How to send HTTP request in java? [duplicate]

Answer 1

I know others will recommend Apache's http-client, but it adds complexity (i.e., more things that can go wrong) that is rarely warranted. For a simple task, java.net.URL will do.

URL url = new URL("http://www.y.com/url");
InputStream is = url.openStream();
try {
  /* Now read the retrieved document from the stream. */
  ...
} finally {
  is.close();
}

Answer 2

This will help you. Don't forget to add the JAR HttpClient.jar to the classpath.

import java.io.FileOutputStream;
import java.io.IOException;

import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.HttpStatus;
import org.apache.commons.httpclient.NameValuePair;
import org.apache.commons.httpclient.methods.PostMethod;

public class MainSendRequest {

     static String url =
         "http://localhost:8080/HttpRequestSample/RequestSend.jsp";

    public static void main(String[] args) {

        //Instantiate an HttpClient
        HttpClient client = new HttpClient();

        //Instantiate a GET HTTP method
        PostMethod method = new PostMethod(url);
        method.setRequestHeader("Content-type",
                "text/xml; charset=ISO-8859-1");

        //Define name-value pairs to set into the QueryString
        NameValuePair nvp1= new NameValuePair("firstName","fname");
        NameValuePair nvp2= new NameValuePair("lastName","lname");
        NameValuePair nvp3= new NameValuePair("email","email@email.com");

        method.setQueryString(new NameValuePair[]{nvp1,nvp2,nvp3});

        try{
            int statusCode = client.executeMethod(method);

            System.out.println("Status Code = "+statusCode);
            System.out.println("QueryString>>> "+method.getQueryString());
            System.out.println("Status Text>>>"
                  +HttpStatus.getStatusText(statusCode));

            //Get data as a String
            System.out.println(method.getResponseBodyAsString());

            //OR as a byte array
            byte [] res  = method.getResponseBody();

            //write to file
            FileOutputStream fos= new FileOutputStream("donepage.html");
            fos.write(res);

            //release connection
            method.releaseConnection();
        }
        catch(IOException e) {
            e.printStackTrace();
        }
    }
}

Answer 3

Apache HttpComponents. The examples for the two modules - HttpCore and HttpClient will get you started right away.

Not that HttpUrlConnection is a bad choice, HttpComponents will abstract a lot of the tedious coding away. I would recommend this, if you really want to support a lot of HTTP servers/clients with minimum code. By the way, HttpCore could be used for applications (clients or servers) with minimum functionality, whereas HttpClient is to be used for clients that require support for multiple authentication schemes, cookie support etc.

Answer 4

From Oracle's java tutorial

import java.net.*;
import java.io.*;

public class URLConnectionReader {
    public static void main(String[] args) throws Exception {
        URL yahoo = new URL("http://www.yahoo.com/");
        URLConnection yc = yahoo.openConnection();
        BufferedReader in = new BufferedReader(
                                new InputStreamReader(
                                yc.getInputStream()));
        String inputLine;

        while ((inputLine = in.readLine()) != null) 
            System.out.println(inputLine);
        in.close();
    }
}

Answer 5

Google java http client has nice API for http requests. You can easily add JSON support etc. Although for simple request it might be overkill.

import com.google.api.client.http.GenericUrl;
import com.google.api.client.http.HttpRequest;
import com.google.api.client.http.HttpResponse;
import com.google.api.client.http.HttpTransport;
import com.google.api.client.http.javanet.NetHttpTransport;
import java.io.IOException;
import java.io.InputStream;

public class Network {

    static final HttpTransport HTTP_TRANSPORT = new NetHttpTransport();

    public void getRequest(String reqUrl) throws IOException {
        GenericUrl url = new GenericUrl(reqUrl);
        HttpRequest request = HTTP_TRANSPORT.createRequestFactory().buildGetRequest(url);
        HttpResponse response = request.execute();
        System.out.println(response.getStatusCode());

        InputStream is = response.getContent();
        int ch;
        while ((ch = is.read()) != -1) {
            System.out.print((char) ch);
        }
        response.disconnect();
    }
}

Answer 6

You can use java.net.HttpUrlConnection.

Example (from here), with improvements. Included in case of link rot:

public static String executePost(String targetURL, String urlParameters) {
  HttpURLConnection connection = null;

  try {
    //Create connection
    URL url = new URL(targetURL);
    connection = (HttpURLConnection) url.openConnection();
    connection.setRequestMethod("POST");
    connection.setRequestProperty("Content-Type", 
        "application/x-www-form-urlencoded");

    connection.setRequestProperty("Content-Length", 
        Integer.toString(urlParameters.getBytes().length));
    connection.setRequestProperty("Content-Language", "en-US");  

    connection.setUseCaches(false);
    connection.setDoOutput(true);

    //Send request
    DataOutputStream wr = new DataOutputStream (
        connection.getOutputStream());
    wr.writeBytes(urlParameters);
    wr.close();

    //Get Response  
    InputStream is = connection.getInputStream();
    BufferedReader rd = new BufferedReader(new InputStreamReader(is));
    StringBuilder response = new StringBuilder(); // or StringBuffer if Java version 5+
    String line;
    while ((line = rd.readLine()) != null) {
      response.append(line);
      response.append('\r');
    }
    rd.close();
    return response.toString();
  } catch (Exception e) {
    e.printStackTrace();
    return null;
  } finally {
    if (connection != null) {
      connection.disconnect();
    }
  }
}