How to send HTTP POST requests on Java?

0 votes

Consider the following 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

May 30, 2018 in Java by developer_1
• 3,220 points
9,292 views

10 answers to this question.

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0 votes

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);
}
answered May 30, 2018 by Rishabh
• 3,520 points
0 votes

Here is the code I used, It works fine.

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();
}
answered Dec 10, 2018 by Niraj
0 votes

This could be a simple implimentation, 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();
answered Dec 10, 2018 by Bhavya
0 votes
You can go through this article for better understanding

http://hc.apache.org/httpcomponents-client-ga/tutorial/html/fluent.html
answered Dec 10, 2018 by Trisha
0 votes

Use http-request that is 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();
}
answered Dec 10, 2018 by Abhinav
0 votes
See this git http-request, this is the easiest method i found

https://github.com/jsunsoftware/http-request
answered Dec 10, 2018 by kriti
0 votes

Use this to send http post requests

HttpRequest<String> httpRequest = HttpRequestBuilder.createPost("https://www.example.com/", String.class)
                                                .responseDeserializer(ResponseDeserializer.toStringDeserializer()).build();
String responseBody = httpRequest.execute(requestParameters).get(); 
answered Dec 10, 2018 by ravi
0 votes

You can use the below function implementation

Request.Post("http://somehost/some-form")
        .addHeader("X-Custom-header", "stuff")
        .viaProxy(new HttpHost("myproxy", 8080))
        .bodyForm(Form.form().add("username", "vip").add("password", "secret").build())
        .execute().saveContent(new File("result.dump"));
answered Dec 10, 2018 by richard072
0 votes

You can call HttpURLConnection.setRequestMethod("POST") and HttpURLConnection.setDoOutput(true).
Actually only the function is needed as POST then becomes the default method.

answered Dec 10, 2018 by Pranav
0 votes

You can use the function as HttpURLConnection.setRequestMethod()

answered Dec 10, 2018 by simplecode

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