CodeGym/Help with Java Tasks/Getting frustrated...

Level 29

Flensburg

12.05.2021
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11comments

Getting frustrated...

Question about the task

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I googled different ways how to unzip a file etc, but the information i find is so different and I really don't know how to make this work. On my computer the program seems to work, I end up getting a zip file which contains a.txt , b.txt and a new/ folder with result.mp3 . I am so frustrated, because I can't really find a way how to do this properly, just a bunch of different ways which I don't understand. The little info provided by CG doesn't help at all. The out-commented lines are different approaches which didn't work as well. Another point which I don't understand - in the Conditions it states " don't use the file system" . But when I used another way of adding the file to the archive (e.g. putting another ZipEntry to the List), the validator said: "To write a file to the archive stream, use the Files.copy(Path, OutputStream) method." Can someone help me or give me a proper source of information for this topic? Thanks in advance.

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.nio.file.Path; import java.nio.file.Paths; import java.util.ArrayList; import java.util.List; 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 { /*Path fileName = Paths.get(args[0]); // /Users/test/Desktop/Folder/result.mp3 Path zipFile = Paths.get(args[1]); // /Users/test/Desktop/Folder/test.zip */ String fileName = args[0]; String zipFile = args[1]; List<ZipEntry> zipFiles = new ArrayList<>(); //zipFiles.add(new ZipEntry("new/" + Paths.get(fileName).getFileName())); ZipInputStream zipInputStream = new ZipInputStream(new FileInputStream(zipFile)); ZipEntry zipEntry = null; while((zipEntry = zipInputStream.getNextEntry()) != null){ zipFiles.add(zipEntry); } zipInputStream.close(); FileOutputStream outStream = new FileOutputStream(zipFile); ZipOutputStream zipOutputStream = new ZipOutputStream(outStream); File file = new File(fileName); zipOutputStream.putNextEntry(new ZipEntry("new/" + file.getName())); Files.copy(file.toPath(), zipOutputStream); //zipOutputStream.putNextEntry(new ZipEntry("new/" + Paths.get(fileName).getFileName())); for(ZipEntry z : zipFiles){ zipOutputStream.putNextEntry(z); } zipOutputStream.close(); } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

12 May 2021, 14:22

useful

In the above code I see that zip entries from the original zip file are saved, and then put back in the output stream; however the actual bytes for those files are not saved and subsequently written back. If all the files are 0 bytes then the code will complete, however any file larger than that will crash the code with an invalid entry size error. The other problem is that the new file will not be saved correctly if there is already an existing file with the same name in the original zip file.

Guadalupe Gagnon

Level 37 , Tampa, United States

12 May 2021, 14:38

useful

Some tips: - you can't get the individual file size of each file in a zip file for purposes of reading - ByteArrayOutputStream is a great object to store bytes. When adding bytes it will automatically resize as necessary - you need to store each file's bytes individually, you can't just read the whole zip file and write the whole zip file. You need to first add the entry then write the bytes - make sure to check, when writing the old bytes, that you don't write "new\<filename>" is it exists.

Jurij Thmsn

Level 29 , Flensburg, Germany

17 May 2021, 06:23

Thanks for your answers, Guadalupe. I'll give it another try. I always try to solve the tasks just with the information provided by CodeGym, but this approach doesn't seem to work on this task. I don't really get why I have to write a code wich writes the bytes "on my own", I thought the ZipOutPutStream does this job. Even after re-reading the information provided by CG I don't seem to understand what a ZipEntry really is and what the ZipStreams do. I'll just check another Tutorial on writing zip files and try to reproduce the code. I guess sometimes I just can't understand everything and just have to accept how it's done.

Jurij Thmsn

Level 29 , Flensburg, Germany

17 May 2021, 07:36

I found about 5 tutorials with totally different approaches on the topic but I am really unsure about which one to use, because every on of them uses "File" to a certain extent. This confuses me: in the Conditions it states " don't use the file system" . But when I used another way of adding the file to the archive, the validator said: "To write a file to the archive stream, use the Files.copy(Path, OutputStream) method." so what do they mean by " don't use the file system"? am I allowed to use Files but not File? Just Paths? I don't want to end up just copying a code from the help section, but every tutorial I found uses File...

Guadalupe Gagnon

Level 37 , Tampa, United States

17 May 2021, 13:22

solution

ZipEntry is the actual path name for files in a ZipFile. Because the data for them is compressed you need to do 2 operations when reading/writing using a Zip(In/Out)putStream. That is first get/set the next ZipEntry, then read/write the bytes for that ZipEntry. You do not need to use the Files class for this, you just need to store the ZipEntries and byte[] arrays from the zip file(a Map<ZipEntry, byte[]> works for this). You can read all the data into the map, then write it all back including the new file. Just make sure to check that the new file is unique; or if it is already contained in the zip file then it replaces that one. Here is one implementation (what I was explaining above) of how the code should generally look: ---------------------------------- create Map<ZE, byte[]> //reading loop: loop all the ZEs in zipfile{ get next zip entry, save to variable set a buffer loop to read the bytes{ read all the bytes into the buffer } add the zip entry and bytes into the map } //writing loop loop all the entries in map{ check first if the ZE is the same as the file being added, skip if true put next zipEntry write all bytes } // add file put zipEntry for the file write the bytes ----------------------------------

Jurij Thmsn

Level 29 , Flensburg, Germany

18 May 2021, 06:26

Thanks a lot Guadalupe! This makes it clear. The key for understanding this is the need to do the two operations when reading/writing.

Jurij Thmsn

Level 29 , Flensburg, Germany

18 May 2021, 12:38

solved it finally 🤯 couldn't make it work with ZipEntries in Map, so I took the Names.

HashMap<String, byte[]> map = new HashMap<>();

        ZipInputStream zipIn = new ZipInputStream(new FileInputStream(zipFile)); //inputStream from the zipFile
        ZipEntry zipEntry;
        byte[] buffer = new byte[1024];
        while((zipEntry = zipIn.getNextEntry()) != null){ //if there is a next entry in zipfile,
            ByteArrayOutputStream outStream = new ByteArrayOutputStream();
            int length;
            while((length = zipIn.read(buffer)) > 0){ //while there are bytes read from the zipEntry to the buffer,
                outStream.write(buffer, 0, length); //write bytes of buffer to the outStream
            }
            map.put(zipEntry.getName(), outStream.toByteArray()); //put bytes of buffer and name of ZipEntry into map
        }
        zipIn.close();

Jurij Thmsn

Level 29 , Flensburg, Germany

18 May 2021, 12:39

I tried to explain my thoughts in the comments, maybe it helps someone:

ZipOutputStream zipOut = new ZipOutputStream(new FileOutputStream(zipFile)); //outputStream to the Zip File
File file = new File(fileName);
zipOut.putNextEntry(new ZipEntry("new/"+ file.getName())); //new entry inside zip file, just the name of the entry, no content. name is name of "fileName" file
Files.copy(file.toPath(), zipOut); // copying the content to the recently named zipEntry, can be copied like this because its no ZIP ( no need to decrompress)

for(Map.Entry<String, byte[]> pair : map.entrySet()){
    String tempName = pair.getKey();
    byte[] tempBytes = pair.getValue();
    if(!tempName.equals("new/" + file.getName())){ //if file has same name like "fileName" file, skip.
        zipOut.putNextEntry(new ZipEntry(tempName)); //put new ZipEntry with name of recent ZipEntry in archive of zipFile
        zipOut.write(tempBytes); //write contents
        zipOut.closeEntry();
    }
}
zipOut.close();

What I still don't understand: why do I need to create the "file"? it already exists on my drive

File file = new File(fileName);

Guadalupe Gagnon

Level 37 , Tampa, United States

18 May 2021, 13:10

useful

Great solve! You don't need the File class specifically, though the functionality it provides makes the code easier. In this code you are using it to get the name of just the file, not its full qualified path. So if the full path were: C:\users\default\desktop\testFile.txt getName() would return: "testFile.txt" You could do the same thing with just the String class (substring() and lastindexof("/")), but the File is already set up to do this. One tip, and this isn't a performance fix, just a thing I do to keep the code structure clean. Think in a way to remove indents from the code. Take lines 9-13 of the writing loop you linked above. At line 9, instead of checking if the fileName does not equal, check that it does equal and continue the looping if it does:

if(tempName.equals("new/" + file.getName())) continue;
/*no need to indent lines*/

Jurij Thmsn

Level 29 , Flensburg, Germany

18 May 2021, 13:19

Thanks for your explanations! I'll try do to this in the future!

Andrei

Level 41

16 June 2021, 08:44

Thanks for sharing your solution! 👍