How to get a BufferredReader from a zip file in resource directory?

Question

public static BufferedReader fileReaderAsResource(String filePath) throws IOException {
            InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
            if (is == null) {
                throw new FileNotFoundException(" Not found: " + filePath);
            }
            return new BufferedReader(new InputStreamReader(is, DEFAULT_ENCODING));
        }

This works for non-zip file. But for zip file, how to return a BufferedReader? The following doesn't work since 'fileName' is a relative path under my 'resources' directory:

public static BufferedReader fileZipReader(String fileName) throws IOException {
        ZipFile zip = new ZipFile(fileName);
        for(Enumeration e = zip.entries(); e.hasMoreElements();){
            ZipEntry zipEntry = (ZipEntry) e.nextElement();
            if(!zipEntry.isDirectory()){
                return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
            }
        }
        throw new FileNotFoundException("File not found: " + fileName);
    }

How to change the 'fileZipReader' to make it work?

*"is a relative path under my 'resources' directory"* - Do you mean it's embedded within you application binary to exists as a stand alone directory from within the working directory context? — MadProgrammer, Apr 05 '18 at 04:24
Out so it's an embedded resource, this basically makes it a Zip file in a Zip. You will need to use `Class#getResource` or `Class#getResourceAsStream` and extract the file to a known location — MadProgrammer, Apr 05 '18 at 06:08
Could you give an answer on this? Thank you. I always got null pointer based on Shailesh suggestion. — user697911, Apr 05 '18 at 06:10
[As a conceptual example](https://stackoverflow.com/questions/30430508/java-how-to-open-a-file-located-in-a-jar-file/30430577#30430577) ... you'd have to use `/main/resources` as your path — MadProgrammer, Apr 05 '18 at 06:14

Shailesh Pratapwar · Accepted Answer · 2018-04-05T06:43:19.300

0

Create a URL object giving a relative path to zip file.

public class IOUtil {

    public BufferedReader fileZipReader(String fileName) throws IOException, URISyntaxException {
        URL zipUrl = IOUtils.class.getClassLoader().getResource(fileName);
        File zipFile = new File(zipUrl.toURI());
        ZipFile zip = new ZipFile(zipFile);
        for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
            ZipEntry zipEntry = (ZipEntry) e.nextElement();
            if (!zipEntry.isDirectory()) {
                return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
            }
        }
        throw new FileNotFoundException("File not found: " + fileName);
    }

    public static void main(String[] args) throws Exception {
        IOUtil util = new IOUtil();

        BufferedReader br = util.fileZipReader("dia/test.txt.zip");
    }
}

edited Apr 05 '18 at 06:43

answered Apr 05 '18 at 04:30

Shailesh Pratapwar

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What's the 'Main' here? – user697911 Apr 05 '18 at 05:47
Use your starting or main class name. If not having main class, use any public class or interface name in your project. – Shailesh Pratapwar Apr 05 '18 at 05:47
But I got a null pointer exception for "zipUrl", which is null. I used "URL zipUrl = IOUtils.class.getResource(fileName);" 'fileName' is the relative path of src/main/resources/test.txt.zip – user697911 Apr 05 '18 at 05:55
1

You meant "/resources/...". My relative path doesn't include this '/resources/' part. – user697911 Apr 05 '18 at 05:57
If you file is at location src/main/resources/myfile.zip, you should just give "myfile.zip" to this function. Are you doing the same ? – Shailesh Pratapwar Apr 05 '18 at 05:57
Let me add this part "src/main/resources" to try, – user697911 Apr 05 '18 at 05:58
Strange, I still got null pointer exception even if used the full path, "/Users/cong/nl/web/src/main/resources/dt/test.txt.zip" – user697911 Apr 05 '18 at 06:01
Is that because my IOUtils is a utilities, and my 'fileZipReader' is a static method? I always got null pointer even if I used the full path. – user697911 Apr 05 '18 at 06:03
if you can show me project structure, that will be helpful – Shailesh Pratapwar Apr 05 '18 at 06:05
Since it seems that the file is a embedded resource, you won't be able reference it as a `File` - it's a Zip file in a Zip file – MadProgrammer Apr 05 '18 at 06:08
It's a standard Java maven structure. – user697911 Apr 05 '18 at 06:08
ok. Don't use full path. Only add the path ahead from src/main/resources. Try changing the method to nonstatic, not sure whether it will work. – Shailesh Pratapwar Apr 05 '18 at 06:15
@ShaileshPratapwar Please see my full example, as editted. – user697911 Apr 05 '18 at 06:26
Finally, "IOUtils.class.getClassLoader().getResource(fileName);" works! – user697911 Apr 05 '18 at 06:32
Good. I will just update the answer so that it will benefit others. – Shailesh Pratapwar Apr 05 '18 at 06:41

How to get a BufferredReader from a zip file in resource directory?

1 Answers1