How to extract JSON from this HTTP request in Java

Question

The method below is hit or miss if it parses the request properly...

Here is what the request looks like:

POST /addEvent/ HTTP/1.1
Host: localhost:1234
Content-Type: multipart/form-data; boundary=Boundary+0xAbCdEfGbOuNdArY
Accept-Encoding: gzip, deflate
Content-Length: 201
Accept-Language: en;q=1, fr;q=0.9, de;q=0.8, ja;q=0.7, nl;q=0.6, it;q=0.5
Accept: application/json
Connection: keep-alive
User-Agent: XXXXXXXXXXXXXXXXXX

--Boundary+0xAbCdEfGbOuNdArY
Content-Disposition: form-data; name="userInfo"

{   "user_id" : 1,   "value" : "Water",   "typeCode" : "Searched" }

Here is how we are extracting it now...

//Key where the request begins
String keyString = "\"userInfo\"";

//Get the index of the key
int end = bufferedJson.lastIndexOf("\"userInfo\"");

//Create substring at beginning of the json
String json = bufferedJson.substring(end+keyString.length(), bufferedJson.length());

//Convert json to feed item
Gson gson = new Gson();
Event eventItem = gson.fromJson(json, Event.class);

I get this error pretty often:

Expected BEGIN_OBJECT but was STRING at line 1 column 1

How can we parse this efficiently?

If it's really a POST request it shouldn't be all on one line like that. — Jim Garrison, Jul 02 '13 at 23:43
@waf I've just found out in other post better and more robust solutions for you to get your JSON data: http://stackoverflow.com/questions/3337056/convenient-way-to-parse-incoming-multipart-form-data-parameters-in-a-servlet — morgano, Jul 03 '13 at 00:46

score 1 · Accepted Answer · answered Jul 03 '13 at 00:37

Use Apache HTTP Client 4 to read Http response body in a convenient way. If you need to marshall your json further to a java object then make use of jackson. Here is the sample code:

import org.apache.http.client.ResponseHandler;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.BasicResponseHandler;
import org.apache.http.impl.client.DefaultHttpClient;

/**
 * This example demonstrates the use of the {@link ResponseHandler} to simplify
 * the process of processing the HTTP response and releasing associated resources.
 */
public class ClientWithResponseHandler {

    public final static void main(String[] args) throws Exception {

        HttpClient httpclient = new DefaultHttpClient();
        try {
            HttpGet httpget = new HttpGet("http://www.google.com/");

            System.out.println("executing request " + httpget.getURI());

            // Create a response handler
            ResponseHandler<String> responseHandler = new BasicResponseHandler();
            // Body contains your json stirng
            String responseBody = httpclient.execute(httpget, responseHandler);
            System.out.println("----------------------------------------");
            System.out.println(responseBody);
            System.out.println("----------------------------------------");

        } finally {
            // When HttpClient instance is no longer needed,
            // shut down the connection manager to ensure
            // immediate deallocation of all system resources
            httpclient.getConnectionManager().shutdown();
        }
    }

}

score 0 · Answer 2 · answered Jul 03 '13 at 00:19

To parse this better:

First, it seems like you're taking a "raw" HTTP POST request, and then reading it line by line using BufferedReader (your comment suggests this), this way you'll lose the new line chars; if you are going to do so, add a new line ("\n") every time you read a line to your final String, this way it doesn't lose the new lines and facilitates the things for the next step.

Now, with this final String, you can use this:

String json = null;

Pattern pattern = Pattern.compile("\n\n");
Matcher matcher = pattern.matcher(myString); // myString is the String you built from your header
if(matcher.find() && matcher.find()) {
    json = myString.substring(matcher.start() + 2);
} else {
    // Handle error: json string wasn't found
}

CAVEATS: this works if:

POST Request will always be multipart/form-data
There are not other parameters in the request
you STOP reading the request as soon as you find your json data
you included "\n" every time you read a line as I said in the first step

Personally I wouldn't read the raw HTTP header, I'd rather use Apache commons FileUpload or the like, but if your are going to do it this way, I think this is the less terrible solution.

@Junaid it answers the question at the end of the post: "How can we parse this efficiently?" — morgano, Jul 03 '13 at 00:23

score 0 · Answer 3 · answered Jan 15 '14 at 17:44

0

You can use

gson().fromJson(request.getReader(), Event.class);

or

String json = request.getReader().readLine();
gson().fromJson(json, Event.class);

answered Jan 15 '14 at 17:44

Diego Plentz

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How to extract JSON from this HTTP request in Java

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