How to send HTTP request in Java?

Question 1

In Java, How to compose an HTTP request message and send it to an HTTP web server?

Question 2

java.sun.com/javase/6/docs/api/java/net/HttpURLConnection.html In particular, getHeaderField, getHeaderFieldKey, and getContent

Question 3

There's a mini tutorial here at SO.

Question 4

You could use the JSoup lib (jsoup.org) . It does exactly what you ask! Document doc = Jsoup.connect("en.wikipedia.org"). get(); (from the site). A more pythonic way for java.

Question 5

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();
 }
 }
}

Question 6

Putting some actual code into this answer will help avoid link rot...

Question 7

Since Java 9, creating HTTP request has become much easier.

Question 8

Yes, a lot has changed in the ten years since this answer was given. Not everyone has moved on from JDK8 to 9 and beyond.

Question 9

How to put some header content in the request with this way of doing, please ?

Question 10

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();
 }
}

Question 11

The strange thing is that some servers will reply you back with strange ? characters (which seems like an encoding error related to request headers but not) if you don't open an output stream and flush it first. I have no idea why this happens but will be great if someone can explain why?

Question 12

@Gorky: Make a new question

Question 13

This is way too much line noise to send an HTTP request imo. Contrast to Python's requests library: response = requests.get('http://www.yahoo.com/'); something of similar brevity should be possible in Java.

Question 14

@leo-the-manic that's because Java is supposed to be a lower level language (than python) and allows (forces) the programmer to handle the details underneath rather than assuming "sane" defaults (i.e. buffering, character encoding, etc.). It is possible to get something as succinct, but then you lose lots of the flexibility of the more barebones approach.

Question 15

@fortran Python has equally low-level options to accomplish the same thing as above.

Question 16

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();
}

Question 17

That doesn't help if you want to monkey with request headers, something that's particularly useful when dealing with sites that will only respond a certain way to popular browsers.

Question 18

You can monkey with request headers using URLConnection, but the poster doesn't ask for that; judging from the question, a simple answer is important.

Question 19

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.

Question 20

FWIW, our code started with java.net.HttpURLConnection, but when we had to add SSL and work around some of the weird use cases in our screwy internal networks, it became a real headache. Apache HttpComponents saved the day. Our project currently still uses an ugly hybrid, with a few dodgy adapters to convert java.net.URLs to the URIs HttpComponents uses. I refactor those out regularly. The only time HttpComponents code turned out significantly more complicated was for parsing dates from a header. But the solution for that is still simple.

Question 21

It would be helpful to add a code snippet here

Question 22

Here's a complete Java 7 program:

class GETHTTPResource {
 public static void main(String[] args) throws Exception {
 try (java.util.Scanner s = new java.util.Scanner(new java.net.URL("http://example.com/").openStream())) {
 System.out.println(s.useDelimiter("\\A").next());
 }
 }
}

The new try-with-resources will auto-close the Scanner, which will auto-close the InputStream.

Question 23

@Ska There is no unhandled exception. main() throws Exception, which encompasses the MalformedURLException and the IOException.

Question 24

Scanner actually is not very optimized when it comes to performance.

Question 25

If you are using Java 11 or newer (except on Android), instead of the legacy HttpUrlConnection class, you can use Java 11 new HTTP Client API.

An example GET request:

var uri = URI.create("https://httpbin.org/get?age=26&isHappy=true");
var client = HttpClient.newHttpClient();
var request = HttpRequest
 .newBuilder()
 .uri(uri)
 .header("accept", "application/json")
 .GET()
 .build();
var response = client.send(request, HttpResponse.BodyHandlers.ofString());
System.out.println(response.statusCode());
System.out.println(response.body());

The same request executed asynchronously:

var responseAsync = client
 .sendAsync(request, HttpResponse.BodyHandlers.ofString())
 .thenApply(HttpResponse::body)
 .thenAccept(System.out::println);
// responseAsync.join(); // Wait for completion

An example POST request:

var request = HttpRequest
 .newBuilder()
 .uri(uri)
 .version(HttpClient.Version.HTTP_2)
 .timeout(Duration.ofMinutes(1))
 .header("Content-Type", "application/json")
 .header("Authorization", "Bearer fake")
 .POST(BodyPublishers.ofString("{ title: 'This is cool' }"))
 .build();
var response = client.send(request, HttpResponse.BodyHandlers.ofString());

For sending form data as multipart (multipart/form-data) or url-encoded (application/x-www-form-urlencoded) format, see this solution.

See this article for examples and more information about HTTP Client API.

For Java standard library HTTP server, see this post.

Question 26

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();
 }
}

Question 27

What do you mean with 'transport'?

Question 28

Sorry, that should have been HTTP_TRANSPORT, I've edited the answer.

Question 29

why is HttpResponse not AutoClosable? What is the difference from this and to working with Apache's CloseableHttpClient?

Question 30

The benefit is the API, which makes it personal preference really. Google's library uses Apache's library internally. That said, I like Google's lib.

Question 31

Simple or complex requests doesn't matter, readability is the king.

Question 32

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 protected]");
 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();
 }
 }
}

Question 33

Seriously, I really like Java, but what's the matter with that stupid NameValuePair list or array. Why not a simple Map<String, String>? So much boilerplate code for such simple use cases...

Question 34

@Joffrey Maps by definition have 1 key per value, means: A map cannot contain duplicate keys ! But HTTP Parameters can have duplicate keys.

Question 35

You may use Socket for this like

String host = "www.yourhost.com";
Socket socket = new Socket(host, 80);
String request = "GET / HTTP/1.0\r\n\r\n";
OutputStream os = socket.getOutputStream();
os.write(request.getBytes());
os.flush();
InputStream is = socket.getInputStream();
int ch;
while( (ch=is.read())!= -1)
 System.out.print((char)ch);
socket.close();

Question 36

@laksys why it should be \r\n instead of \n?

Question 37

@CuriousGuy look at this link programmers.stackexchange.com/questions/29075/…

Question 38

It seems to be even easier and more straight forward than the other solutions. Java makes things more complicated than it should be.

Question 39

There's a great link about sending a POST request here by Example Depot::

try {
 // Construct data
 String data = URLEncoder.encode("key1", "UTF-8") + "=" + URLEncoder.encode("value1", "UTF-8");
 data += "&" + URLEncoder.encode("key2", "UTF-8") + "=" + URLEncoder.encode("value2", "UTF-8");
 // Send data
 URL url = new URL("http://hostname:80/cgi");
 URLConnection conn = url.openConnection();
 conn.setDoOutput(true);
 OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
 wr.write(data);
 wr.flush();
 // Get the response
 BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
 String line;
 while ((line = rd.readLine()) != null) {
 // Process line...
 }
 wr.close();
 rd.close();
} catch (Exception e) {
}

If you want to send a GET request you can modify the code slightly to suit your needs. Specifically you have to add the parameters inside the constructor of the URL. Then, also comment out this wr.write(data);

One thing that's not written and you should beware of, is the timeouts. Especially if you want to use it in WebServices you have to set timeouts, otherwise the above code will wait indefinitely or for a very long time at least and it's something presumably you don't want.

Timeouts are set like this conn.setReadTimeout(2000); the input parameter is in milliseconds

Question 40

New HTTP Client in Java 11+

Java 11 brought a new HTTP client. Contained in the new module java.net.http. For more info, see:

JEP 321: HTTP Client API
JEP 110: HTTP/2 Client (Incubator)
Java HTTP Client project page, with links to tutorials & presentations.

Using the new HTTP Client is easy. The framework provides convenient fluent syntax.

First create the client object.

HttpClient client = HttpClient.newBuilder( )
 .version( HttpClient.Version.HTTP_1_1 )
 .build( ) ;

Set up the request.

HttpRequest request = HttpRequest.newBuilder( )
 .uri( URI.create( "http://worldtimeapi.org/api/timezone/UTC" ) )
 .timeout( Duration.ofMinutes( 1 ) )
 .GET( )
 .build( );

Hit the server.

HttpResponse < String > response = client.send( request , HttpResponse.BodyHandlers.ofString( ) );

Extract the response parts.

response.statusCode( )
response.body( )

Full example code:

package work.basil.example.web;
import java.io.IOException;
import java.net.URI;
import java.net.http.HttpClient;
import java.net.http.HttpRequest;
import java.net.http.HttpResponse;
import java.time.Duration;
public class ExHttpClient
{
 public static void main ( String[] args )
 {
 // JEP 321: HTTP Client API
 // https://openjdk.org/jeps/321
 try (
 HttpClient client = HttpClient.newBuilder( )
 .version( HttpClient.Version.HTTP_1_1 )
 .build( ) ;
 )
 {
 // Example HTTP server: http://worldtimeapi.org/
 HttpRequest request = HttpRequest.newBuilder( )
 .uri( URI.create( "http://worldtimeapi.org/api/timezone/UTC" ) )
 .timeout( Duration.ofMinutes( 1 ) )
 .GET( )
 .build( );
 HttpResponse < String > response = client.send( request , HttpResponse.BodyHandlers.ofString( ) );
 System.out.println( response.statusCode( ) );
 System.out.println( response.body( ) );
 } catch ( IOException | InterruptedException e )
 {
 throw new RuntimeException( e );
 }
 }
}

When run:

200

{"utc_offset":"+00:00","timezone":"UTC","day_of_week":3,"day_of_year":248,"datetime":"2024年09月04日T18:51:19.336832+00:00","utc_datetime":"2024年09月04日T18:51:19.336832+00:00","unixtime":1725475879,"raw_offset":0,"week_number":36,"dst":false,"abbreviation":"UTC","dst_offset":0,"dst_from":null,"dst_until":null,"client_ip":"71.212.7.38"}

HTTP Server

By the way, Java 18+ comes with a basic implementation of a HTTP Web server. See JEP 408: Simple Web Server. Caveat: This implementation is not meant for use in production; its purpose is learning and experimenting.

This simple HTTP server can be run from command-line in a console using a tool named jwebserver. Or you can run the server from Java code.

So you could create your own little server to test your client code.

duffymo duffymo 310k46 gold badges376 silver badges570 bronze badges · Accepted Answer · 2009-08-31 22:35:25Z

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();
 }
 }
}

Putting some actual code into this answer will help avoid link rot...
Yes, a lot has changed in the ten years since this answer was given. Not everyone has moved on from JDK8 to 9 and beyond.
How to put some header content in the request with this way of doing, please ?

CollectivesTM on Stack Overflow

How to send HTTP request in Java?

11 Answers 11

4 Comments

11 Comments

2 Comments

2 Comments

2 Comments

An example GET request:

An example POST request:

For Java standard library HTTP server, see this post.

Comments

5 Comments

2 Comments

3 Comments

Comments

New HTTP Client in Java 11+

HTTP Server

Comments

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