reader.close();
            } catch (IOException err) {
                err.printStackTrace();
            }
        }

    }
}

Answer 1

抛出异常时，执行会跳转到catch块。在你拥有的内容中，这是在循环之后，因此循环不会继续，只需在try附近添加parseInt。

try {
   String currentLine;

   reader = new BufferedReader(new FileReader("QuizScores.txt"));
   while ((currentLine = reader.readLine()) != null) {

       int sum = 0;
       String[] nums = currentLine.split("\\s+");
       for (int i = 0; i < nums.length; i++) {
           try{
               int num = Integer.parseInt(nums[i]);
               if (num != -1) {
                   sum += num;
               }
           } catch( NumberFormatException nfe )
           {
              // maybe log it?
           }
       }

       System.out.println(sum / nums.length);
   }
} catch (IOException err) {
    err.printStackTrace();
} 
// catch (NumberFormatException err) {}
finally {
    try {
       if (reader != null){
           reader.close();
    } catch (IOException err) {
        err.printStackTrace();
    }
}

另请注意，您使用的Integer.parseInt会在输入"34.7"处抛出异常，因此您可能希望使用Double.parseDouble

Answer 2

使用正则表达式怎么样？例如：

 if (currentLine.matches(".*[a-zA-Z].*")) {
     //letters contained.
 } else {
     //no letters contained.
 }

请参阅正则表达式演示：http://regex101.com/r/rQ6oR1

Answer 3

你可以试试这个：

 import java.io.*;

    public class ScoreReader {

        public static void main(String[] args) {
            BufferedReader reader = null;

            try {
                String currentLine;

                reader = new BufferedReader(new FileReader("QuizScores.txt"));
                while ((currentLine = reader.readLine()) != null) {

                    int sum = 0;

                    String[] nums = currentLine.split("\\s+");
                    for (int i = 0; i < nums.length; i++) {

                        if (isInt(num)) {
                            sum += num;
                        }
                    }

                    System.out.println(sum / nums.length);
                }
            } catch (IOException err) {
                err.printStackTrace();
            } 


            finally {
                try {
                    if (reader != null)
                        reader.close();
                } catch (IOException err) {
                    err.printStackTrace();
                }
            }

        }

public boolean isInt(String num)

    {
        boolean flag=false;


        try
        {
            int i=Integer.parseInt(num);
            flag=true;
        }

        catch(NumberFormatException e)
        {
            e.printStackTrace();

        }

        return flag;
    }

    }

Answer 4

根据您的评论。如果您的文件每行包含一个数字。那么这将是最简单的方法。

Scanner sc = new Scanner(new File("QuizScores.txt"));
        int sum = 0;
        int count =0;
        while( sc.hasNext()){
            String tmpNum = sc.next();
            if (isNumeric(tmpNum)){
                sum = sum + (int) Double.parseDouble(tmpNum); // if you want t capture in double use Double instead.
                count++;
            }
        }
        System.out.println(sum/count);


public static boolean isNumeric(String str)
    {
        return str.matches("-?\\d+(\\.\\d+)?");  //match a number with optional '-' and decimal.
    }

Java NumberFormatException跳过线

4 个答案: