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Spring REST - create .zip file and send it to the client

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

like image 981
azalut Avatar asked Jan 14 '15 21:01

azalut


People also ask

How do I zip a spring boot project?

While downloading multiple files, we can create a zip file in spring boot and download that zip file alone rather then downloading multiple files individually. For this purpose, we first need to create a zip file in spring boot and then set the content type as application/zip to download the zip file.

How do I create a ZIP file from a project?

Press and hold (or right-click) the file or folder, select (or point to) Send to, and then select Compressed (zipped) folder. A new zipped folder with the same name is created in the same location.

How do I zip a file in Java?

Steps to Compress a File in Java Open a ZipOutputStream that wraps an OutputStream like FileOutputStream. The ZipOutputStream class implements an output stream filter for writing in the ZIP file format. Put a ZipEntry object by calling the putNextEntry(ZipEntry) method on the ZipOutputStream.


4 Answers

seems to be solved. I replaced:

response.setContentType("application/zip"); 

with:

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

And now I get 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.

like image 82
azalut Avatar answered Sep 20 '22 07:09

azalut


@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();
}
like image 39
denov Avatar answered Sep 21 '22 07:09

denov


@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();
            });
}
like image 43
cesar Avatar answered Sep 21 '22 07:09

cesar


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.

like image 41
Raf Avatar answered Sep 22 '22 07:09

Raf