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How to encode Post Data JSON in CloseableHttpClient APi

I have used the CloseableHttpClient APi for a Post call and Basic Auth for authorisation
private CloseableHttpClient httpclient = HttpClients.createDefault();
HttpPost httppost = new HttpPost("https://example.com");
MyJson myJson = new MyJson(); //custom java object to be posted as Request Body
Gson gson = new Gson();
String param = gson.toJson(myJson);
StringEntity urlparam = new StringEntity(param);
String credentials = username + ":" + passwprd;
String base64Credentials = new String(Base64.getencoder().encode(credentials.getBytes()));
String authorizartionHeader = "Basic" + base64Credentials;
httppost.setHeader("Content-Type", "application/Json");
httppost.setHeader("Authorization", authorizartionHeader);
urlparam.setContentEncoding("UTF-8");
httppost.setEntity(urlparam);
httpclient.execute(httppost);
I am getting error
"Invalid UTF-8 middle byte"
I have encoded the JSON still the encoding is not working for other locales except English. How to encode the Post data.
I tried using the method
httppost.setEntity(new URLEncodedFormEntity(namevaluePair, "UTF-8")) but I don't have any Namevaluepair and if the add the Username-pswd in that then getting Null pointer response.

You should try to set everything as UTF-8
StringEntity urlparam = new StringEntity(param, StandardCharsets.UTF_8);
And add proper header
httppost.setHeader("Content-Type", "application/json;charset=UTF-8");

HttpClient 4.5.3 cant get the response toString [duplicate]

I have this code on my Android phone.
URI uri = new URI(url);
HttpPost post = new HttpPost(uri);
HttpClient client = new DefaultHttpClient();
HttpResponse response = client.execute(post);
I have a asp.net webform application that has in the page load this
Response.Output.Write("It worked");
I want to grab this Response from the HttpReponse and print it out. How do I do this?
I tried response.getEntity().toString() but it just seems to print out the address in memory.
Thanks

Use ResponseHandler. One line of code. See here and here for sample Android projects using it.
public void postData() {
// Create a new HttpClient and Post Header
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://www.yoursite.com/user");
try {
// Add your data
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("id", "12345"));
nameValuePairs.add(new BasicNameValuePair("stringdata", "AndDev is Cool!"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
// Execute HTTP Post Request
ResponseHandler<String> responseHandler=new BasicResponseHandler();
String responseBody = httpclient.execute(httppost, responseHandler);
JSONObject response=new JSONObject(responseBody);
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
} catch (IOException e) {
// TODO Auto-generated catch block
}
}
add combination of this post and complete HttpClient at - http://www.androidsnippets.org/snippets/36/

I would just do it the old way. It's a more bulletproof than ResponseHandler, in case you get different content types in the response.
ByteArrayOutputStream outstream = new ByteArrayOutputStream();
response.getEntity().writeTo(outstream);
byte [] responseBody = outstream.toByteArray();

I used the following code
BufferedReader r = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
StringBuilder total = new StringBuilder();
String line = null;
while ((line = r.readLine()) != null) {
total.append(line);
}
r.close();
return total.toString();

The simplest approach is probably using org.apache.http.util.EntityUtils:
String message = EntityUtils.toString(response.getEntity());
It reads the contents of an entity and returns it as a String. The content is converted using the character set from the entity (if any), failing that, "ISO-8859-1" is used.
If necessary, you can pass a default character set explicitly - e.g.
String message = EntityUtils.toString(response.getEntity(). "UTF-8");
It gets the entity content as a String, using the provided default character set if none is found in the entity. If the passed default character set is null, the default "ISO-8859-1" is used.

This code will return the entire response message in respond as a String, and status code in rsp, as an int.
respond = response.getStatusLine().getReasonPhrase();
rsp = response.getStatusLine().getStatusCode();`

Space between http url in java

This code working fine
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet("http://localhost:8090/Servlet/ServletFirst?to=1234&from=567&text=testtest");
If i use space between parameter value. It throws exception
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet("http://localhost:8090/Servlet/ServletFirst?to=1234&from=567&textParam=test test");
space between test test throws error. How to resolve?

You must URL encode the parameter in your URL; use %20 instead of the space.
HttpGet request = new HttpGet("http://localhost:8090/Servlet/ServletFirst?to=1234&from=567&textParam=test%20test");
Java has a class to do do URL encoding for you, URLEncoder:
String param = "test test";
String enc = URLEncoder.encode(param, "UTF-8");
String url = "http://...&textParam=" + enc;

Just use a %20 to represent a space.
This is all part of the URL encoding: http://www.w3schools.com/tags/ref_urlencode.asp
So you would want:
HttpGet request = new HttpGet("http://localhost:8090/Servlet/ServletFirst?to=1234&from=567&text=test%20test");

Use
URLEncoder.encode("test test","UTF-8")
So change your code to
HttpGet request = new HttpGet("http://localhost:8090/Servlet/ServletFirst?to=1234&from=567&textParam="+URLEncoder.encode("test test","UTF-8"));
Note Don't Encode Whole url
URLEncoder.encode("http://...test"); // its Wrong because it will also encode the // in http://

Use %20 to indicate space in the URL, as space is not an allowed character. See the Wikipedia entry for character data in a URL.
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet(
"http://localhost:8090/Servlet/ServletFirst?to=1234&from=567&textParam=test%20test");

URL Encoding with httpclient

I have a list of URLs which I need to get the content of.
The URL is with special characters and thus needs to be encoded.
I use Commons HtpClient to get the content.
when I use:
GetMethod get = new GetMethod(url);
I get a " Invalid "illegal escape character" exception.
when I use
GetMethod get = new GetMethod();
get.setURI(new URI(url.toString(), false, "UTF-8"));
I get 404 when trying to get the page, because a space is turned to %2520 instead of just %20.
I've seen many posts about this problem, and most of them advice to build the URI part by part. The problem is that it's a given list of URLs, not a one that I can handle manually.
Any other solution for this problem?
thanks.

What if you create a new URL object from it's string like URL urlObject = new URL(url), then do urlObject.getQuery() and urlObject.getPath() to split it right, parse the Query Params into a List or a Map or something and do something like:
EDIT: I just found out that HttpClient Library has a URLEncodedUtils.parse() method which you can use easily with the code provided below. I'll edit it to fit, however is untested.
With Apache HttpClient it would be something like:
URI urlObject = new URI(url,"UTF-8");
HttpClient httpclient = new DefaultHttpClient();
List<NameValuePair> formparams = URLEncodedUtils.parse(urlObject,"UTF-8");
UrlEncodedFormEntity entity;
entity = new UrlEncodedFormEntity(formparams);
HttpPost httppost = new HttpPost(urlObject.getPath());
httppost.setEntity(entity);
httppost.addHeader("Content-Type","application/x-www-form-urlencoded");
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity2 = response.getEntity();
With Java URLConnection it would be something like:
// Iterate over query params from urlObject.getQuery() like
while(en.hasMoreElements()){
String paramName = (String)en.nextElement(); // Iterator over yourListOfKeys
String paramValue = yourMapOfValues.get(paramName); // replace yourMapOfNameValues
str = str + "&" + paramName + "=" + URLEncoder.encode(paramValue);
}
try{
URL u = new URL(urlObject.getPath()); //here's the url path from your urlObject
URLConnection uc = u.openConnection();
uc.setDoOutput(true);
uc.setRequestProperty("Content-Type","application/x-www-form-urlencoded");
PrintWriter pw = new PrintWriter(uc.getOutputStream());
pw.println(str);
pw.close();
BufferedReader in = new BufferedReader(new
InputStreamReader(uc.getInputStream()));
String res = in.readLine();
in.close();
// ...
}

If you need to manipulate with request URIs it is strongly advisable to use URIBuilder shipped with Apache HttpClient.

try it out
GetMethod get = new GetMethod(url.replace(" ","%20")).toASCIIString());

Please use the URLEncoder class.
I used it in an exact scenario and it worked just fine for me.
What I did is to use the URL class, to get the part that comes after the host
(for example - at www.bla.com/mystuff/bla.jpg this would be "mystuff/bla.jpg" - you should URLEncode only this part, and then consturct the URL again.
For example, if the orignal string is "http://www.bla.com/mystuff/bla foo.jpg" then:
Encode - "mystuff/bla foo.jpg" and get "mystuff/bla%20foo.jpg" and then attach this to the host and protocol parts:
"http://www.bla.com/mystuff/bla%20foo.jpg"
I hope this helps

IllegalStateException: Content has been consumed - How to resolve?

I have the following code:
HttpGet httpGet = new HttpGet(serverAdress + "/rootservices");
httpGet.setHeader("Accept", "text/xml");
HttpResponse response = client.execute(httpGet, localContext);
String projectURL = XMLDocumentParser.parseDocument(response.getEntity().getContent(), "oslc_scm:scmServiceProviders", "rdf:resource");
String workItemURL = XMLDocumentParser.parseDocument(response.getEntity().getContent(), "oslc_cm:cmServiceProviders", "rdf:resource");
The problem here is that I read two times the HttpResponse object. So the second time I get the exception. But although I know the problem, I can´t find an easy solution. So what is a good way to solve that problem?

Read the input stream returned by response.getEntity().getContent() into a byte[], stored in a local variable. See Convert InputStream to byte array in Java.
byte[] content = IOUtils.toByteArray(response.getEntity().getContent());
String projectURL = XMLDocumentParser.parseDocument(new ByteArrayInputStream(content), "oslc_scm:scmServiceProviders", "rdf:resource");
String workItemURL = XMLDocumentParser.parseDocument(new ByteArrayInputStream(content), "oslc_cm:cmServiceProviders", "rdf:resource");

Read the response content only once, and copy it to some kind of buffer compatible with XMLDocumentParser.parseDocument. Then parse the data directly from your buffer, as many times as you want.

Why don't you try it this way?
HttpGet httpGet = new HttpGet(serverAdress + "/rootservices");
httpGet.setHeader("Accept", "text/xml");
InputStream in = null;
HttpResponse response = client.execute(httpGet, localContext);
in = response.getEntity().getContent();
String projectURL = XMLDocumentParser.parseDocument(in, "oslc_scm:scmServiceProviders", "rdf:resource");
String workItemURL = XMLDocumentParser.parseDocument(in, "oslc_cm:cmServiceProviders", "rdf:resource");