How to open a txt file and read numbers in Java

Question

How can I open a .txt file and read numbers separated by enters or spaces into an array list?

Answer 1

File file = new File("file.txt");   
Scanner scanner = new Scanner(file);
List<Integer> integers = new ArrayList<>();
while (scanner.hasNext()) {
    if (scanner.hasNextInt()) {
        integers.add(scanner.nextInt());
    } 
    else {
        scanner.next();
    }
}
System.out.println(integers);

Answer 2

Read file, parse each line into an integer and store into a list:

List<Integer> list = new ArrayList<Integer>();
File file = new File("file.txt");
BufferedReader reader = null;

try {
    reader = new BufferedReader(new FileReader(file));
    String text = null;

    while ((text = reader.readLine()) != null) {
        list.add(Integer.parseInt(text));
    }
} catch (FileNotFoundException e) {
    e.printStackTrace();
} catch (IOException e) {
    e.printStackTrace();
} finally {
    try {
        if (reader != null) {
            reader.close();
        }
    } catch (IOException e) {
    }
}

//print out the list
System.out.println(list);

Answer 3

A much shorter alternative is below:

Path filePath = Paths.get("file.txt");
Scanner scanner = new Scanner(filePath);
List<Integer> integers = new ArrayList<>();
while (scanner.hasNext()) {
    if (scanner.hasNextInt()) {
        integers.add(scanner.nextInt());
    } else {
        scanner.next();
    }
}

A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace. Although default delimiter is whitespace, it successfully found all integers separated by new line character.

Answer 4

import java.io.*;  
public class DataStreamExample {  
     public static void main(String args[]){    
          try{    
            FileWriter  fin=new FileWriter("testout.txt");    
            BufferedWriter d = new BufferedWriter(fin);
            int a[] = new int[3];
            a[0]=1;
            a[1]=22;
            a[2]=3;
            String s="";
            for(int i=0;i<3;i++)
            {
                s=Integer.toString(a[i]);
                d.write(s);
                d.newLine();
            }

            System.out.println("Success");
            d.close();
            fin.close();    



            FileReader in=new FileReader("testout.txt");
            BufferedReader br=new BufferedReader(in);
            String i="";
            int sum=0;
            while ((i=br.readLine())!= null)
            {
                sum += Integer.parseInt(i);
            }
            System.out.println(sum);
          }catch(Exception e){System.out.println(e);}    
         }    
        }  

OUTPUT:: Success 26

Also, I used array to make it simple.... you can directly take integer input and convert it into string and send it to file. input-convert-Write-Process... its that simple.

Answer 5

Good news in Java 8 we can do it in one line:

List<Integer> ints = Files.lines(Paths.get(fileName))
                          .map(Integer::parseInt)
                          .collect(Collectors.toList());

Answer 6

   try{

    BufferedReader br = new BufferedReader(new FileReader("textfile.txt"));
    String strLine;
    //Read File Line By Line
    while ((strLine = br.readLine()) != null)   {
      // Print the content on the console
      System.out.println (strLine);
    }
    //Close the input stream
    in.close();
    }catch (Exception e){//Catch exception if any
      System.err.println("Error: " + e.getMessage());
    }finally{
     in.close();
    }

This will read line by line,

If your no. are saperated by newline char. then in place of

 System.out.println (strLine);

You can have

try{
int i = Integer.parseInt(strLine);
}catch(NumberFormatException npe){
//do something
}  

If it is separated by spaces then

try{
    String noInStringArr[] = strLine.split(" ");
//then you can parse it to Int as above
    }catch(NumberFormatException npe){
    //do something
    }