Adding header for HttpURLConnection
JavaHttpJava Problem Overview
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);
}
Java Solutions
Solution 1 - Java
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
Solution 2 - Java
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);
Solution 3 - Java
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);
Solution 4 - Java
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));
}
Solution 5 - Java
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
Solution 6 - Java
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
Solution 7 - Java
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