Why this is not working?

Question about the task

Under discussion

I don't get it. I look to some other solutions. It is more or less doing the same thing, but just saving the content somewhere else first, but it is kind of the same at the end, no? Also, why don't we simply skip the entries to add the new one?

The main method accepts a list of arguments.
The first argument is the full path to the fileName file.
The second argument is the path to a ZIP file.
Add the fileName file to the archive inside the "new" directory.
If the archive already has a file with the same name in the "new" directory, then replace it.

Example input:
C:/result.mp3 C:/pathToTest/test.zip

Files inside test.zip:
a.txt b.txt

After running Solution.main, the test.zip archive must contain the following:
new/result.mp3
a.txt b.txt

Hint: you need to first save the contents of all of the zip entries somewhere,
and then write all the entities together with the added file to the archive.
Don't use the file system.

Requirements:

In the main method, create a ZipInputStream for the archive (main's second argument). You need to read all of its contents.
In the main method, create a ZipOutputStream for the archive (main's second argument).
You need to write the contents of the file passed to main as the first argument to the ZipOutputStream.
You need to write all of the other content read from the ZipInputStream to the ZipOutputStream.
The archive streams must be closed.

package com.codegym.task.task31.task3105; import java.io.*; import java.nio.file.Files; import java.util.zip.ZipEntry; import java.util.zip.ZipInputStream; import java.util.zip.ZipOutputStream; /* Adding a file to an archive */ public class Solution { public static void main(String[] args) throws IOException { FileInputStream input = new FileInputStream(args[1]); ZipInputStream inputZip = new ZipInputStream(input); FileOutputStream output = new FileOutputStream(args[1]); ZipOutputStream outputZip = new ZipOutputStream(output); ZipEntry entry; ByteArrayOutputStream stream = new ByteArrayOutputStream(); int len; byte[] buffer = new byte[1024]; while ((entry = inputZip.getNextEntry()) != null) { while ((len = inputZip.read(buffer)) > 0) { stream.write(buffer, 0, len); } outputZip.putNextEntry(entry); outputZip.write(stream.toByteArray()); } inputZip.close(); File file = new File(args[0]); outputZip.putNextEntry(new ZipEntry("new/" + file.getName())); Files.copy(file.toPath(), outputZip); outputZip.close(); } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

11 November 2020, 14:37

I don't think you should be reading and writing to the same file at the same time (lines 24-30) Consider saving the zip entry and byte[] to use in an output loop as soon as reading is complete. The biggest issue with this is that when you open up a new OutputStream (and don't flag it to append), it will over write any files (including zip files) that already exist at that location. Make sure to move lines 18-19 to after line 31 to prevent that.