CodeGym/Help with Java Tasks/My code works as described in the task. What is wrong?

Level 22

Seversk

05.04.2021
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9comments

My code works as described in the task. What is wrong?

Question about the task

Iterating through a file tree

Java Collections, Level 1, Lesson 2

Archived

I implemented recursive file collection from all inner folders, but my code do not pass. What is wrong?

1. The main method is passed two arguments as inputs.
The first argument is path, which is the path to the directory; the second is resultFileAbsolutePath, which is the name (full path) of an existing file that will store the result.
2. For each file in the path directory and in all of its subdirectories, do the following:
For each file whose size in bytes is NOT greater than 50, do the following:
2.1. Sort them by file name in ascending order. Don't include the path when sorting.
2.2. Rename resultFileAbsolutePath to "allFilesContent.txt" (use the FileUtils.renameFile method and FileUtils.isExist if necessary).
2.3. Sequentially write the contents of each file from step 2.1 to allFilesContent.txt. After the body of each file, write "\n".
All files have the TXT extension.
Use "/" as the path separator.

Requirements:

The file passed as the second argument to main must be renamed to allFilesContent.txt.
You need to create a stream to write to the renamed file.
The contents of every file that is 50 bytes or smaller must be written to the allFilesContent.txt file.
The output stream to the file must be closed.
Don't use static variables.

package com.codegym.task.task31.task3101; /* Iterating through a file tree */ import java.io.*; import java.nio.file.Path; import java.nio.file.Paths; import java.util.*; public class Solution { public static void main(String[] args) throws IOException { String fullName = args[0] + "/" + args[1]; // args[0] - path, args[1] - name File container = new File(fullName); Path pathToContainer = Paths.get(container.getParent()); File newContainer = new File(pathToContainer + "/" + "allFilesContent.txt"); FileUtils.renameFile(container, newContainer); File path = new File(newContainer.getParent()); File[] files = path.listFiles(); List<File> allFiles = new ArrayList<>(); getAllFiles(allFiles, files); FileOutputStream output = new FileOutputStream(newContainer); sortFiles(allFiles); //System.out.println(allFiles); for (File file : allFiles) { if (file.equals(newContainer)) { continue; } if (file.length() <= 50) { FileInputStream input = new FileInputStream(file); int i=-1; while((i = input.read()) != -1) { output.write(i); //System.out.println((char) i); } output.write('\n'); output.flush(); input.close(); } } output.close(); } private static void sortFiles(List<File> allFiles) { for (int i = 0; i < allFiles.size(); i++) { for (int j = i + 1; j < allFiles.size(); j++) { if (allFiles.get(i).getName().compareTo(allFiles.get(j).getName()) > 0) { File temp = allFiles.get(i); allFiles.set(i, allFiles.get(j)); allFiles.set(j, temp); } } } } private static void getAllFiles(List<File> allFiles, File[] files) { for (File file : files) { if (file.isDirectory()) { File[] names = file.listFiles(); getAllFiles(allFiles, names); } else { allFiles.add(file); } } } }

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Nouser

Level 36 , Germany

5 April 2021, 07:59

I've experienced a lot of problems here and could just solve that task with try and fail. Therer were 2 points where validation drove me mad. 1. I've created the set and then renamed the file... the collection needs to be crated afterwards or you fail. 2. I used a try-catch block for the output stream ans had to declare the set inside the try block or it won't validate (don't ask me why CG asks for that) 3. renaming is critical and the first thing you have to do (if not, guess what) is to check, if the destination file exists, if yes, delete it. Then rename source -> dest

Dmitri

Level 22 , Seversk, Russia

5 April 2021, 09:13

I added your recommendations to the Solution class but it dad not help:

public class Solution {
    public static void main(String[] args) {
        String fullName = args[0] + "/" + args[1]; // args[0] - path, args[1] - name
        File container = new File(fullName);
        Path pathToContainer = Paths.get(container.getParent());
        File newContainer = new File(pathToContainer + "/" + "allFilesContent.txt");
        if (newContainer.exists()) {
            if (newContainer.delete()) {
                System.out.println("New container exists and was deleted.");
            }
        }
        FileUtils.renameFile(container, newContainer);
        try {
            File path = new File(newContainer.getParent());
            File[] files = path.listFiles();
            List<File> allFiles = new ArrayList<>();
            getAllFiles(allFiles, files);
            FileOutputStream output = new FileOutputStream(newContainer);
            sortFiles(allFiles);
            for (File file : allFiles) {
                if (file.equals(newContainer)) {
                    continue;
                }
                if (file.length() <= 50) {
                    FileInputStream input = new FileInputStream(file);
                    int i;
                    while ((i = input.read()) != -1) {
                        output.write(i);
                    }
                    output.write('\n');
                    output.flush();
                    input.close();
                }
            }
            output.close();
        } catch (IOException e) {
            System.out.println(e.getMessage());
        }
    }

Nouser

Level 36 , Germany

5 April 2021, 10:11

Your fullName, container setup seems also not to be correct. Both arguments are pathes. The first is a pah to the directory you should examine. The second is a path to a file that you should use to write the result into. That second paths file name you should rename to allFilesContent.txt That's how I made it

File sourceDir = new File(args[0]);
File container = new File(args[1]);

//create a new destination File
File dest = new File(container.getParent() + "/" + "allFilesContent.txt");
renameFile(container, dest);

the renameFile method is using only FileUtil methods if (exists) delete rename then the try catch block (I've used try with resources here)

try (BufferedOutputStream out = new BufferedOutputStream(new FileOutputStream(dest))) {

    //instantiate a TreeSet with sort order by filename (excl. path), ascending
    Set<File> allFiles = new TreeSet<>((File o1, File o2) -> o1.getName().compareTo(o2.getName()));

    //fill the set with Files matching the task criteria (files root dir and all subdirs, txt and <= 50b)
    //getFiles returns the Set again
    for (File file : getFiles(allFiles, sourceDir)) {
        out.write(Files.readAllBytes(file.toPath()));
        out.write('\n');
    }
} catch (IOException ioe) {/* ignore */}

Dmitri

Level 22 , Seversk, Russia

6 April 2021, 11:46

Thank you! It improved, now it is only - "The contents of every file that is 50 bytes or smaller must be sorted by file name and written to the allFilesContent.txt file." Though it looks my code sorts files pretty good...

Nouser

Level 36 , Germany

6 April 2021, 13:36

how's your code looking atm? The getAllFiles method looks good. If you call that from line 8 above and add a check for the 50kb limit in the else clause, then it should work. The sort method you won't need anymore if you use the set.

Dmitri

Level 22 , Seversk, Russia

7 April 2021, 07:27

sortFiles method just sort them by name:

private static void sortFiles(List<File> allFiles) {
        for (int i = 0; i < allFiles.size(); i++) {
            for (int j = i + 1; j < allFiles.size(); j++) {
                if (allFiles.get(i).getName().compareTo(allFiles.get(j).getName()) > 0) {
                    File temp = allFiles.get(i);
                    allFiles.set(i, allFiles.get(j));
                    allFiles.set(j, temp);
                }
            }
        }
    }

ok, I shall try set.

Dmitri

Level 22 , Seversk, Russia

7 April 2021, 07:34

Actually, if I used set that would sort files by absolute name ( with path) which probably is not what is needed.

Nouser

Level 36 , Germany

7 April 2021, 07:43

Try the TreeSet I posted above. It does a perfect job and sorts by file name. Then drop your sort method and just adapt the getAllFiles method.

Dmitri

Level 22 , Seversk, Russia

7 April 2021, 08:47

Thank you for your tips, but I think that it is enough guessing for this task, I just move to the next task.