We Keep Coding

Parsing a HttpResponse JSON in Java

I am trying to do a HTTP post request to a web API and then parse the received HttpResponse and access the key value pairs in the body. My code is like this:
public class access {
// http://localhost:8080/RESTfulExample/json/product/post
public static void main(String[] args) {
HttpClient httpclient = HttpClients.createDefault();
HttpPost httppost = new HttpPost("https://XXXXXXX/RSAM_API/api/Logon");
// Request parameters and other properties.
List<NameValuePair> urlParameters = new ArrayList<NameValuePair>(2);
urlParameters.add(new BasicNameValuePair("UserId", "XXXXX"));
urlParameters.add(new BasicNameValuePair("Password", "XXXXXX"));
try {
httppost.setEntity(new UrlEncodedFormEntity(urlParameters));
//Execute and get the response.
HttpResponse response = httpclient.execute(httppost);
BufferedReader rd = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
String line;
while(null !=(line=rd.readLine())){
System.out.println(line);
}
System.out.println(response);
String resp = EntityUtils.toString(response.getEntity());
JSONObject obj = new JSONObject(resp);
}
catch (Exception e){
e.printStackTrace();
}
}
}
I am trying to access the body by converting it to a JSONObject with these 2 lines of code:
String resp = EntityUtils.toString(response.getEntity());
JSONObject obj = new JSONObject(resp);
But I get an error in the second line saying:
JSONObject
(java.util.Map)
in JSONObject cannot be applied
to
(java.lang.String)
Not sure if this is the correct approach. Is there a way to do what I am trying to do?
Any help would be appreciated, Thank you.
EDIT:
So when I try to print the response body using the following lines,
String resp = EntityUtils.toString(response.getEntity());
System.out.println(resp);
I get the result: {"APIKey":"xxxxxxxxxxxxxx","StatusCode":0,"StatusMessage":"Y‌​ou have been successfully logged in."}
I am looking for a way to parse this result and then access each element. Is there a way to do this?

According to JsonSimple's JsonObject documentation it takes map in the constructor but not a String. So the error you are getting what it says.
You should use JSONParser to parse the string first.
Its also better to provide the encoding as part of EntityUtils.toString say UTF-8 or 16 based off your scenario.
IOUtils.toString() from Apache Commons IO would be a better choice to use too.

Try the below line to parse the JSON:
JSONParser parser = new JSONParser();
JSONObject obj = (JSONObject) parser.parse(resp);
The above lines will vaildate the JSON and through exception if the JSON is invalid.

You don't need to read the response in the extra while loop. EntityUtils.toString(response.getEntity()); will do this for you. As you read the response stream before, the stream is already closed when comming to response.getEntity().

Extract string from http response in java client

I want to extract the string returned from java web service in java client. The string returned from java web service is as follows:
{"Name":"Raj Johri","Email":"mailraj#server.com","status":true}
Which is a Json string format. I have written client code to extract this string as follows:
public static void main(String[] args) throws Exception{
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost post = new HttpPost("http://localhost:8080/JsonWebService/services/JsonWebService/getData");
post.setHeader("Content-Type", "application/xml");
HttpResponse httpres = httpClient.execute(post);
HttpEntity entity = httpres.getEntity();
String json = EntityUtils.toString(entity).toString();
System.out.println("json:" + json);
}
I am getting following print on the console for json as:
json:<ns:getDataResponse xmlns:ns="http://ws.jsonweb.com"><ns:return>{"Name":"Raj Johri","Email":"mailraj#server.com","status":true}</ns:return></ns:getDataResponse>
Please tell me how to extract the string
{"Name":"Raj Johri","Email":"mailraj#server.com","status":true}
which is the actual message. Thanks in advance...

Well, The respons is as type of xml, and your json is in the <ns:return> node , so i suggest you to enter in depth of the xml result and simply get your json from the <ns:return> node.
Note:
I suggest you to try to specifying that you need the response as JSON type:
post.setHeader("Content-type", "application/json");
post.setHeader("Accept", "application/json");

There is a dirty way to do this (beside the xml parsing way)
if you are getting the same XML every time,
you can use split()
String parts[] = json.split("<ns:return>");
parts = parts[1].split("</ns:return>");
String jsonPart = parts[0];
now jsonPart should contain only {"Name":"Raj Johri","Email":"mailraj#server.com","status":true}

How to parse a large amount of data from a JSON webservice

I have a URL which returns a large amount of data in response, but I can't get a full response from the webservice URL. The webservice responds as follows:
\"},
{\"minute_usage_end_time\":\"11:59\",\"minute_usage_start_time\":\"11:00\",\"kwh_usage\":\"0\",\"meter_reading_date\":\"08-02-2011\"},{...
What should I do?

StringEntity entity = new StringEntity(jsonObject.toString(),HTTP.UTF_8);
httpPost.setEntity(entity);
HttpResponse response = httpClient.execute(httpPost);
BufferedReader reader = new BufferedReader(new InputStreamReader(response.getEntity().getContent(), "UTF-8"));
Log.v("mas\n", reader.readLine());
String jsonResultStr = reader.readLine();

HttpGet in Android/Java with GZip encoding

I am running MVC 4 on my server and to save a bit of data for my users I figured I would enable GZip encoding, to do this I simply used:
(C#)
Response.AddHeader("Content-Encoding", "gzip");
Response.Filter = new GZipStream(Response.Filter, CompressionMode.Compress);
In my android application I use:
(Java)
String response = "";
DefaultHttpClient client = new DefaultHttpClient();
HttpGet httpGet = new HttpGet(url);
try {
HttpResponse execute = client.execute(httpGet);
InputStream content = execute.getEntity().getContent();
BufferedReader buffer = new BufferedReader(
new InputStreamReader(content));
String s = "";
while ((s = buffer.readLine()) != null) {
response += s;
}
} catch (Exception e) {
e.printStackTrace();
}
return response;
When I use GZip the Java code nuts out and causes GC to run, I was never patient enough to wait for it to return.
When I took off GZip from the server it ran perfectly fine. The function to get the response returns straight away with no problem.
I tried adding this to the java code:
httpGet.addHeader("Accept-Encoding", "gzip");
With no success.
Question is, is there something I'm not getting? Can I not put the response in a stream if it is using GZip? Am I meant to use the stream and uncompress it after?
What am I doing wrong?

Instead of using
DefaultHttpClient client = new DefaultHttpClient();
you can use
ContentEncodingHttpClient client = new ContentEncodingHttpClient();
which is a subclass of DefaultHttpClient and supports GZIP content.
You need Apache HttpClient 4.1 for this.
If you have Apache HttpClient 4.2, you should use
DecompressingHttpClient client = new DecompressingHttpClient();
if you have Apache HttpClient 4.3, you should use the HttpClientBuilder

Getting URL after a redirect using HttpClient.Execute(HttpGet)

I have searched for a while and I am not finding a clear answer. I am trying to log into a webstie.
https://hrlink.healthnet.com/
This website redirects to a login page that is not consitent. I have to post my login credentials to the redirected URL.
Im am trying to code this in Java but I do not understand how to get the URL from the response. It may look a bit messy but I have it this way while I am testing.
HttpGet httpget = new HttpGet("https://hrlink.healthnet.com/");
HttpResponse response = httpclient.execute(httpget);HttpEntity entity = response.getEntity();
String redirectURL = "";
for(org.apache.http.Header header : response.getHeaders("Location")) {
redirectURL += "Location: " + header.getValue()) + "\r\n";
}
InputStream is;
is = entity.getContent();
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
String result = sb.toString();
I know i get redirected because my result string shows be the actual login page but I am not able to get the new URL.
In FireFox I am using TamperData. When I navigate to this website https://hrlink.healthnet.com/ I have a GET with a 302 - Found and the Location of the Login Page. Then another GET to the actual Login Page
Any help is greatly appreciated thank you.

Check out w3c documentation:
10.3.3 302 Found
The temporary URI SHOULD be given by the Location field in the response. Unless the request method was HEAD, the entity of the response SHOULD contain a short hypertext note with a hyperlink to the new URI(s).
If the 302 status code is received in response to a request other than GET or HEAD, the user agent MUST NOT automatically redirect the request unless it can be confirmed by the user, since this might change the conditions under which the request was issued.
One solution is to use POST method to break auto-redirecting at client side:
HttpPost request1 = new HttpPost("https://hrlink.healthnet.com/");
HttpResponse response1 = httpclient.execute(request1);
// expect a 302 response.
if (response1.getStatusLine().getStatusCode() == 302) {
String redirectURL = response1.getFirstHeader("Location").getValue();
// no auto-redirecting at client side, need manual send the request.
HttpGet request2 = new HttpGet(redirectURL);
HttpResponse response2 = httpclient.execute(request2);
... ...
}
Hope this helps.