CodeGym/Help with Java Tasks/Help Me

Level 31

Vijayawada

21.03.2020
1007views
10comments

Help Me

Question about the task

Archiver (part 2)

Java Multithreading, Level 7, Lesson 15

Resolved

It is correct but not verifying

Now let's implement the createZip(Path source) method, which will zip the file specified by the source argument.
Java has a special ZipOutputStream class in the java.util.zip package. It compresses (zips)
the data passed to it. To keep multiple files from blending together when they are compressed into one archive, we create a special entity called a ZipEntry for each of them. In other words, we first put a ZipEntry in the ZipOutputStream, and then we write the file's contents. When the file is written, it is automatically compressed, and when it is read, it is automatically decompressed. A ZipEntry doesn't have to be a file; it can also be a folder.

To zip a file (create a new archive and add the file to it):
1. Create a new ZipOutputStream using the zipFile variable and the Files class's newOutputStream method.
2. Create a new ZipEntry. Pass a String containing the new entry's name to the ZipEntry constructor.
You need to get the name from the full path source. Get only the file name and convert it to a String. 3. Add the created zip entry to the zip stream.
4. Copy the data from the file we are zipping to the zip stream. To do this:
4.1. Create an InputStream for the source file using the Files class's newInputStream method.
4.2. Make a loop that will read data from the InputStream (created in 4.1) as long as it has data and write that data to the ZipOutputStream (created in step 1)
4.3. Close the InputStream using a try-with-resources
5. Close the zip stream's zip entry
6. Close the zip stream using a try-with-resources
7. Run the program and verify that a file can be zipped

Requirements:

The createZip method must create a ZipOutputStream using the zipFile field and the Files.newOutputStream method.
The createZip method must create a ZipEntry using the file name obtained from the source parameter.
The created ZipEntry must be added to the ZipOutputStream.
An InputStream must be created for the source parameter using the Files.newInputStream method.
Data from the InputStream needs to be copied to the ZipOutputStream.
Close the current ZipOutputStream object's current entry.
The InputStream for the source must be closed.
The ZipOutputStream must be closed.

package com.codegym.task.task31.task3110; import java.io.FileInputStream; import java.io.IOException; import java.io.InputStream; import java.nio.file.Files; import java.nio.file.Path; import java.util.zip.ZipEntry; import java.util.zip.*; public class ZipFileManager { private Path zipFile; private Path source; public ZipFileManager(Path zipFile){ this.zipFile=zipFile; } public void createZip(Path source) throws Exception{ ZipOutputStream out = new ZipOutputStream(Files.newOutputStream(zipFile)); InputStream in = new ZipInputStream(Files.newInputStream(source)); ZipEntry ze = new ZipEntry(source.toString()); out.putNextEntry(ze); while(in.available()>0){ int count = in.read(); out.write(count); } in.close(); out.close(); } }

Comments (10)

Popular
New
Old

You must be signed in to leave a comment

Seb

Level 41 , Crefeld, Germany

21 March 2020, 09:30

It looks to me as if your Archiver and ZipFileManager classes have the same content. Your Archiver class should look different.

Satya Vath

Level 31 , Vijayawada, India

21 March 2020, 09:33

package com.codegym.task.task31.task3110;


import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.nio.file.Path;
import java.nio.file.Paths;

public class Archiver {
    public static void main(String[] args) {
        BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));
        Path p = null;
        try {
            p = Paths.get(buf.readLine());
            ZipFileManager zipFileManager = new ZipFileManager(p);
            zipFileManager.createZip(p);
            String fileToZip = buf.readLine();
            Path path2 = Paths.get(fileToZip);
            zipFileManager.createZip(path2);
        }
        catch(Exception r){

        }
    }
}

Satya Vath

Level 31 , Vijayawada, India

21 March 2020, 09:33

This is My Archiever Class

Seb

Level 41 , Crefeld, Germany

21 March 2020, 09:50

useful

Without really running your code, it's a little tricky to pinpoint the exact problem, but you can try the following: Adjust line 21 of your ZipFileManager class as follows: ZipEntry ze = new ZipEntry(new File(source.toString()).getName()); Delete line 16 of your Archiver class. Close the BufferedReader in your Archiver class. I'm not entirely sure whether that's it, but I hope it'll help.

Satya Vath

Level 31 , Vijayawada, India

21 March 2020, 10:06

Infinite loop

Seb

Level 41 , Crefeld, Germany

21 March 2020, 20:39

solution

Task conditions 4.3 and 6: Try with resources needs to be used... Something like: try (ZipOutputStream zipOutputStream = new ZipOutputStream(Files.newOutputStream(zipFile))) and try (InputStream inputStream = Files.newInputStream(source))

Satya Vath

Level 31 , Vijayawada, India

22 March 2020, 05:47

Can You Give Eloberated Code On It

Satya Vath

Level 31 , Vijayawada, India

22 March 2020, 06:03

Still Infinite Loop

Seb

Level 41 , Crefeld, Germany

22 March 2020, 08:11

Check your inbox.

Henrique

Level 41 , São Paulo, Brazil

12 August 2020, 14:47

Hey, Seb, can you help me here? I just don't understand ANYTHING about this thing, and I swear that I'm trying. My code is here: https://codegym.cc/help/12194