CodeGym/Help with Java Tasks/I must not understand the task conditions

Level 41

Chandler

11/15/22
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9comments

I must not understand the task conditions

Question about the task

Resolved

It seems to work perfectly when I test it. -Reads file names until the String end is entered. -Creates multiple different names for files and gives each name a different .part number. -Creates the actual files -Then writes each file's bytes/contents to the Main file. It will create the files and add all their contents to the main file without the part. I failed verification 2-5

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.

Hello world the sky is falling towards the earth.

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Jaime Padilla

Level 41 , Chandler, United States

Expert

15 November 2022, 22:39

I thought about it and figured maybe they want the user to be able to determine which part to set and I still fail the same requirements

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String fileName = reader.readLine();
        TreeSet<File> list = new TreeSet<>();
        File dest = new File("C:\\Users\\jaypa\\Documents\\Codegym\\CodeGymTasks\\2.JavaCore\\src\\com\\codegym\\task\\task18\\task1825\\Lion.avi");

        while (!fileName.equals("end")) {
            File file = new File("C:\\Users\\jaypa\\Documents\\Codegym\\CodeGymTasks\\2.JavaCore\\src\\com\\codegym\\task\\task18\\task1825\\" + fileName);
            list.add(file);
            file.createNewFile();
            fileName = reader.readLine();
        }

        for (File file : list) {
            try (BufferedReader reader2 = new BufferedReader(new FileReader(file)); BufferedWriter writer = new BufferedWriter(new FileWriter(dest, true))) {
                while (reader2.ready()) {
                    String data = reader2.readLine();
                    writer.write(data);
                    writer.close();
                }
            }
        }
    }
}

Thomas

Level 41 , Bayreuth, Germany

16 November 2022, 08:57

solution

Lion is an example and should not be hardcoded. Same for the path to the files You ask for files (incl. path) till the user enter 'end'

c:\movies\Titanic.avi.part1
c:\movies\Titanic.avi.part2
c:\movies\Titanic.avi.part12
end

from that input you extract the output path c:\movies\ Then you sort that input. Use my above examples to verify that your code sorts it exactly that given way. (hint: it doesn't - proof:)

new TreeSet<>(Arrays.asList("c:\\movies\\Titanic.avi.part1", "c:\\movies\\Titanic.avi.part2", "c:\\movies\\Titanic.avi.part12"))
        .forEach(System.out::println);

Read the content of the files and save it (in the correct order) in the output file c:\movies\Titanic.avi That's all ;) (Another hint: you're using reader/ writer classes. These are excellent for working with character streams. But the example talks about *.avi files. And that's a binary format. So I'm not sure if Codegym will accept the stream classes you've chosen).

Jaime Padilla

Level 41 , Chandler, United States

Expert

18 November 2022, 06:37

I had to get rid of the try resource statements. And I had to change everything with file in it to String. The only way it would pass. It took me the most tries out of any task I've done. Honestly, I'm really disappointed with Codegym's task descriptions.

public class Solution {
    public static void main(String[] args) throws IOException {
        HashMap<File, Integer> map = new HashMap<>();
        LinkedHashMap<File, Integer> sortedSet = new LinkedHashMap<>();

        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String fileName = reader.readLine();
        File file;
        String destin = fileName.substring(0,fileName.lastIndexOf('.'));

        while (!fileName.equals("end")) {
            file = new File(fileName);
            String[] numSort = fileName.split("\\.part");
            map.put(file, Integer.valueOf(numSort[1]));
            fileName = reader.readLine();
        }
        ArrayList<Integer> list = new ArrayList<>(map.values());
        Collections.sort(list);
        for (int num : list) {
            for (Entry<File, Integer> entry : map.entrySet()) {
                if (entry.getValue().equals(num)) {
                    sortedSet.put(entry.getKey(), num);
                }
            }
        }
//        File file1 = new File(destin);
//        for (File x : sortedSet.keySet()) {
//            System.out.println(x);
//        }
//        System.out.println(destin);

        BufferedOutputStream writer = new BufferedOutputStream(new FileOutputStream(destin, true));

        for (File files : sortedSet.keySet()) {
            BufferedInputStream reader2 = new BufferedInputStream(new FileInputStream(files));
                while (reader2.available() > 0)
                    writer.write(reader2.read());

                reader2.close();
        }
        writer.close();
    }
}

Thomas

Level 41 , Bayreuth, Germany

18 November 2022, 09:16

useful

That was also a really difficult task and solving it shows that you have already learned a lot. I've coded this one several times in the meanwhile and can tell you, that try-with-resources is not a problem. Also your approach using a TreeSet was excellent thinking. The only thing missing was that you need to change the way the TreeSet sorts (Comparator is the keyword). That Comparator again is not easy and you probably haven't heared of it before. But a TreeMap (Integer, String or Path/ File) is also possible. Or, as you did it a HashMap and then you just sort (depending on how you designed the code) the keySet or values. For the streams... they asked you to use a buffer and imho you nearly always should do that if working with files. You either use an already coded class or a byte buffer. To know how this works is useful every now and then. Take care

Jaime Padilla

Level 41 , Chandler, United States

Expert

18 November 2022, 18:01

Thanks Thomas, I definitely did learn a lot. It took me practically 2 days because I was confused about what the task was asking and also because I was being stubborn and wanted it to work with the files instead of String. I tried the comparator route and learned a lot about it, but in the end, I didn't understand how to use one comparator to sort by String then by Integer, and then link everything together; I couldn't grasp that just yet. How does the comparator work when passed into a TreeSet? Does it sequentially pass each element through our comparator and sort? Where is the sorting code? What method is taking care of the sort? My mind was fried at many points during this task.... probably still is lol. I feel like if I fully understood the comparator the task would have been much easier.

Jaime Padilla

Level 41 , Chandler, United States

Expert

18 November 2022, 18:10

Why would this not pass? This is my original code before I changed data types to String and got rid of the try resource statement

public class Solution {
    public static void main(String[] args) throws IOException {
        HashMap<File, Integer> map = new HashMap<>();
        LinkedHashMap<File, Integer> sortedSet = new LinkedHashMap<>();

        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String fileName = reader.readLine();
        File file;
        String destin = fileName.substring(0,fileName.lastIndexOf('.'));

        while (!fileName.equals("end")) {
            file = new File(fileName);
            String[] numSort = fileName.split("\\.part");
            map.put(file, Integer.valueOf(numSort[1]));
            fileName = reader.readLine();
        }
        ArrayList<Integer> list = new ArrayList<>(map.values());
        Collections.sort(list);
        for (int num : list) {
            for (Entry<File, Integer> entry : map.entrySet()) {
                if (entry.getValue().equals(num)) {
                    sortedSet.put(entry.getKey(), num);
                }
            }
        }
        File file1 = new File(destin);
//        for (File x : sortedSet.keySet()) {
//            System.out.println(x);
//        }
//        System.out.println(destin);


        for (File files : sortedSet.keySet()) {
            try (BufferedInputStream reader2 = new BufferedInputStream(new FileInputStream(files)); BufferedOutputStream writer = new BufferedOutputStream(new FileOutputStream(file1, true))) {
                while (reader2.available() > 0)
                    writer.write(reader2.read());
            }
        }
    }
}

Thomas

Level 41 , Bayreuth, Germany

18 November 2022, 18:54

Haven't tested your code, just had a quick look. The problem probably is the usage of file append in your FileOutputStream. Validation doesn't like that you you should avoid it like hell (unless stated in the conditions to use it). Validation does not use real files on disks. It probably uses some virtualization, some class that acts like a file. And once it's closed, it's gone.

Jaime Padilla

Level 41 , Chandler, United States

Expert

18 November 2022, 21:55

Oddly enough it somehow passed with the append true. I will try to stay away from using it in the future. Although, if I am trying to put all bytes from multiple files into one file, wouldn't I need to append? Otherwise, it just erases everything from the previous file. Thanks for the responses, I am learning a lot.

Thomas

Level 41 , Bayreuth, Germany

18 November 2022, 22:26

I hope it's understandable

try (OutputStream out = new FileOutputStream("out.txt")) {
    out.write("abc".getBytes());
    // as long as you do not close the out stream, subsequent writes will be appended
    out.write("def".getBytes());
    // doesn't matter if you open an input stream and then write from it to the out stream
    try (InputStream in = new FileInputStream("in.txt")) {
        for (int data; (data = in.read()) != -1; out.write(data)) ;
    }
}
// now the out file has been closed. If you open it again for writing it'll be deleted (except of course
// you open it in append mode)
try (OutputStream out = new FileOutputStream("out.txt")) {

}

Then it'll look for that task similar to (if files is eg. a TreeSet<String>)

try (OutputStream out = new BufferedOutputStream(new FileOutputStream(baseName))) {
    for (String file : files) {
        try (InputStream in = new BufferedInputStream(new FileInputStream(file))) {
            while (in.available() > 0) {
                out.write(in.read());
            }
        }
    }
}
// alternative using a byte buffer
try (OutputStream out = new FileOutputStream(baseName)) {
    byte[] buffer = new byte[8 * 1024];
    for (String file : files) {
        try (InputStream in = new FileInputStream(file)) {
            int len;
            while ((len = in.read(buffer)) > 0) {
                out.write(buffer, 0, len);
            }
        }
    }
}