CodeGym/Help with Java Tasks/I have no idea what's wrong, can anybody help?

Level 25

Gdańsk

24.07.2020
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I have no idea what's wrong, can anybody help?

Question about the task

Archiver (part 2)

Java Multithreading, Level 7, Lesson 15

Resolved

Can you give me at least a hint what I should do? Thx.

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.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.nio.file.Path; import java.nio.file.Paths; public class Archiver { public static void main(String[] args) throws Exception { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); System.out.println("Enter full path to archive (.zip):"); String sArchive = br.readLine(); Path pArchive = Paths.get(sArchive); System.out.println("Enter path to file u wanna compress:"); String sFile = br.readLine(); Path pFile = Paths.get(sFile); ZipFileManager z = new ZipFileManager(pArchive); z.createZip(pFile); } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

22 October 2020, 17:49

Your code looks like it has the correct idea, however there is one important change to make: *hint* Data from the InputStream (note: not ZipInputStream) needs to be copied to the ZipOutputStream.

Guadalupe Gagnon

Level 37 , Tampa, United States

22 October 2020, 18:06

solution

Also: 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

BlueJavaBanana

Level 37

22 October 2020, 16:50

Do you still need help?

WIDMO

Level 25 , Gdańsk, Poland

Expert

22 October 2020, 17:22

sure, have you got any tips for me?

BlueJavaBanana

Level 37

23 October 2020, 11:46

solution

Yes. Lose the buffers before your while loop. You are also going to need 2 try with resources, one inside another. The first one will close the out stream, the second will close the in stream. This is how the start of he first try with resources might look. Let me know if you need any help.

try (ZipOutputStream out = new ZipOutputStream(Files.newOutputStream(zipFile))) {

  //Second try goes here with your while loop inside

}

Henrique

Level 41 , São Paulo, Brazil

12 August 2020, 14:36

Did you pass?

WIDMO

Level 25 , Gdańsk, Poland

Expert

12 August 2020, 22:25

nope, but maybe tommorow (or rather today) I'll get back to it.

Henrique

Level 41 , São Paulo, Brazil

13 August 2020, 14:01

The biggest problem with this task is that it's extremely poorly written. It's so confusing. Only at the end of the steps they say that you must "Close the InputStream using a try-with-resources". So I tried to use the try-with-resources construction only at the end, and only to CLOSE the stream. But later I discovered that I had to use the try-with-resources since the beginning. So you need to do something like this and put the code inside it (I don't remember if it's the whole code or not that you put inside):

try (ZipOutputStream zipOutputStream = new ZipOutputStream(Files.newOutputStream(zipFile))) {
//code
}

I don't have the code for this one anymore, but if I remember correctly, I used one try-with-resources inside the other (one for InputStream and one for ZipOutputStream), but I don't remember which one was outer and which was inner. Hope it helps.