如何在Java中使用扫描仪读取.txt文件时删除inputMismatchException

Question

import java.io.*;
import java.util.*;

public class FrequencyCount
{
    public static void main(String...args) throws IOException
    {
        int count0=0;
        int count1=0;
        int x=0;
        //System.out.println(new File(".").getCanonicalPath());
        Scanner scan=new Scanner(new File("F:\\DUCAT COCHNG\\java progs\\IO\\str.txt.txt"));
        while(scan.hasNext())
        {
            x=scan.nextInt();
            if(x==0)
                count0++;
            else if(x==1)
                count1++;
        }   

        System.out.println("Frequency of 0's is: "+count0);
        System.out.println("Frequency of 1's is: "+count1);

    }
}

output shown is:

F:\\DUCAT COCHNG\\java progs\\IO>java FrequencyCount Exception in thread "main" java.util.InputMismatchException: For input string: " 00000111010010011001010010100101010001000101110111110101010011101001001010010010 1" at java.util.Scanner.nextInt(Scanner.java:2165) at java.util.Scanner.nextInt(Scanner.java:2118) at FrequencyCount.main(FrequencyCount.java:15)

F:\\DUCAT COCHNG\\java progs\\IO>

str.txt file is:

000001110100100110010100101001010100010001011101111101010100111010010010100100101

Answer 1

I think this may help:

try{
        Scanner scan = new Scanner(new File(filename));
        while (scan.hasNext())
        {
           String str = scan.next(); 
            char[] myChar = str.toCharArray();
            for(char c: myChar){
                x=Character.getNumericValue(c);
                if(x==0) 
                    count0++; 
                else if(x==1) 
                    count1++; 
            }
        }
       System.out.println("Frequency of 0's is: "+count0);
       System.out.println("Frequency of 1's is: "+count1);
   }
  catch(Exception e)
  { e.printStackTrace();}

Answer 2

You want to use scan.hasNextInt() instead of scan.hasNext() . The latter checks for any token, and I suspect you have a new line at the end of your file that scanner is parsing. This obviously can't be converted to an integer, and so the exception is thrown.

You may also want to close your scanner as well. Just use scan.close() . Not strictly necessary here in such a small program, but good practice.