Absolutely no idea of what to do. HELP PLS

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idk if it's just me, but I think these requirements and conditions are EXTREMELY confusing i cannot even test the code, because if won't even compile... so, please, help?

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.InputStreamReader; import java.nio.file.Path; import java.nio.file.Paths; public class Archiver { public static void main(String[] args) throws Exception { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String fileName = reader.readLine(); String fileToZip = reader.readLine(); Path path = Paths.get(fileName); Path pathToZip = Paths.get(fileToZip); System.out.println("Please enter the full path to the archive (don't forget that the file name is also part of the full path!)."); ZipFileManager zipFileManager = new ZipFileManager(path); System.out.println("Please enter the path to the file to be zipped (don't confuse this with the archive file, which we already entered!)."); zipFileManager.createZip(pathToZip); } }

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catalinene

Level 29 , Greven, Germany

12 August 2020, 23:59

You are not using a try-with-resources correctly, I would advise you to review the concept. 1.Move the InputStream inside the try-with like this:

try(InputStream inputStream = Files.newInputStream(source)) {
}

2.Remove line 30. The InputStream is going to be clossed automatically after the execution of the try-block 3.Finally you need to make sure that the OutputStream is closed no matter what happens in the try-block 4.Delete line 35-37. Nothing is happening there Hope it helps!

Henrique

Level 41 , São Paulo, Brazil

13 August 2020, 14:06

Hi, catalinene. Thanks for your answer. For me, 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. If I remember correctly, to solve this task I used one try-with-resources inside the other (one for InputStream and one for ZipOutputStream). I don't know if it's the best way, though. Anyway, I appreciate your help!

catalinene

Level 29 , Greven, Germany

13 August 2020, 22:08

Hope you passed! I know it can be confusing sometimes, I had a second Java course this semester at the University and every homework was like this, aka the order of the implementation could have differed from the order of the task requirements. What I ussualy do is read the complete task, make an idea and overview of what is required and then start programming. Even if it takes more time, it can save a lot of frustration