CODEWITHSUNDEEP
java
Marc
Marc

Reputation: 219

How do I create a zipfile from a non-physical file?

I have a project called "MyApp". MyApp will use a java library I am creating called "MyLibrary". I am writing a function in "MyLibrary" to unzip a zip file in "MyApp"(or whatever app is using "MyLibrary") "resources" dir.

Reading https://community.oracle.com/blogs/kohsuke/2007/04/25/how-convert-javaneturl-javaiofile I cannot create a File via a path because it is not a "physical file". I was using zip4j, but its constructor takes a File or String rather than an InputStream. So I cannot do this:

ZipFile zipfile = new    
        ZipFile("src/main/resources/compressed.zip");
downloadedZipfile.extractAll("src/main/resources");

java.io.File javadoc and http://www.mkyong.com/java/how-to-convert-inputstream-to-file-in-java/ indicate there isn't a way to convert an InputStream to a File.

Is there another way to access the zip file in the project that is using my library? Thank you in advance.

UPDATE: The zipIn lacks entries, so the while loop won't extract the file out. 

InputStream in = getInputStream("", JSON_FILENAME);
    ZipInputStream zipIn = new ZipInputStream(in);

    ZipEntry entry;
    try {
        while ((entry = zipIn.getNextEntry()) != null) {
            String filepath = entry.getName();
            if(!entry.isDirectory()) {
                extractFile(zipIn, filepath);
            }
            else {
                File dir = new File(filepath);
                dir.mkdir();
            }
            zipIn.closeEntry();
        }
        zipIn.close();
    } catch (IOException e) {
        e.printStackTrace();
    }

private void extractFile(ZipInputStream zipIn, String filepath) {
    BufferedOutputStream bos = null;
    try {
        bos = new BufferedOutputStream(new FileOutputStream(filepath));
    } catch (FileNotFoundException e) {
        e.printStackTrace();
    }
    byte[] bytesIn = new byte[BUFFER_SIZE];
    int read = 0;
    try {
        while ((read = zipIn.read(bytesIn)) != -1) {
            bos.write(bytesIn, 0, read);
        }
        bos.close();
    } catch (IOException e) {
        e.printStackTrace();
    }
}

Will the file be extracted into the home directory of "MyApp" when using this code that belongs to "MyLibrary"?

Upvotes: 0

Views: 95

Answers (1)

wero
wero

Reputation: 32980

If you have a InputStream to your virtual ZIP file you can use java.util.zip.ZipInputStream to read the ZIP entries:

InputStream in = ...
ZipInputStream zipIn = new ZipInputStream(in);

ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
     // handle the entry
}

Upvotes: 2

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