Adding header for HttpURLConnection

Question

I'm trying to add header for my request using HttpUrlConnection but the method setRequestProperty() doesn't seem working. The server side doesn't receive any request with my header.

HttpURLConnection hc;
    try {
        String authorization = "";
        URL address = new URL(url);
        hc = (HttpURLConnection) address.openConnection();


        hc.setDoOutput(true);
        hc.setDoInput(true);
        hc.setUseCaches(false);

        if (username != null && password != null) {
            authorization = username + ":" + password;
        }

        if (authorization != null) {
            byte[] encodedBytes;
            encodedBytes = Base64.encode(authorization.getBytes(), 0);
            authorization = "Basic " + encodedBytes;
            hc.setRequestProperty("Authorization", authorization);
        }

Works for me, how do you tell the header was sent and not received? — Tomasz Nurkiewicz
– Tomasz Nurkiewicz, Commented Oct 4, 2012 at 17:26
sorry if this sounds dumb, but where are you calling connect() on the URLConnection? — Vikdor
– Vikdor, Commented Oct 4, 2012 at 17:26
I'm not sure if this has an effect but you can try adding connection.setRequestMethod("GET"); (or POST or whatever you want)? — noobed
– noobed, Commented Oct 4, 2012 at 18:20
You initialise authorization to the empty string. If either username or password is null, then authorization will be the empty string, not null. Therefore, the final if will get executed, but the "Authorization" property will be set to empty, seems to me. — zerzevul
– zerzevul, Commented Apr 12, 2019 at 12:07

Mike · Accepted Answer · 2018-09-20 12:16:18Z

516

I have used the following code in the past and it had worked with basic authentication enabled in TomCat:

URL myURL = new URL(serviceURL);
HttpURLConnection myURLConnection = (HttpURLConnection)myURL.openConnection();

String userCredentials = "username:password";
String basicAuth = "Basic " + new String(Base64.getEncoder().encode(userCredentials.getBytes()));

myURLConnection.setRequestProperty ("Authorization", basicAuth);
myURLConnection.setRequestMethod("POST");
myURLConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
myURLConnection.setRequestProperty("Content-Length", "" + postData.getBytes().length);
myURLConnection.setRequestProperty("Content-Language", "en-US");
myURLConnection.setUseCaches(false);
myURLConnection.setDoInput(true);
myURLConnection.setDoOutput(true);

You can try the above code. The code above is for POST, and you can modify it for GET

edited Sep 20, 2018 at 12:16

Mike

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answered Mar 21, 2013 at 19:00

Cyril Beschi

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8 Comments

Denis Gladkiy Over a year ago

A little addition for android developers (on API >= 8 a.k.a 2.2): android.util.Base64.encode(userCredentials.getBytes(), Base64.DEFAULT); Base64.DEFAULT tells to use RFC2045 for base64 encoding.

Pankaj Nimgade Over a year ago

@Denis, would you please tell me why should one use headers. I have to validate some credentials from android I am using php on xammp. how should i go for it. as i don't know how to write php code with headers

GlenPeterson Over a year ago

Where did the variable postData come from in your example?

Philip Rego Over a year ago

Why are they called "RequestProperty" when everyone calls them Headers??

stuparm Over a year ago

One addition for Java8 version : Base64 class is a little bit changed. Decoding should be done using: String basicAuth = "Basic " + java.util.Base64.getEncoder().encodeToString(userCredentials.getBytes());

|

AlexW · Accepted Answer · 2017-04-10 00:00:49Z

20

Just cause I don't see this bit of information in the answers above, the reason the code snippet originally posted doesn't work correctly is because the encodedBytes variable is a byte[] and not a String value. If you pass the byte[] to a new String() as below, the code snippet works perfectly.

encodedBytes = Base64.encode(authorization.getBytes(), 0);
authorization = "Basic " + new String(encodedBytes);

edited Apr 10, 2017 at 0:00

AlexW

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answered Mar 16, 2015 at 20:19

Joseph Cozad

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Comments

Parag Jadhav · Accepted Answer · 2017-06-06 18:18:55Z

15

If you are using Java 8, use the code below.

URLConnection connection = url.openConnection();
HttpURLConnection httpConn = (HttpURLConnection) connection;

String basicAuth = Base64.getEncoder().encodeToString((username+":"+password).getBytes(StandardCharsets.UTF_8));
httpConn.setRequestProperty ("Authorization", "Basic "+basicAuth);

edited Jun 6, 2017 at 18:18

Parag Jadhav

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answered Jun 6, 2017 at 17:48

Chandubabu

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Comments

gori · Accepted Answer · 2020-06-25 20:03:11Z

6

Finally this worked for me

private String buildBasicAuthorizationString(String username, String password) {

    String credentials = username + ":" + password;
    return "Basic " + new String(Base64.encode(credentials.getBytes(), Base64.NO_WRAP));
}

edited Jun 25, 2020 at 20:03

answered Apr 4, 2015 at 20:38

gori

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1 Comment

Yuriy N. Over a year ago

@d3dave. String was created from byte array and concatenated with "Basic ". The problem in OP code was that he concatenated "Basic " with byte[] and sends it as header.

pintu · Accepted Answer · 2015-12-01 12:06:49Z

Your code is fine.You can also use the same thing in this way.

public static String getResponseFromJsonURL(String url) {
    String jsonResponse = null;
    if (CommonUtility.isNotEmpty(url)) {
        try {
            /************** For getting response from HTTP URL start ***************/
            URL object = new URL(url);

            HttpURLConnection connection = (HttpURLConnection) object
                    .openConnection();
            // int timeOut = connection.getReadTimeout();
            connection.setReadTimeout(60 * 1000);
            connection.setConnectTimeout(60 * 1000);
            String authorization="xyz:xyz$123";
            String encodedAuth="Basic "+Base64.encode(authorization.getBytes());
            connection.setRequestProperty("Authorization", encodedAuth);
            int responseCode = connection.getResponseCode();
            //String responseMsg = connection.getResponseMessage();

            if (responseCode == 200) {
                InputStream inputStr = connection.getInputStream();
                String encoding = connection.getContentEncoding() == null ? "UTF-8"
                        : connection.getContentEncoding();
                jsonResponse = IOUtils.toString(inputStr, encoding);
                /************** For getting response from HTTP URL end ***************/

            }
        } catch (Exception e) {
            e.printStackTrace();

        }
    }
    return jsonResponse;
}

Its Return response code 200 if authorizationis success

Eyal Sooliman · Accepted Answer · 2017-02-26 14:50:57Z

1

With RestAssurd you can also do the following:

String path = baseApiUrl; //This is the base url of the API tested
    URL url = new URL(path);
    given(). //Rest Assured syntax 
            contentType("application/json"). //API content type
            given().header("headerName", "headerValue"). //Some API contains headers to run with the API 
            when().
            get(url).
            then().
            statusCode(200); //Assert that the response is 200 - OK

edited Feb 26, 2017 at 14:50

answered Jul 19, 2016 at 11:38

Eyal Sooliman

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2 Comments

Nathaniel Ford Over a year ago

Do you mind formatting the code here a bit more cleanly? Also, what is given() supposed to be?

Eyal Sooliman Over a year ago

Hi, This is the base usage for rest-Assurd (testing rest Api). I added explains to the code.

ivvasch · Accepted Answer · 2022-08-19 21:03:21Z

It work for me. I had to send request to another hand, and transfer header "Authorization" + jwt and some params via POST. By another side we formed jettyRequest with params and headers. If I send this sequence of code:

URL url = new URL(serviceURL);
HttpURLConnection myURLConnection = (HttpURLConnection)url.openConnection();
myURLConnection.setRequestMethod("POST");
myURLConnection.setRequestProperty ("Authorization", jwt); // <---- this place
// some code add params

then I received only params in a body. If I send this:

URL url = new URL(serviceURL);
HttpURLConnection myURLConnection = (HttpURLConnection)url.openConnection();
myURLConnection.setRequestProperty ("Authorization", jwt); // <---- this place
myURLConnection.setRequestMethod("POST");
// some code add params

then I received headers Authorization and params.

yuriyni · Accepted Answer · 2023-12-05 06:47:48Z

if you are sure that the server is trustworthy and you are only connecting to it for testing purposes, you can use this option.

You will need to modify your Java code to use a custom TrustManager that does not validate the server's certificate.

additional code snippet for json body. And also send request and get repsponse examples:

//Create json body 
// option1
            JSONObject json = new JSONObject();
            json.put("param1Name", param1);
            json.put("param2Name", param2); 

//Create json body 
// option2
//String jsonInputString = "{"param1Name": param1, "param2Name": param2}";

            String jsonInputString = json.toJSONString();
//Create the Request Body
            try(OutputStream os = myURLConnection.getOutputStream()) {
                byte[] input = jsonInputString.getBytes("utf-8");
                os.write(input, 0, input.length);
            }

//Read the Response From Input Stream
            try(BufferedReader br = new BufferedReader(
                    new InputStreamReader(myURLConnection.getInputStream(), "utf-8"))) {

                String responseLine = null;
                while ((responseLine = br.readLine()) != null) {
                    response.append(responseLine.trim());
                }
                System.out.println(response.toString());

            }

Here is an full example of such a code is here

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