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, Oct 04 '12 at 17:26
sorry if this sounds dumb, but where are you calling `connect()` on the URLConnection? — Vikdor, Oct 04 '12 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, Oct 04 '12 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, Apr 12 '19 at 12:07

score 505 · Accepted Answer · edited Sep 20 '18 at 12:16

505

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 '18 at 12:16

Mike

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answered Mar 21 '13 at 19:00

Cyril Beschi

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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. – Denis Gladkiy Dec 27 '13 at 10:08
@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 – Pankaj Nimgade Feb 17 '15 at 08:00
14

Where did the variable `postData` come from in your example? – GlenPeterson Jan 11 '16 at 19:28
android.util.Base64 is not accessible in latest APIs, you should use org.apache.commons.codec.binary.Base64 or else you can follow this link http://androidcodemonkey.blogspot.in/2010/03/how-to-base64-encode-decode-android.html, to download Base64 class and use it. Hope it helps... – praveenb Feb 29 '16 at 14:44
had to replace `myURLConnection.setRequestMethod("POST");` with `myURLConnection.setRequestProperty("Request Method", "POST");` - rest worked fine! (Probably because of newer version of package?) – Samuel Jul 08 '16 at 14:54
40

Why are they called "RequestProperty" when everyone calls them Headers?? – Philip Rego Jun 21 '17 at 19:58
2

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());``` – c3R1cGFy Jan 18 '18 at 10:53
I don't understand how this answers the question. He said he sends the header and it doesn't arrive at the server. – Douglas Held Oct 11 '18 at 17:28

score 18 · Answer 2 · edited Apr 10 '17 at 00:00

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);

score 14 · Answer 3 · edited Jun 06 '17 at 18:18

14

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 06 '17 at 18:18

Parag Jadhav

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answered Jun 06 '17 at 17:48

Chandubabu

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gori · Answer 4 · 2020-06-25T20:03:11.830

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 '20 at 20:03

answered Apr 04 '15 at 20:38

gori

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@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. – Yuriy N. Nov 16 '15 at 12:39

score 5 · Answer 5 · answered Dec 01 '15 at 12:06

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 · Answer 6 · 2017-02-26T14:50:57.633

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 '17 at 14:50

answered Jul 19 '16 at 11:38

Eyal Sooliman

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Do you mind formatting the code here a bit more cleanly? Also, what is `given()` supposed to be? – Nathaniel Ford Feb 23 '17 at 00:51
Hi, This is the base usage for rest-Assurd (testing rest Api). I added explains to the code. – Eyal Sooliman Feb 26 '17 at 14:48

ivvasch · Answer 7 · 2022-08-19T21:03:21.877

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.

score -1 · Answer 8 · answered Dec 28 '19 at 15:18

Step 1: Get HttpURLConnection object

URL url = new URL(urlToConnect);
HttpURLConnection httpUrlConnection = (HttpURLConnection) url.openConnection();

Step 2: Add headers to the HttpURLConnection using setRequestProperty method.

Map<String, String> headers = new HashMap<>();

headers.put("X-CSRF-Token", "fetch");
headers.put("content-type", "application/json");

for (String headerKey : headers.keySet()) {
    httpUrlConnection.setRequestProperty(headerKey, headers.get(headerKey));
}

Reference link

Adding header for HttpURLConnection

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