CodeGym/Pomoc z zadaniami Java/What I missed?

Poziom 20

Debica

08.09.2024
90widoki
13komentarze

What I missed?

Pytania dotyczące zadania

Rozwiązane

Hello, Mates. And I don't know what I do wrong? 2nd req. is in line 35 and there is whole path. 3rd. req all data should be in TreeSet list ??? Or I should use a different List. 4th req. Idk they want to use byte array or BufferedInputStream or for write BufferedOutputStream? 5th req. both are closed. line 30 and 42

Stwórzmy plik złożony z różnych elementów.
Odczytuj nazwy plików z konsoli.
Każdy plik ma swoją nazwę: <someName>.partN.

Na przykład, Lion.avi.part1, Lion.avi.part2, ..., Lion.avi.part37.

Nazwy plików są podawane w kolejności losowej. Słowo "end" jest używane do zatrzymywania odczytu nazw plików.
W folderze, w którym znajdują się wszystkie pliki, utwórz plik bez rozszerzenia "part", tzn. bez ".<partN>".

Na przykład, Lion.avi.

Użyj bufora, aby skopiować wszystkie bajty z plików częściowych do utworzonego pliku.
Skopiuj pierwszy w odpowiedniej kolejności, najpierw pierwszą część, potem drugą, ..., w końcu - część ostatnią.
Zamknij strumienie.

Wymagania:

Program musi odczytywać nazwy plików z konsoli do momentu wprowadzenia słowa "end".
Utwórz strumień do zapisu do pliku bez sufiksu "part" (".<partN>") w folderze z wszystkimi plikami "part".
Skopiuj wszystkie bajty z plików *.partN do nowego pliku.
Do odczytu i zapisu należy używać bufora.
Strumienie pliku muszą zostać zamknięte.
Nie używaj zmiennych statycznych.

package pl.codegym.task.task18.task1825; import java.io.*; import java.util.Comparator; import java.util.Set; import java.util.TreeSet; /* Tworzenie pliku */ public class Solution { public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String nameFile = ""; TreeSet<Integer> mySet = new TreeSet<>(); while (true) { nameFile = reader.readLine(); if (nameFile.equals("end")) { break; } FileInputStream fileInputStream = new FileInputStream(nameFile); byte[] inputBuffer = new byte[fileInputStream.available()]; while (fileInputStream.available() > 0) { mySet.add(fileInputStream.read(inputBuffer)); } fileInputStream.close(); } String[] splitedNameFile = nameFile.split("."); // splitedNameFile[0] should contain whole path with file name, and splitedNameFile[1] should attach an extension to name. String newNameFile = splitedNameFile[0] + "." + splitedNameFile[1]; FileOutputStream fileOutputStream = new FileOutputStream(newNameFile); fileOutputStream.flush(); for (Integer i : mySet) { fileOutputStream.write(i); } fileOutputStream.close(); } }

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Thomas

Poziom 41 , Bayreuth, Germany

9 września, 06:50

Using a TreeSet or TreeMap is the way to go here. Going the TreeSet route is far more difficult as you need to write a custom comparator. That said let's look at the reason why it does not validate. You read filenames from the console (in no particular order) and inside that loop you right away read in the content of the files theses filenames point to. Now you have a resulting array and I assume??? that you want to add these arrays to the set. Here we have two problems a) You don't add the array to the set. This aa) would not work as you can not add byte[] arrays to a Set<Integer> bb) the sort order would be? At least nothing related to the filename cc) instead you add the filesize to the Set. That's returned as an int from the read(byte[]) method and therefore you get no exceptions. dd) You could use a map<int, byte[]> and put the array as the payload and the number after part as the key b) so you don't add the content of the file to the set as it does not sort the way you need it. Usually you put the path to the files inside the set. But then again you have the order problem. Just try it out yourself and add following entries to a Set<String>: part1, part2, part3, part10, part11 That's clearly not what you want. As mentioned above, to use set you'd have to delve deeper into Comparator and write your own one that takes this into account. - As said, you could try it the way I mentioned under a). However that uses a lot of memory. You need to load all parts into memeory. - The second (easier) approach is also to use a TreeMap and instead to use the byte[] array as payload you put the path and as key the number after part. That way you avoid sorting problems Next little problem I'd mention is the way you get the filename. This eventually may be sufficient for the tests codegym runs but I'm not sure about that. If CG tests some pathes with additional dots then your target filename is wrong. What you can relay on is a filename with two dots end

Thomas

Poziom 41 , Bayreuth, Germany

9 września, 06:56

For the streams I'd in each case use try-with-resources. Each time. No exceptions. At least not if a stream needs to be closed in the same method it is opened and it supporst auto close. Imho the exact stream type does not matter. InputStream in = new FileInputStream(...) should also do. Using buffered streams is just an optimization (that you of course would use in real projects).

Robert

Poziom 20 , Debica, Poland

9 września, 21:07

So this TreeMap<Integer, byte[]> I should put last number from name of .part1 namely 1. Because it is not saying I put file name to console in order. I can read in random file names, but after that I should keep it in order.

while (true)
        {
            nameFile = reader.readLine();
            if (nameFile.equals("end")) {
                break;
            }
            FileInputStream fileInputStream = new FileInputStream(nameFile);
            byte[] inputBuffer = new byte[fileInputStream.available()];
            while (fileInputStream.available() > 0) {
                myMap.put(Integer.parseInt(String.valueOf(nameFile.charAt(nameFile.length()-1))), inputBuffer);
            }
            fileInputStream.close();
        }

code Tried that way, and gives me time out error. Now I'm trying with this path. You suggest complete path with file name? After that, I will be having a TreeMap <String part, String path> Without the data. Then there will be no problem with long time load? But how then write the data when my reference is TreeMap? It will be tricky. I will try tomorrow now is late - brain off.

Thomas

Poziom 41 , Bayreuth, Germany

10 września, 06:42

rozwiązanie

The dropbox code is task1821 The code part here has the problem that you do not read in the file and that getting the part number is not working correctly. Here is a made up path that should not error out and the result should be 1381. That result is your key.

c:\Temp\movie.parts\lion\lion.avi.part1381

Skopiuj pierwszy w odpowiedniej kolejności, najpierw pierwszą część, potem drugą, ..., w końcu - część ostatnią. From that I understand you are supposed to sort the part files and then the (correct) composition of that files should result in a working movie. So here in short again what I'd do (very pseude code) if you absolutely insist on using byte arrays in containers.

define TreeMap<Integer, String>

loop:
    read filename
    put(partNumber, filename)

define target filename (filename minus .partxxx)

open output stream to target
loop over map, get value
    reconstruct path from value
    open input stream to path
    write input to output
    close input
close output

I've written three versions of that now. The first as outlined here. A second one using a class (record) as holder, a TreeSet and a Comparator to sort the set by content of the Holder class. The third as I suggested, using a TreeMap with the part number as key and the part path as value. That is by far the easiest approach and I recommend to go that route. It avoids some pitfalls and makes (programming) life easier.

Robert

Poziom 20 , Debica, Poland

10 września, 20:39

repaired code from up Ok there is my new code, but still have issue. new code task1825_2 I will try to find some sample.avi and use it to check. What I've found so far is my fileName stores “end” as the last data and that can't be used as a save path, so I'll probably have to take some value from the map.

Robert

Poziom 20 , Debica, Poland

10 września, 20:57

I added in line 32:

nameFile = myMap.get("1");

Thomas, it passes validation 💪

Thomas

Poziom 41 , Bayreuth, Germany

11 września, 07:23

Excellent, that was one very difficult task, Here's my approach using a map that stores the read content (as you did in the beginning, no optimization and input checking)

TreeMap<Integer, byte[]> fileContents = new TreeMap<>();
String fullPath = "";

try (BufferedReader console = new BufferedReader(new InputStreamReader(System.in))) {

    for (String path; !"end".equals(path = console.readLine());) {
        fullPath = path;

        try (InputStream in = new FileInputStream(path)) {
            byte[] bytes = in.readAllBytes();
            String key = path.substring(path.lastIndexOf("t") + 1);
            fileContents.put(Integer.parseInt(key), bytes);
        }
    }
}

String target = fullPath.substring(0, fullPath.lastIndexOf("."));

try (OutputStream out = new FileOutputStream(target)) {
    for (byte[] value : fileContents.values()) {
        out.write(value);
    }
}

Thomas

Poziom 41 , Bayreuth, Germany

11 września, 07:35

Did that work with a TreeMap<String, String>? When I coded the task that won't validate as it gave wrong sorting, hence disallowing an easy approach using a TreeSet. If they changed that, then you could use a Set, too, without having to write a complicated Comparator (however then only parts 0 - 9 sort correctly).

Thomas

Poziom 41 , Bayreuth, Germany

11 września, 07:43

TreeSet and Comparator

TreeSet<String> files = new TreeSet<>(Comparator.comparingInt(String::length).thenComparing(Comparator.naturalOrder()));

try (BufferedReader console = new BufferedReader(new InputStreamReader(System.in))) {
   for (String fileName; !"end".equals(fileName = console.readLine()); files.add(fileName)) ;
}

String baseName = files.first().substring(0, files.first().lastIndexOf('.'));

try (OutputStream out = new FileOutputStream(baseName)) {
    for (String file : files) {
        try (InputStream in = new FileInputStream(file)) {
            out.write(in.readAllBytes());
        }
    }
}

If they changed validation, then you could try just a TreeSet without the Comparator. That'll be a lot easier.

Robert

Poziom 20 , Debica, Poland

11 września, 17:35

Yes, it worked with TreeMap <String, String> task1825_2 file At first, I wasn't sure if it would sort the numbers in String, but after all it's ASCII, and it flew in sequence like a table, so it's cool. I was thinking to parse into Integer and switch to <Integer, String>. In the end, there was no need. Unfortunately I won't be able to test validate with Comparator, unfortunately I don't have the right subscription to revalidate the task.

Robert

Poziom 20 , Debica, Poland

11 września, 17:45

Thomas, I want to ask. You use try catch block different. In previous task, we're using

Try {
} catch {
} finally {
}

You're using

Try (...) {
}

Can you explain your notation?

Thomas

Poziom 41 , Bayreuth, Germany

11 września, 17:59

As mentioned, when I did this task I wasn't able to validate with Map<String, String> or Set<String> without Comparator. And that is correct imho as the Strings sort that way part1 part10 part11 part2 part3 And that is not the order you want. Back then I thought that was what we were supposed to learn. Now this seems to have changed and the task is somewhat easier ;) The try catch block thingie. This is called try-with-resources (you may want to google it). All resources that need to be closed after usage (like streams) and that implement AutoCloseable can be used that way. When leaving the try block the resource gets closed automatically. No need anymore for closing it in the finally block and no need for another try-catch block for the close() statement there anymore (if you want to do it correctly). You of course can (and should) use catch blocks. But here you used throws IOException in the main method definition. And if you just rethrow or just print the stacktrace then the catch block is adding no value to the code.

Robert

Poziom 20 , Debica, Poland

11 września, 18:49

You have right, my bad. I should convert it. Table ASCII have 0-9. They simplified it. The right way will be use <Integer, String> to read larger parts like avi.part4329