Question

JAVA

Write a program that prompts the user to enter a file name, then opens the file in text mode and reads it. The input files are assumed to be in CSV format. The input files contain a list of integers on each line separated by commas. The program should read each line, sort the numbers and print the comma separated list of integers on the console. Each sorted list of integers from the same line should be printed together on a separate line on the console (see below for test cases).

While entering the file name, the program should allow the user to type quit to exit the program.

If the file with a given name does not exist, then display a message and allow the user to re-enter the file name.

On each line the input file may contain any number of integers.

You can use any number of stacks, queues, arrays, or ArrayList objects in your program.

You may choose to use any of the sorting algorithms (provided in above class files for your convenience). All classes are uploaded in advance and there is no need to upload them with your submission. You should only upload one file: FileSorting.java. You are able to download the sorting algorithms in their respective files to use in your eclipse project for testing.

Input validation:

a) If the file does not exist, then you should display a message "File somefile.txt is not found." and allow the user to re-enter the file name.

b) If the file is empty, then display a message "File somefile.txt is empty." and exit the program.

Hints:

a) Perform file name input validation immediately after the user entry and use a while loop.

b) You can use either StringTokenizer class or String.split() method to separate each line into tokens.

c) For each line, create an array of integers and populate with the numbers from that line.

d) Use either one of the InsertionSort.sort(), SelectionSort.sort(), or BubbleSort.sort() methods to sort the integers.

Answer 1

I was not sure if user is entering a '.txt' file or a '.csv' file so i wrote the code on the basis that user enters the extension with the file name. If you know the extension and does not want the user to enter the extension in the input then you can update the code and uncomment the line csvFile = csvFile + ".csv"; if the extension is '.csv' otherwise uncomment the line csvFile = csvFile + ".txt"; if the extension is '.txt' just below the 'try{' statement.

import java.io.BufferedReader;

import java.io.File;

import java.io.FileReader;

import java.io.IOException;

import java.util.Scanner;

public class FileSorting

{

    public static int fun()

    {

        //creating object of Scanner

        final Scanner s=new Scanner(System.in);

        System.out.print("Enter file name:");

        //taking file name input

        String csvFile = s.nextLine();

        //checking if user entered quit

        if(csvFile.equals("quit")||csvFile.equals("Quit")||csvFile.equals("QUIT"))

        {

            //returning -1 if user entered quit

            return -1;

        }

        else

        {

            try {

            //csvFile = csvFile + ".csv";    

            //csvFile = csvFile + ".txt";    

            //creating object of file

            final File file = new File(csvFile);

            //checking if file exists and if file is not empty

            if(file.exists()&&file.length()!=0)

            {

                final FileReader fr = new FileReader(file);

                final BufferedReader br = new BufferedReader(fr);

                String line = "";

                //creating a string array

                String[] tempArr;

                //reading the input line by line and spliting when ',' arise

                //and storing each element in tempArr

                while((line = br.readLine()) != null) {

                    tempArr = line.split(",");

                    //calculating the length of each row

                    final int len=tempArr.length;

                    //defining a int array

                    final int arr[]=new int[len];

                    //converting string array to int array

                    for(int i=0;i<len;i++)

                    arr[i]=Integer.parseInt(tempArr[i]);

                    //applying insertion sort on arr

                    for (int j = 1; j < len; j++) {  

                        final int key = arr[j];  

                        int i = j-1;  

                        while ( (i > -1) && ( arr[i] > key ) ) {  

                            arr[i+1] = arr[i];  

                            i--;  

                        }  

                        arr[i+1] = key;  

                    }

                    //displaying the output after sorting

                    for(int i=0;i<len;i++)

                    System.out.print(arr[i]+",");

                    System.out.println();

                }

                br.close();

            }

            //checking if file exists but is empty

            else if(file.length()==0&&file.exists())

            {

                System.out.println("File "+file+" is empty");

                return 0;

            }

            //when file is not found

            else

            {

                System.out.println("File "+file+" not found");

                return 0;

            }

        }

        catch(final IOException ioe) {

            ioe.printStackTrace();

        }

    }

        return 0;

    }

    public static void main(final String args[])

    {

        int a;

        //calling the function fun until the user enters quit

        do

        {

            a=fun();

        }while(a!=-1); //if user enters quit we return -1 from fun()

        //if we get a=-1 it means user entered quit

        if(a==-1)

        System.out.println("You entered quit");

    }

}

Output:

In the above pic you can see the content of 'file.csv' and the output after running the code. There was no file named 'abc.csv' so the output was 'File abc.csv was not found'.

File Edit Selection View Go Run Terminal Help file.csv - java - Visual Studio Code Х FileSorting.java 3 file.csv X C:4 Command Prompt file.csv 1 1,2,3,4 2 5,3,2,7 3 9,4,1,8 4 20,56,28,29 5 28,56,32,89 Microsoft Windows [Version 10.0.18363.836] (c) 2019 Microsoft Corporation. All rights reserved. go C:\Users\Aasmeen>cd Desktop C:\Users\Aasmeen\Desktop>cd java C:\Users\Aasmeen\Desktop\java>javac FileSorting.java A C:\Users\Aasmeen\Desktop\java>java FileSorting Enter file name:file.csv 1,2,3,4, 2,3,5,7, 1,4,8,9, 20,28,29,56, 28,32,56,89, Enter file name: abc.csv File abc.csv not found Enter file name: quit You entered quit C:\Users\Aasmeen\Desktop\java> 14 A 31 Ln 5, Col 12 Spaces: 4 UTF-8 CRLF Plain Text Go Live O Search for anything DI O 23 E a со R ? a » ENG 12:34 AM 07-06-2020 6