Java - sending HTTP parameters via POST method easily

Question

I am successfully using this code to send HTTP requests with some parameters via GET method

void sendRequest(String request)
{
            


        

Answer 1

I couldn't get Alan's example to actually do the post, so I ended up with this:

String urlParameters = "param1=a&param2=b&param3=c";
URL url = new URL("http://example.com/index.php");
URLConnection conn = url.openConnection();

conn.setDoOutput(true);

OutputStreamWriter writer = new OutputStreamWriter(conn.getOutputStream());

writer.write(urlParameters);
writer.flush();

String line;
BufferedReader reader = new BufferedReader(new InputStreamReader(conn.getInputStream()));

while ((line = reader.readLine()) != null) {
    System.out.println(line);
}
writer.close();
reader.close();         

Answer 2

I see some other answers have given the alternative, I personally think that intuitively you're doing the right thing ;). Sorry, at devoxx where several speakers have been ranting about this sort of thing.

That's why I personally use Apache's HTTPClient/HttpCore libraries to do this sort of work, I find their API to be easier to use than Java's native HTTP support. YMMV of course!

Answer 3

GET and POST method set like this... Two types for api calling 1)get() and 2) post() . get() method to get value from api json array to get value & post() method use in our data post in url and get response.

 public class HttpClientForExample {

    private final String USER_AGENT = "Mozilla/5.0";

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

        HttpClientExample http = new HttpClientExample();

        System.out.println("Testing 1 - Send Http GET request");
        http.sendGet();

        System.out.println("\nTesting 2 - Send Http POST request");
        http.sendPost();

    }

    // HTTP GET request
    private void sendGet() throws Exception {

        String url = "http://www.google.com/search?q=developer";

        HttpClient client = new DefaultHttpClient();
        HttpGet request = new HttpGet(url);

        // add request header
        request.addHeader("User-Agent", USER_AGENT);

        HttpResponse response = client.execute(request);

        System.out.println("\nSending 'GET' request to URL : " + url);
        System.out.println("Response Code : " + 
                       response.getStatusLine().getStatusCode());

        BufferedReader rd = new BufferedReader(
                       new InputStreamReader(response.getEntity().getContent()));

        StringBuffer result = new StringBuffer();
        String line = "";
        while ((line = rd.readLine()) != null) {
            result.append(line);
        }

        System.out.println(result.toString());

    }

    // HTTP POST request
    private void sendPost() throws Exception {

        String url = "https://selfsolve.apple.com/wcResults.do";

        HttpClient client = new DefaultHttpClient();
        HttpPost post = new HttpPost(url);

        // add header
        post.setHeader("User-Agent", USER_AGENT);

        List<NameValuePair> urlParameters = new ArrayList<NameValuePair>();
        urlParameters.add(new BasicNameValuePair("sn", "C02G8416DRJM"));
        urlParameters.add(new BasicNameValuePair("cn", ""));
        urlParameters.add(new BasicNameValuePair("locale", ""));
        urlParameters.add(new BasicNameValuePair("caller", ""));
        urlParameters.add(new BasicNameValuePair("num", "12345"));

        post.setEntity(new UrlEncodedFormEntity(urlParameters));

        HttpResponse response = client.execute(post);
        System.out.println("\nSending 'POST' request to URL : " + url);
        System.out.println("Post parameters : " + post.getEntity());
        System.out.println("Response Code : " + 
                                    response.getStatusLine().getStatusCode());

        BufferedReader rd = new BufferedReader(
                        new InputStreamReader(response.getEntity().getContent()));

        StringBuffer result = new StringBuffer();
        String line = "";
        while ((line = rd.readLine()) != null) {
            result.append(line);
        }

        System.out.println(result.toString());

    }

}

Answer 4

here i sent jsonobject as parameter //jsonobject={"name":"lucifer","pass":"abc"}//serverUrl = "http://192.168.100.12/testing" //host=192.168.100.12

  public static String getJson(String serverUrl,String host,String jsonobject){

    StringBuilder sb = new StringBuilder();

    String http = serverUrl;

    HttpURLConnection urlConnection = null;
    try {
        URL url = new URL(http);
        urlConnection = (HttpURLConnection) url.openConnection();
        urlConnection.setDoOutput(true);
        urlConnection.setRequestMethod("POST");
        urlConnection.setUseCaches(false);
        urlConnection.setConnectTimeout(50000);
        urlConnection.setReadTimeout(50000);
        urlConnection.setRequestProperty("Content-Type", "application/json");
        urlConnection.setRequestProperty("Host", host);
        urlConnection.connect();
        //You Can also Create JSONObject here 
        OutputStreamWriter out = new OutputStreamWriter(urlConnection.getOutputStream());
        out.write(jsonobject);// here i sent the parameter
        out.close();
        int HttpResult = urlConnection.getResponseCode();
        if (HttpResult == HttpURLConnection.HTTP_OK) {
            BufferedReader br = new BufferedReader(new InputStreamReader(
                    urlConnection.getInputStream(), "utf-8"));
            String line = null;
            while ((line = br.readLine()) != null) {
                sb.append(line + "\n");
            }
            br.close();
            Log.e("new Test", "" + sb.toString());
            return sb.toString();
        } else {
            Log.e(" ", "" + urlConnection.getResponseMessage());
        }
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } catch (JSONException e) {
        e.printStackTrace();
    } finally {
        if (urlConnection != null)
            urlConnection.disconnect();
    }
    return null;
}

Answer 5

I higly recomend http-request built on apache http api.

For your case you can see example:

private static final HttpRequest<String.class> HTTP_REQUEST = 
      HttpRequestBuilder.createPost("http://example.com/index.php", String.class)
           .responseDeserializer(ResponseDeserializer.ignorableDeserializer())
           .build();

public void sendRequest(String request){
     String parameters = request.split("\\?")[1];
     ResponseHandler<String> responseHandler = 
            HTTP_REQUEST.executeWithQuery(parameters);

   System.out.println(responseHandler.getStatusCode());
   System.out.println(responseHandler.get()); //prints response body
}

If you are not interested in the response body

private static final HttpRequest<?> HTTP_REQUEST = 
     HttpRequestBuilder.createPost("http://example.com/index.php").build();

public void sendRequest(String request){
     ResponseHandler<String> responseHandler = 
           HTTP_REQUEST.executeWithQuery(parameters);
}

For general sending post request with http-request: Read the documentation and see my answers HTTP POST request with JSON String in JAVA, Sending HTTP POST Request In Java, HTTP POST using JSON in Java

Answer 6

Appears that you also have to callconnection.getOutputStream() "at least once" (as well as setDoOutput(true)) for it to treat it as a POST.

So the minimum required code is:

    URL url = new URL(urlString);
    HttpURLConnection connection = (HttpURLConnection) url.openConnection();
    //connection.setRequestMethod("POST"); this doesn't seem to do anything at all..so not useful
    connection.setDoOutput(true); // set it to POST...not enough by itself however, also need the getOutputStream call...
    connection.connect();
    connection.getOutputStream().close(); 

You can even use "GET" style parameters in the urlString, surprisingly. Though that might confuse things.

You can also use NameValuePair apparently.