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Sending HTTP POST Request In Java

lets assume this URL...

http://www.example.com/page.php?id=10            

(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();

But I still can't figure out how to send it via POST


S
Stephen Ostermiller

###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<NameValuePair> params = new ArrayList<NameValuePair>(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 :)
@Juan (and Martin Lyne) thank you for the comments. I just updated the answer.
you should add the imported libs
And also giving an error that cannot resolve getEntity()
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!
C
Community

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<String,String> arguments = new HashMap<>();
arguments.put("username", "root");
arguments.put("password", "sjh76HSn!"); // This is a fake password obviously
StringJoiner sj = new StringJoiner("&");
for(Map.Entry<String,String> 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 = "{\"username\":\"root\",\"password\":\"password\"}" .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) {
    String o = "Content-Disposition: form-data; name=\"" 
             + URLEncoder.encode(name,"UTF-8") + "\"\r\n\r\n";
    out.write(o.getBytes(StandardCharsets.UTF_8));
    out.write(URLEncoder.encode(field,"UTF-8").getBytes(StandardCharsets.UTF_8));
    out.write("\r\n".getBytes(StandardCharsets.UTF_8));
}

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()

This post is useful, but quite flawed. It took me 2 days to get it working. So to get it working you have to replace StandartCharsets.UTF8 with StandardCharsets.UTF_8 . boundaryBytes and finishBoundaryBytes need to get two additional hyphens which are NOT transmitted in the Content-Type, so boundaryBytes = ("--" + boundary + "\r\n").get... You also need to transmit the boundaryBytes once BEFORE the first field or the first field will be ignored!
Why out.write(finishBoundaryBytes); line need? http.connect(); will perform sending POST, isn't it?
It's relatively easier than what I expected considering it's Java :)
enigmatic \r\n\r\n means CRLF CRLF (carriage return + line feed). It creates 2x new line. First new line is to finish current line. Second line is to distinguish http header from http body in a request. HTTP is ASCII based protocol. This is the rule for inserting \r\n.
"Easy" In other languages this is like a one-line call. Why is it 8-12 lines in Java? qr.ae/TWAQA6
L
Line
String rawData = "id=10";
String type = "application/x-www-form-urlencoded";
String encodedData = URLEncoder.encode( rawData, "UTF-8" ); 
URL u = new URL("http://www.example.com/page.php");
HttpURLConnection conn = (HttpURLConnection) u.openConnection();
conn.setDoOutput(true);
conn.setRequestMethod("POST");
conn.setRequestProperty( "Content-Type", type );
conn.setRequestProperty( "Content-Length", String.valueOf(encodedData.length()));
OutputStream os = conn.getOutputStream();
os.write(encodedData.getBytes());

Important to notice: using anything other then String.getBytes() does not seem to work. For example, using a PrintWriter totally fails.
and how to set 2 post data? Separate by colon, comma?
encode(String) is deprecated. You have to use encode(String, String), which specifies the encoding type. Example: encode(rawData, "UTF-8").
You may want to following at the end. This would make sure the request is finished and server gets a chance to process the response: conn.getResponseCode();
dont encode the entire string.. you have to encode just the value of each parameter
M
Mar Cnu

The first answer was great, but I had to add try/catch to avoid Java compiler errors.
Also, I had troubles to figure how to read the HttpResponse with Java libraries.

Here is the more complete code :

/*
 * Create the POST request
 */
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://example.com/");
// Request parameters and other properties.
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("user", "Bob"));
try {
    httpPost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));
} catch (UnsupportedEncodingException e) {
    // writing error to Log
    e.printStackTrace();
}
/*
 * Execute the HTTP Request
 */
try {
    HttpResponse response = httpClient.execute(httpPost);
    HttpEntity respEntity = response.getEntity();

    if (respEntity != null) {
        // EntityUtils to get the response content
        String content =  EntityUtils.toString(respEntity);
    }
} catch (ClientProtocolException e) {
    // writing exception to log
    e.printStackTrace();
} catch (IOException e) {
    // writing exception to log
    e.printStackTrace();
}

Sorry, but you didn't catch any errors, you introduced them. Catching exceptions in a place where you can't handle them is plain wrong and e.printStackTrace() doesn't handle anything.
java.net.ConnectException: Connection timed out: connect
L
Lucky

A simple way using Apache HTTP Components is

Request.Post("http://www.example.com/page.php")
            .bodyForm(Form.form().add("id", "10").build())
            .execute()
            .returnContent();

Take a look at the Fluent API


H
Hani

I suggest using Postman to generate the request code. Simply make the request using Postman then hit the code tab:

https://i.stack.imgur.com/XFkQE.png

https://i.stack.imgur.com/M6v0o.png


Nice suggestions.
c
chandan kumar

simplest way to send parameters with the post request:

String postURL = "http://www.example.com/page.php";

HttpPost post = new HttpPost(postURL);

List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("id", "10"));

UrlEncodedFormEntity ent = new UrlEncodedFormEntity(params, "UTF-8");
post.setEntity(ent);

HttpClient client = new DefaultHttpClient();
HttpResponse responsePOST = client.execute(post);

You have done. now you can use responsePOST. Get response content as string:

BufferedReader reader = new BufferedReader(new  InputStreamReader(responsePOST.getEntity().getContent()), 2048);

if (responsePOST != null) {
    StringBuilder sb = new StringBuilder();
    String line;
    while ((line = reader.readLine()) != null) {
        System.out.println(" line : " + line);
        sb.append(line);
    }
    String getResponseString = "";
    getResponseString = sb.toString();
//use server output getResponseString as string value.
}

s
spok

Easy with java.net:

public void post(String uri, String data) throws Exception {
HttpClient client = HttpClient.newBuilder().build();
HttpRequest request = HttpRequest.newBuilder()
        .uri(URI.create(uri))
        .POST(BodyPublishers.ofString(data))
        .build();

HttpResponse<?> response = client.send(request, BodyHandlers.discarding());
System.out.println(response.statusCode());

Here is more information: https://openjdk.java.net/groups/net/httpclient/recipes.html#post


F
Flair

Using okhttp :

Source code for okhttp can be found here https://github.com/square/okhttp.

If you're writing a pom project, add this dependency

<dependency>
        <groupId>com.squareup.okhttp3</groupId>
        <artifactId>okhttp</artifactId>
        <version>4.2.2</version>
    </dependency>

If not simply search the internet for 'download okhttp'. Several results will appear where you can download a jar.

your code :

import okhttp3.*;
        
import java.io.IOException;

public class ClassName{
        private void sendPost() throws IOException {
        
                // form parameters
                RequestBody formBody = new FormBody.Builder()
                        .add("id", 10)
                        .build();
        
                Request request = new Request.Builder()
                        .url("http://www.example.com/page.php")
                        .post(formBody)
                        .build();


                OkHttpClient httpClient = new OkHttpClient();
        
                try (Response response = httpClient.newCall(request).execute()) {
        
                    if (!response.isSuccessful()) throw new IOException("Unexpected code " + response);
        
                    // Get response body
                    System.out.println(response.body().string());
                }
        }
    }

u
user207421

Call HttpURLConnection.setRequestMethod("POST") and HttpURLConnection.setDoOutput(true); Actually only the latter is needed as POST then becomes the default method.


it it HttpURLConnection.setRequestMethod() :)
B
Beno Arakelyan

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

HttpRequest<String> httpRequest = HttpRequestBuilder.createPost("http://www.example.com/page.php", String.class)
.responseDeserializer(ResponseDeserializer.ignorableDeserializer()).build();

public void send(){
   String response = httpRequest.execute("id", "10").get();
}