Spring REST - create ZIP file and send it to the client

Question

I want to create a ZIP file that contains my archived files that I received from the backend, and then send this file to a user. For 2 days I have been looking for the answer and can't find proper solution, maybe you can help me :)

For now, the code is like this (I know I shouldn't do it all in the Spring controller, but don't care about that, it is just for testing purposes, to find the way to make it works):

    @RequestMapping(value = "/zip")
    public byte[] zipFiles(HttpServletResponse response) throws IOException {
        // Setting HTTP headers
        response.setContentType("application/zip");
        response.setStatus(HttpServletResponse.SC_OK);
        response.addHeader("Content-Disposition", "attachment; filename=\"test.zip\"");

        // Creating byteArray stream, make it bufferable and passing this buffer to ZipOutputStream
        ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
        BufferedOutputStream bufferedOutputStream = new BufferedOutputStream(byteArrayOutputStream);
        ZipOutputStream zipOutputStream = new ZipOutputStream(bufferedOutputStream);

        // Simple file list, just for tests
        ArrayList<File> files = new ArrayList<>(2);
        files.add(new File("README.md"));

        // Packing files
        for (File file : files) {
            // New zip entry and copying InputStream with file to ZipOutputStream, after all closing streams
            zipOutputStream.putNextEntry(new ZipEntry(file.getName()));
            FileInputStream fileInputStream = new FileInputStream(file);

            IOUtils.copy(fileInputStream, zipOutputStream);

            fileInputStream.close();
            zipOutputStream.closeEntry();
        }

        if (zipOutputStream != null) {
            zipOutputStream.finish();
            zipOutputStream.flush();
            IOUtils.closeQuietly(zipOutputStream);
        }
        IOUtils.closeQuietly(bufferedOutputStream);
        IOUtils.closeQuietly(byteArrayOutputStream);

        return byteArrayOutputStream.toByteArray();
    }

But the problem is, that using the code, when I enter the URL localhost:8080/zip, I get a file test.zip.html instead of .zip file.

When I remove .html extension and leave just test.zip it opens correctly. So my questions are:

• How to avoid returning this .html extension?
• Why is it added?

---

I have no idea what else can I do. I was also trying replace ByteArrayOuputStream with something like:

OutputStream outputStream = response.getOutputStream();

and set the method to be void so it returns nothing, but It created .zip file which was damaged?

On my MacBook after unpacking the test.zip I was getting test.zip.cpgz which was again giving me test.zip file and so on.

On Windows the .zip file was damaged as I said and couldn't even open it.

I also suppose, that removing .html extension automatically will be the best option, but how?

Hope it is no as hard as It seems to be :)
Thanks

Backbite

you might find the answer here helpful.. https://mcmap.net/q/412268/-how-to-return-a-zip-file-stream-using-java-springboot – Medium

Answer 1

The problem is solved.

I replaced:

response.setContentType("application/zip");

with:

@RequestMapping(value = "/zip", produces="application/zip")

And now I get a clear, beautiful .zip file.

---

If any of you have either better or faster proposition, or just want to give some advice, then go ahead, I am curious.

Backbite

+1 For non-trivial question and answering your own question! The only question / concern I have with this implementation is the final return statement. I think this would load the entire file into memory because of toByteArray(). Right? I wonder if there is a way to stream it to avoid some OutOfMemoryException errors if the directory is too large. – Mandibular

see this page if you need generic solution for any case: https://mcmap.net/q/410843/-spring-rest-create-zip-file-and-send-it-to-the-client – Marcin

Answer 2

@RequestMapping(value="/zip", produces="application/zip")
public void zipFiles(HttpServletResponse response) throws IOException {

    //setting headers  
    response.setStatus(HttpServletResponse.SC_OK);
    response.addHeader("Content-Disposition", "attachment; filename=\"test.zip\"");

    ZipOutputStream zipOutputStream = new ZipOutputStream(response.getOutputStream());

    // create a list to add files to be zipped
    ArrayList<File> files = new ArrayList<>(2);
    files.add(new File("README.md"));

    // package files
    for (File file : files) {
        //new zip entry and copying inputstream with file to zipOutputStream, after all closing streams
        zipOutputStream.putNextEntry(new ZipEntry(file.getName()));
        FileInputStream fileInputStream = new FileInputStream(file);

        IOUtils.copy(fileInputStream, zipOutputStream);

        fileInputStream.close();
        zipOutputStream.closeEntry();
    }    

    zipOutputStream.close();
}

Capitol

because this solution streams data directly to response output stream. – Extenuatory

This is java server side codes, how to download it in client by javascript?? – Used

@HarveyDent - yous should ask that as a new question. – Capitol

Please, I was stuck into this problem for two whole days. thanks in advance – Used

you just call the url. in the code above it would be http://<host>/zip – Capitol

see this page if you need generic solution for any case: https://mcmap.net/q/410843/-spring-rest-create-zip-file-and-send-it-to-the-client – Marcin

Answer 3

@RequestMapping(value="/zip", produces="application/zip")
public ResponseEntity<StreamingResponseBody> zipFiles() {
    return ResponseEntity
            .ok()
            .header("Content-Disposition", "attachment; filename=\"test.zip\"")
            .body(out -> {
                var zipOutputStream = new ZipOutputStream(out);

                // create a list to add files to be zipped
                ArrayList<File> files = new ArrayList<>(2);
                files.add(new File("README.md"));

                // package files
                for (File file : files) {
                    //new zip entry and copying inputstream with file to zipOutputStream, after all closing streams
                    zipOutputStream.putNextEntry(new ZipEntry(file.getName()));
                    FileInputStream fileInputStream = new FileInputStream(file);

                    IOUtils.copy(fileInputStream, zipOutputStream);

                    fileInputStream.close();
                    zipOutputStream.closeEntry();
                }

                zipOutputStream.close();
            });
}

Tamtama

+1 for helping me figure out how to do this without creating a HttpServletResponse, since this doesn't seem to set a status code on 404s. – Laverne

If there is a run time exception anywhere in the body block, will still return 200 – Hughes

Attaching a relevant post about trouble handling exception with StreamingResponseBody https://mcmap.net/q/412269/-exception-handling-in-streamingresponsebody-not-working/5777189 – Hughes

see this page if you need generic solution for any case: https://mcmap.net/q/410843/-spring-rest-create-zip-file-and-send-it-to-the-client – Marcin

Answer 4

I am using REST Web Service of Spring Boot and I have designed the endpoints to always return ResponseEntity whether it is JSON or PDF or ZIP and I came up with the following solution which is partially inspired by denov's answer in this question as well as another question where I learned how to convert ZipOutputStream into byte[] in order to feed it to ResponseEntity as output of the endpoint.

Anyway, I created a simple utility class with two methods for pdf and zip file download

@Component
public class FileUtil {
    public BinaryOutputWrapper prepDownloadAsPDF(String filename) throws IOException {
        Path fileLocation = Paths.get(filename);
        byte[] data = Files.readAllBytes(fileLocation);

        HttpHeaders headers = new HttpHeaders();
        headers.setContentType(MediaType.parseMediaType("application/pdf"));
        String outputFilename = "output.pdf";
        headers.setContentDispositionFormData(outputFilename, outputFilename);
        headers.setCacheControl("must-revalidate, post-check=0, pre-check=0");

        return new BinaryOutputWrapper(data, headers); 
    }

    public BinaryOutputWrapper prepDownloadAsZIP(List<String> filenames) throws IOException {
        HttpHeaders headers = new HttpHeaders();
        headers.setContentType(MediaType.parseMediaType("application/zip"));
        String outputFilename = "output.zip";
        headers.setContentDispositionFormData(outputFilename, outputFilename);
        headers.setCacheControl("must-revalidate, post-check=0, pre-check=0");

        ByteArrayOutputStream byteOutputStream = new ByteArrayOutputStream();
        ZipOutputStream zipOutputStream = new ZipOutputStream(byteOutputStream);

        for(String filename: filenames) {
            File file = new File(filename); 
            zipOutputStream.putNextEntry(new ZipEntry(filename));           
            FileInputStream fileInputStream = new FileInputStream(file);
            IOUtils.copy(fileInputStream, zipOutputStream);
            fileInputStream.close();
            zipOutputStream.closeEntry();
        }           
        zipOutputStream.close();
        return new BinaryOutputWrapper(byteOutputStream.toByteArray(), headers); 
    }
}

And now the endpoint can easily return ResponseEntity<?> as shown below using the byte[] data and custom headers that is specifically tailored for pdf or zip.

@GetMapping("/somepath/pdf")
public ResponseEntity<?> generatePDF() {
    BinaryOutputWrapper output = new BinaryOutputWrapper(); 
    try {
        String inputFile = "sample.pdf"; 
        output = fileUtil.prepDownloadAsPDF(inputFile);
        //or invoke prepDownloadAsZIP(...) with a list of filenames
    } catch (IOException e) {
        e.printStackTrace();
        //Do something when exception is thrown
    } 
    return new ResponseEntity<>(output.getData(), output.getHeaders(), HttpStatus.OK); 
}

The BinaryOutputWrapper is a simple immutable POJO class I created with private byte[] data; and org.springframework.http.HttpHeaders headers; as fields in order to return both data and headers from utility method.

Pneumothorax

ByteArrayOutputStream stores the bytes in memory. So for large files you can run out of heap space very quickly. – Karleen

Answer 5

Only thing that worked for Spring Boot application (without any hardcoded file paths!)

@GetMapping(value = "/zip-download", produces="application/zip")
public void zipDownload(@RequestParam List<String> name, HttpServletResponse response) throws IOException {

    response.setStatus(HttpServletResponse.SC_OK);
    response.addHeader(HttpHeaders.CONTENT_DISPOSITION, "attachment; filename=\"" + zipFileName + "\"");

    ZipOutputStream zipOut = new ZipOutputStream(response.getOutputStream());

    for (String fileName : name) {

        // resource content length
        int contentLength = 123;
        // resource input stream
        InputStream stream = InputStream.nullInputStream();

        ZipEntry zipEntry = new ZipEntry(fileName);
        zipEntry.setSize(contentLength);

        zipOut.putNextEntry(zipEntry);
        StreamUtils.copy(stream, zipOut);

        zipOut.closeEntry();
    }

    zipOut.finish();
    zipOut.close();
}

Marcin

Unfortunately, it didn't work for me. The client gave the error invalid zip file. – Scarcely

@Scarcely probably because it was empty, that's common thing – Marcin

Yeah that was the issue. Sorry about the down vote. I cannot change it now due to Stackoverflow restrictions. Thanks – Scarcely