How do you read POST data from a servlet into a ZipStream?

Question

The situation is that a ZIP file has been POSTed to a Tomcat server and since it has no parameter name associated with it, we're going right to the request's stream.

  ServletInputStream sis = request.getInputStream()
  ZipInputStream zis = new ZipInputStream(sis)
  ZipEntry zEntry = zis.getNextEntry()
  while (zEntry != null) {
    // do something with zEntry
    zEntry = zis.getNextEntry()
  }

Compellingly simple, but it doesn't work. It never enters the while loop because the first zEntry is null. (The ZIP file is valid, btw)

Any ideas? Thanks

EDIT: Type is multipart/form-data ("multipart/form-data; boundary=---------------------8ce556d90e9deb6")

What's the request body encoding? `multipart/form-data`? – BalusC Oct 10 '11 at 18:03 — BalusC, Oct 10 '11 at 18:03

score 0 · Accepted Answer · edited Jun 20 '20 at 09:12

EDIT: Type is multipart/form-data ("multipart/form-data; boundary=---------------------8ce556d90e9deb6")

You need to use a multipart/form-data parser to extract the uploaded file. You shouldn't feed it raw to the ZipInputStream. It however surprises me that it didn't threw an exception that it's not in ZIP file format.

The getParameter() and consorts are designed for application/www-form-urlencoded requests only. They will all return null on multipart/form-data requests. You need to use getParts() method instead. Or, when you're still on Servlet 2.5 or older, then you need to parse the body with help of Apache Commons Fileupload.

Either way, you should be able to iterate over multipart/form-data items and get the uploaded file(s) as an InputStream which you in turn read using ZipInputStream.

How do you read POST data from a servlet into a ZipStream?

1 Answers1

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