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Home/ Questions/Q 6679557
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T04:22:22+00:00 2026-05-26T04:22:22+00:00

The situation is that a ZIP file has been POSTed to a Tomcat server

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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”)

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  1. Editorial Team
    Editorial Team
    2026-05-26T04:22:23+00:00Added an answer on May 26, 2026 at 4:22 am

    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.

    See also:

    • How to upload files to server using JSP/Servlet?
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