How do I return a zip file to the browser via the response OutputStream?

Question

In this situation, I have created a zip file containing search result files, and am trying to send it to the user. Here is the chunk of code I am currently trying to use.

File[] zippable = new File[files.size()];
File resultFile = ZipCreator.zip(files.toArray(zippable), results);
InputStream result = new FileInputStream(resultFile);
IOUtils.copy(result, response.getOutputStream());

However, this currently doesn't work quite right. Instead of returning the zip file that I have created, it returns an html file. If I manually change the file extension afterwards, I can see that the contents of the file are still the search results that I need. So the problem just lies in returning the proper extension to the response.

Does anyone have any advice for this situation?

Borecole

Answer 1

You need to set the Content-Type response header to the value application/zip (or application/octet-stream, depending on the target browser). Additionally, you may want to send additional response headers indicating attachment status and filename.

Boote

I think I may be misinterpreting you here. I've added a line to the code, so it now reads response.setHeader("Content-Type", "application/zip"); Still no success though. This call is being made just before the call to IOUtils.copy – Borecole

@Bennie: looks good, but again, there is some confusion about the "right" MIME type to use for a ZIP file, so you might want to play with those values. – Boote

Alright, it looks like it's working now, at least for returning the file. It's doing a strange thing though, where the file is called "exportZipSearchResults.htm.zip". Is there a built-in way to change the filename on the response? – Borecole

@Bennie: try using the Content-Disposition header (possibly with value "inline") and a filename of your choice. – Boote

Answer 2

You need to set the content type header to application/octet-stream prior to streaming the results. Depends on what implementation of response you are using on how you actually do this.

Obstinate

Answer 3

Here is some working code, just in case anyone needs it:

protected void doGet(HttpServletRequest request, HttpServletResponse response) {

        // The zip file you want to download
        File zipFile = new File(zipsResourcesPath + zipFileName);

        response.setContentType("application/zip");
        response.addHeader("Content-Disposition", "attachment; filename=" + zipFileName);
        response.setContentLength((int) zipFile.length());

        try {

            FileInputStream fileInputStream = new FileInputStream(zipFile);
            OutputStream responseOutputStream = response.getOutputStream();
            int bytes;
            while ((bytes = fileInputStream.read()) != -1) {
                responseOutputStream.write(bytes);
            }
        } catch (IOException e) {
            logger.error("Exception: " + e);
        }
}

And the HTML:

<a class="btn" href="/path_to_servlet" target="_blank">Download zip</a>

Hope this helps!

Whoa

Please show how you defined the "zipFile" ? How you added files into this "zipFile"? – Cichlid

I've edited the code so you can see how the zipFile is defined. Please reconsider the downvote @Cichlid . I appreciate comments, as this helps me to improve my answers. – Whoa

this is getting already created zip file. But how to get multiple files and make them into zip using Java? and then send to response. – Cichlid

Answer 4

So I found a hack for this : ) Just add ".zip" in your filename and set your content type as application/zip. Works like a charm.

response.setContentType("application/zip");
String licenseFileName = eId;
response.setHeader("Content-disposition", "attachment; filename=\"" + licenseFileName +".zip");

Podophyllin

what is eId in this example? – Albata

@Albata eId was an identifier in my case. It's just the file name you want. It can be anything basically. – Podophyllin