Spring boot rest service to download a zip file which contains multiple file

Question

I am able to download a single file but how I can download a zip file which contain multiple files.

Below is the code to download a single file but I have multiples files to download. Any help would greatly appreciated as I am stuck on this for last 2 days.

@GET
@Path("/download/{fname}/{ext}")
    @Produces(MediaType.APPLICATION_OCTET_STREAM)
public Response  downloadFile(@PathParam("fname") String fileName,@PathParam("ext") String fileExt){
    File file = new File("C:/temp/"+fileName+"."+fileExt);
    ResponseBuilder rb = Response.ok(file);
    rb.header("Content-Disposition", "attachment; filename=" + file.getName());
    Response response = rb.build();
    return response;
}

have you tried something with `response.setContentType("application/zip");`? — Shreyas, Jul 14 '18 at 20:21
I think you will have to use `rb.header("Content-Type", "application/zip");` — Shreyas, Jul 14 '18 at 20:24
So zip file is already there or you wish to create it on the fly? Also, one concern if full file needs to be in memory or needs to be streamed in chunks while downloading ? — Sabir Khan, Jul 16 '18 at 12:39
I have to create the zip file on fly and to answer your second question is that full file needs to be in memory but we are restricting the download if it goes above 100 MB. — abhishek chauhan, Jul 17 '18 at 10:04
@abhishekchauhan: Its not guaranteed that you will have a single concurrent client so as per memory foot print , 100MB might have to get multiplied by number of concurrent allowed connections. Just a thought. — Sabir Khan, Jul 17 '18 at 12:33

score 13 · Answer 1 · edited Jun 19 '19 at 20:00

Here is my working code I have used response.getOuptStream()

@RestController
public class DownloadFileController {

    @Autowired
    DownloadService service;

    @GetMapping("/downloadZip")
    public void downloadFile(HttpServletResponse response) {

        response.setContentType("application/octet-stream");
        response.setHeader("Content-Disposition", "attachment;filename=download.zip");
        response.setStatus(HttpServletResponse.SC_OK);

        List<String> fileNames = service.getFileName();

        System.out.println("############# file size ###########" + fileNames.size());

        try (ZipOutputStream zippedOut = new ZipOutputStream(response.getOutputStream())) {
            for (String file : fileNames) {
                FileSystemResource resource = new FileSystemResource(file);

                ZipEntry e = new ZipEntry(resource.getFilename());
                // Configure the zip entry, the properties of the file
                e.setSize(resource.contentLength());
                e.setTime(System.currentTimeMillis());
                // etc.
                zippedOut.putNextEntry(e);
                // And the content of the resource:
                StreamUtils.copy(resource.getInputStream(), zippedOut);
                zippedOut.closeEntry();
            }
            zippedOut.finish();
        } catch (Exception e) {
            // Exception handling goes here
        }
    }
}

Service Class:-

public class DownloadServiceImpl implements DownloadService {

    @Autowired
    DownloadServiceDao repo;

    @Override
    public List<String> getFileName() {

        String[] fileName = { "C:\\neon\\FileTest\\File1.xlsx", "C:\\neon\\FileTest\\File2.xlsx", "C:\\neon\\FileTest\\File3.xlsx" };

        List<String> fileList = new ArrayList<>(Arrays.asList(fileName));       
        return fileList;
    }
}

How can we give the location to the zip file to be downloaded? — Hüseyin, Feb 02 '21 at 10:14

score 4 · Accepted Answer · answered Jul 17 '18 at 12:53

Use these Spring MVC provided abstractions to avoid loading of whole file in memory. org.springframework.core.io.Resource & org.springframework.core.io.InputStreamSource

This way, your underlying implementation can change without changing controller interface & also your downloads would be streamed byte by byte.

See accepted answer here which is basically using org.springframework.core.io.FileSystemResource to create a Resource and there is a logic to create zip file on the fly too.

That above answer has return type as void, while you should directly return a Resource or ResponseEntity<Resource> .

As demonstrated in this answer, loop around your actual files and put in zip stream. Have a look at produces and content-type headers.

Combine these two answers to get what you are trying to achieve.

thanks for your support. Now I am able to create zip on fly and download it as well. — abhishek chauhan, Jul 19 '18 at 05:47
Your welcome !! Its better that you update your question with your final working code so it helps others too. — Sabir Khan, Jul 19 '18 at 08:52

score 0 · Answer 3 · edited Oct 31 '20 at 15:06

public void downloadSupportBundle(HttpServletResponse response){

      File file = new File("supportbundle.tar.gz");
      Path path = Paths.get(file.getAbsolutePath());
      logger.debug("__path {} - absolute Path{}", path.getFileName(),
            path.getRoot().toAbsolutePath());

      response.setContentType("application/octet-stream");
      response.setHeader("Content-Disposition", "attachment;filename=supportbundle.tar.gz");
      response.setStatus(HttpServletResponse.SC_OK);

      System.out.println("############# file name ###########" + file.getName());

      try (ZipOutputStream zippedOut = new ZipOutputStream(response.getOutputStream())) {

         FileSystemResource resource = new FileSystemResource(file);

         ZipEntry e = new ZipEntry(resource.getFilename());
         e.setSize(resource.contentLength());
         e.setTime(System.currentTimeMillis());
         zippedOut.putNextEntry(e);
         StreamUtils.copy(resource.getInputStream(), zippedOut);
         zippedOut.closeEntry();

         zippedOut.finish();
      } catch (Exception e) {
         
      }
}

Spring boot rest service to download a zip file which contains multiple file

3 Answers3