How to read all ".java" files from a ".zip" file

Question

I want to get all .java files and put them in an ArrayList of files. I have taken a ZipInputStream and then a ZipEntry to iterate through files but can not understand how to get the files from the ZipEntry.

public List<File> getJavaFiles(MultipartFile file){
    List<File> javaFiles = new ArrayList<File>();

    ZipEntry zipEntry;

    log.info("getJavaFiles");
    try {
        ZipInputStream zip;
        try {
            zip = new ZipInputStream( file.getInputStream());
            while((zipEntry = zip.getNextEntry()) != null){
                if(zipEntry.getName().endsWith(".java")){
                    log.info(zipEntry.getName());

                    //How do I put the java file in my array list
                }
            }
            zip.close();
        } catch (IOException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        }


    } catch (Exception e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }


    return javaFiles;
}

How do you want to use returned `List`? What is its purpose? — Pshemo, Sep 14 '17 at 11:32
Please check this way [link](https://stackoverflow.com/a/36890475/1012497) — nikli, Sep 14 '17 at 11:33
I want to get all the comments and identifiers of the java files to check code smells. So I wanted to get all the java files from this class. Then work with those files (i.e get comments and identifiers to check code smells). @Pshemo — Sharafat Ahmed Sabir, Sep 14 '17 at 11:37
`File` describes location of file like zip file, not its entry. Maybe you want URI instead. Visit "[Get a File or URI object for a file inside an archive with Java?](https://stackoverflow.com/q/2049659)" and "[Read directly a file within a Zip file - Java](https://stackoverflow.com/q/13399448)" for more info. — Pshemo, Sep 14 '17 at 11:54
Thanks for the clarifications and the links. This helps a lot. @Pshemo — Sharafat Ahmed Sabir, Sep 14 '17 at 12:15

score 6 · Accepted Answer · answered Sep 14 '17 at 11:47

Here is the code that can read and print the .java files in zip files::

You can also get the file and its content using this code.

import java.io.IOException;
import java.io.InputStream;
import java.nio.charset.StandardCharsets;
import java.util.Enumeration;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;

import org.apache.commons.io.IOUtils;

public class test {
    public static void main(String[] args) throws IOException {
        ZipFile zipFile = new ZipFile("test.zip");
        Enumeration<? extends ZipEntry> entries = zipFile.entries();
        System.out.println(entries);

        while(entries.hasMoreElements()){
            ZipEntry entry = entries.nextElement();
            System.out.println(entry.getName());
            if(entry.getName().contains("java")){
                InputStream stream = zipFile.getInputStream(entry);
                System.out.println(IOUtils.toString(stream, StandardCharsets.UTF_8));
            }
        }
    }
}

But as I am using MultipartFile instead of ZipFile (Though the file is actually a zip file), can you tell me how to get the string from it? — Sharafat Ahmed Sabir, Sep 15 '17 at 05:45
You can refer this link https://stackoverflow.com/questions/24339990/how-to-convert-a-multipart-file-to-file — Abhishek Honey, Sep 15 '17 at 06:10

score 0 · Answer 2 · answered Sep 14 '17 at 11:40

0

Maybe you can directly use File like below:

File javaFile = new File(zipEntry.getName());
javaFiles.add(javaFile);

answered Sep 14 '17 at 11:40

nikli

2,281
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But this will not have the original contents of the file that is in the zip file – Sharafat Ahmed Sabir Sep 14 '17 at 11:43

score 0 · Answer 3 · answered Sep 14 '17 at 11:45

Please check with this code

 ZipFile zipFile = new ZipFile("C:/test.zip");

Enumeration<? extends ZipEntry> entries = zipFile.entries();

while(entries.hasMoreElements()){
    ZipEntry entry = entries.nextElement();
    InputStream stream = zipFile.getInputStream(entry);
}

ssc327 · Answer 4 · 2017-09-14T12:10:06.533

If you want a list of Files that you will later traverse, then first you need to extract the files somewhere. The others answers have examples of how to fetch the inputstream and then write it to disk so I won't repeat that.

Once you extract the files, then you can populate your list with references to them.

p.s. I am guessing that you are doing this for a project or something but in general tools like Sonarqube are great from static code analysis.

How to read all ".java" files from a ".zip" file

4 Answers4