Not sure how many more hours to spend even after I get the desired output

Question about the task

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I feel that the problem validation is very poorly designed. I get the output when I execute but fails validation. Can't spend any more hours. Some one please help me here.

Let's build a file from various pieces.
Read file names from the console.
Each file has a name: <someName>.partN.

For example, Lion.avi.part1, Lion.avi.part2, ..., Lion.avi.part37.

The file names are supplied in random order. The word "end" is used to stop reading in file names.
In the folder where all the files are located, create a file without the "part" suffix, i.e. without ".<partN>".

For example, Lion.avi.

Use a buffer to copy all the bytes from the partial files to the created file.
Copy the first in the proper order, first the first part, then the second, ..., finally - the last part.
Close the streams.

Requirements:

The program must read file names from the console until the word "end" is entered.
Create a stream to write to the file without the "part" suffix (".<partN>") in the folder with all the "part" files.
Copy all the bytes from the *.partN files to the new file.
You should use a buffer for the reading and writing.
The file streams must be closed.
Don't use static variables.

package com.codegym.task.task18.task1825; import java.io.*; import java.util.*; /* Building a file */ public class Solution { public static void main(String[] args) throws IOException { Scanner scanner = new Scanner(System.in); String file; List<String> filenames = new ArrayList<>(); while (true){ file = scanner.nextLine(); if (file.equals("end")){ break; } else { filenames.add(file); } } String originalFile = filenames.get(0); originalFile = originalFile.replaceAll("\\.part[0-9]+", ""); BufferedOutputStream bfo = new BufferedOutputStream(new FileOutputStream(originalFile,true));; List<Integer> partNumbers = new ArrayList<>(); for (String filename : filenames){ partNumbers.add(Integer.parseInt(filename.replaceAll(".*Lion\\.avi\\.part", ""))); } Collections.sort(partNumbers); for (Integer partNumber : partNumbers){ String partFile = originalFile + ".part" + partNumber; BufferedInputStream bf = new BufferedInputStream(new FileInputStream(partFile)); byte[] buffer = new byte[bf.available()]; byte read; while ((read = (byte)bf.read(buffer, 0, buffer.length)) > 0) { bfo.write(buffer, 0, read); } bf.close(); } bfo.close(); } }

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Surya

Level 33 , Newark, United States

6 June 2020, 07:41

worked, Thank you.

Surya

Level 33 , Newark, United States

6 June 2020, 04:32

Noted. Thank you I'll update the code to take any generic file name and let you know.

Thomas

Level 41 , Bayreuth, Germany

5 June 2020, 17:55

I'd say you have to code more universal. Lion was meant as an example. If the filename is somewhat different, your solution won't work. I wrote a little helper method that returns the number. Maybe there are more elegant solutions...

public static int parseFileName(String s) {
    return Integer.parseInt(s.substring(s.lastIndexOf(".")).replaceAll("\\D+", ""));
}

As a sidenote, I'd declare the BufferedInputStream bf outside the for each loop. Resourcewise that seems to be far better and may help for getting the last red cross switch to green.