How can I read octet-stream to plain string/text by using java?

Question

I have some content coming from a byte[] which stands for the data from the request which gets saved to a .txt file.

try {
        while ((bytesRead = streamFromClient.read(currentRequest)) != -1) {

                LOG.info("Received request...");
                addInvoiceRequestOnly();
                streamToServer.write(currentRequest, 0, bytesRead);
                streamToServer.flush();
                dumpTrafficToFile();
            }

        } catch (IOException e) {
            LOG.severe("Could not read/write from/to the request...");
            e.getStackTrace();
        }

POST /domibus/services/...
Host: domibusbackend
Connection: close
Content-Length: 16189
Content-Type: multipart/related; type="application/soap+xml"; boundary="uuid:6b5b42a6-ea2f-4830-84c3-c799f38ca32a"; start="<root.message@cxf.apache.org>"; start-info="application/soap+xml"
Accept: */*
User-Agent: Apache-CXF/3.3.2
Cache-Control: no-cache
Pragma: no-cache

.........

Content-Type: application/octet-stream
Content-Transfer-Encoding: binary
Content-ID: <53c6399c-a6b1-4ffa-9c85-d8b7bb337f28@www.someDomain.com>
CompressionType: application/gzip
MimeType: application/xml

†’^tÜ:–jq®Z{€Üş˝`cWłÓx˘ěxĐWé"v«8-ňBStÂá›Ë+•jnCćcv‰v2—ťř‘÷ż÷ ĺůéűI˝sJá@Vzľ¸…“ßĹ¸¤Ž2]§yÁbů,m ĺgŮ±ťŠ¸áĐĽ<í.ÖÚeGü®î…Č>
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Can somebody tell me how can I convert this data to human readable data? It sais in the header that it should be binary, but it is not, binary data looks different. I got a lot of readable data from the request but the last part, the one that you see is that alien thing which I don't know how to decode it... If somebody can help me with this I'd appreciate it. Thank you!

This request looks malformed to me. `MimeType` is not a standard header, and it appears to conflict with the `Content-Type` which should probably be `application/xml` (or `text/xml`). It also seems strange that there's no `Content-Length` header; without it, there's no way to validate that the message is complete. — dnault, Feb 13 '20 at 18:27
"binary data looks different" -- how so? Isn't this just what happens when your text editor tries to interpret binary data as text? — dnault, Feb 13 '20 at 18:28
@dnault I haven't displayed all the content from the request; I added the header as well — Irinel, Feb 13 '20 at 18:30
Thanks for updating your question. It looks like you might be trying to implement a SOAP server from scratch. Instead, you might consider using one of the existing frameworks (google "java soap server"). — dnault, Feb 13 '20 at 18:43
@dnault I am trying to get all the requests that are coming from the client by following an example like this http://www.java2s.com/Code/Java/Network-Protocol/Asimpleproxyserver.htm — Irinel, Feb 13 '20 at 18:46

Govinda Sakhare · Answer 1 · 2020-02-13T18:34:28.003

2

I am assuming you are using Servlet-API if you are using any framework you need to modify first code snippet to extract the request body and pass it to extract().

public void doPost(HttpServletRequest request, HttpServletResponse response) {
    InflaterInputStream is = InflaterInputStream(request.getInputStream()); 
    String readableString = extract(is);
    // do the processing
}

Then convert compressed-binary data into readable String

private String extract(InputStream is) throws IOException {
    StringBuffer sb = new StringBuffer();
    BufferedReader in = new BufferedReader(new InputStreamReader(is));
    String inputLine = "";
    while ((inputLine = in.readLine()) != null) {
        sb.append(inputLine);
    }
    return sb.toString();
}

edited Feb 13 '20 at 18:34

answered Feb 13 '20 at 18:31

Govinda Sakhare

5,009
6
33
74

no, I am not using that, I am taking the request from a byte[] – Irinel Feb 13 '20 at 18:34
What is `currentRequest`? – Govinda Sakhare Feb 13 '20 at 18:52
currentRequest is a byte array byte[] ...the byte[] holds the data from the request – Irinel Feb 13 '20 at 18:53

score 1 · Answer 2 · answered Feb 13 '20 at 18:30

1

This suggest compression : CompressionType: application/gzip

So feed body into java.util.zip.GZIPInputStream

answered Feb 13 '20 at 18:30

rkosegi

14,165
5
50
83

it throws ZipException - not zip format...I tried different ways of reading that encoded data which said that would be gzip compressed but using GZIPInputStream is not a solution... – Irinel Feb 17 '20 at 08:53

How can I read octet-stream to plain string/text by using java?

2 Answers2