Sending HTTP POST Request In Java
(Here id needs to be sent in a POST request) I want to send the id = 10 to the server’s page.php , which accepts it in a POST method. How can i do this from within Java? I tried this :
URL aaa = new URL("http://www.example.com/page.php"); URLConnection ccc = aaa.openConnection(); 12 Answers 12
Updated answer
Since some of the classes, in the original answer, are deprecated in the newer version of Apache HTTP Components, I’m posting this update.
By the way, you can access the full documentation for more examples here.
HttpClient httpclient = HttpClients.createDefault(); HttpPost httppost = new HttpPost("http://www.a-domain.example/foo/"); // Request parameters and other properties. List params = new ArrayList(2); params.add(new BasicNameValuePair("param-1", "12345")); params.add(new BasicNameValuePair("param-2", "Hello!")); httppost.setEntity(new UrlEncodedFormEntity(params, "UTF-8")); //Execute and get the response. HttpResponse response = httpclient.execute(httppost); HttpEntity entity = response.getEntity(); if (entity != null) < try (InputStream instream = entity.getContent()) < // do something useful >> Original answer
I recommend to use Apache HttpClient. its faster and easier to implement.
HttpPost post = new HttpPost("http://jakarata.apache.org/"); NameValuePair[] data = < new NameValuePair("user", "joe"), new NameValuePair("password", "bloggs") >; post.setRequestBody(data); // execute method and handle any error responses. . InputStream in = post.getResponseBodyAsStream(); // handle response. for more information check this URL: http://hc.apache.org/
After trying for a while to get my hands on PostMethod it seems its actually now called HttpPost as per stackoverflow.com/a/9242394/1338936 — just for anyone finding this answer like I did 🙂
For anyone getting the same issue as @AdarshSingh, I found a solution after looking at this provided example. Just change HttpClient to CloseableHttpClient, and HttpResponse to CloseableHttpResponse!
Sending a POST request is easy in vanilla Java. Starting with a URL , we need t convert it to a URLConnection using url.openConnection(); . After that, we need to cast it to a HttpURLConnection , so we can access its setRequestMethod() method to set our method. We finally say that we are going to send data over the connection.
URL url = new URL("https://www.example.com/login"); URLConnection con = url.openConnection(); HttpURLConnection http = (HttpURLConnection)con; http.setRequestMethod("POST"); // PUT is another valid option http.setDoOutput(true); We then need to state what we are going to send:
Sending a simple form
A normal POST coming from a http form has a well defined format. We need to convert our input to this format:
Map arguments = new HashMap<>(); arguments.put("username", "root"); arguments.put("password", "sjh76HSn!"); // This is a fake password obviously StringJoiner sj = new StringJoiner("&"); for(Map.Entry entry : arguments.entrySet()) sj.add(URLEncoder.encode(entry.getKey(), "UTF-8") + "=" + URLEncoder.encode(entry.getValue(), "UTF-8")); byte[] out = sj.toString().getBytes(StandardCharsets.UTF_8); int length = out.length; We can then attach our form contents to the http request with proper headers and send it.
http.setFixedLengthStreamingMode(length); http.setRequestProperty("Content-Type", "application/x-www-form-urlencoded; charset=UTF-8"); http.connect(); try(OutputStream os = http.getOutputStream()) < os.write(out); >// Do something with http.getInputStream() Sending JSON
We can also send json using java, this is also easy:
byte[] out = "" .getBytes(StandardCharsets.UTF_8); int length = out.length; http.setFixedLengthStreamingMode(length); http.setRequestProperty("Content-Type", "application/json; charset=UTF-8"); http.connect(); try(OutputStream os = http.getOutputStream()) < os.write(out); >// Do something with http.getInputStream() Remember that different servers accept different content-types for json, see this question.
Sending files with java post
Sending files can be considered more challenging to handle as the format is more complex. We are also going to add support for sending the files as a string, since we don’t want to buffer the file fully into the memory.
For this, we define some helper methods:
private void sendFile(OutputStream out, String name, InputStream in, String fileName) < String o = "Content-Disposition: form-data; name=\"" + URLEncoder.encode(name,"UTF-8") + "\"; filename=\"" + URLEncoder.encode(filename,"UTF-8") + "\"\r\n\r\n"; out.write(o.getBytes(StandardCharsets.UTF_8)); byte[] buffer = new byte[2048]; for (int n = 0; n >= 0; n = in.read(buffer)) out.write(buffer, 0, n); out.write("\r\n".getBytes(StandardCharsets.UTF_8)); > private void sendField(OutputStream out, String name, String field) We can then use these methods to create a multipart post request as follows:
String boundary = UUID.randomUUID().toString(); byte[] boundaryBytes = ("--" + boundary + "\r\n").getBytes(StandardCharsets.UTF_8); byte[] finishBoundaryBytes = ("--" + boundary + "--").getBytes(StandardCharsets.UTF_8); http.setRequestProperty("Content-Type", "multipart/form-data; charset=UTF-8; boundary=" + boundary); // Enable streaming mode with default settings http.setChunkedStreamingMode(0); // Send our fields: try(OutputStream out = http.getOutputStream()) < // Send our header (thx Algoman) out.write(boundaryBytes); // Send our first field sendField(out, "username", "root"); // Send a seperator out.write(boundaryBytes); // Send our second field sendField(out, "password", "toor"); // Send another seperator out.write(boundaryBytes); // Send our file try(InputStream file = new FileInputStream("test.txt")) < sendFile(out, "identification", file, "text.txt"); >// Finish the request out.write(finishBoundaryBytes); > // Do something with http.getInputStream()