I am facing a problem where I am not able to set the "Authorization" Header.
I am Able to set the rest of the headers but when I am using the particular Key I am not able to set any data. Please help.
URL myURL = new URL(url);
HttpURLConnection myURLConnection = (HttpURLConnection)myURL.openConnection();
String basicAuth = "Bearer 6f6b06fe-131e-314b-9ef8-42f2cbdcfc18";
myURLConnection.setRequestMethod("GET");
myURLConnection.setRequestProperty("Accept-Language", "en-US,en;q=0.5");
myURLConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
myURLConnection.setRequestProperty("Content-Language", "en-US");
myURLConnection.setRequestProperty("Authorization", "basicAuth");
myURLConnection.setUseCaches(false);
myURLConnection.setDoInput(true);
myURLConnection.setDoOutput(true);
System.out.println(myURLConnection.getRequestProperties());
Hoping to hear soon. Thank you.
The statement
myURLConnection.getRequestProperties()
does not list all headers.
By looking at the source of HttpURLConnection you notice Authorization is part of the headers excluded by HttpURLConnection#getRequestProperties.
http://hg.openjdk.java.net/jdk8/jdk8/jdk/file/484e16c0a040/src/share/classes/sun/net/www/protocol/http/HttpURLConnection.java
This does not mean the header isn't set.
I think You made a little mistake there, The value for the key Authorization must not be "basicAuth". So please replace the code with :
myURLConnection.setRequestProperty("Authorization", basicAuth);
Or try this :
String basicAuth = "Bearer 6f6b06fe-131e-314b-9ef8-42f2cbdcfc18";
String encodedAuth= Base64.encode(basicAuth.getBytes());
myURLConnection.setRequestProperty("Authorization", encodedAuth);
Try with following code:
public void sendPost(String URL, String jsonData, String authUrl) throws Exception {
post = new HttpPost(URL);
// add header
post.setHeader("Authorization", accessToken);
post.setHeader("User-Agent", USER_AGENT);
if (!jsonData.isEmpty()) {
post.setEntity(new StringEntity(jsonData, ContentType.create("application/json")));
}
client = HttpClientBuilder.create().build();
response = client.execute(post);
outputFile = new File("path of file");
fos = new FileOutputStream(outputFile);
headers = response.getAllHeaders();
bw = new BufferedWriter(new OutputStreamWriter(fos));
for (Header header : headers) {
bw.write(header.getName() + ": " + header.getValue() + "\n");
}
bw.write("Response Code : " + response.getStatusLine());
bw.close();
}
Related
I have an issue regarding OData querying with an Java Client.
If I use Postman, everything works as expected and I'm receiving a response from the web service with the metadata. But in my Java Client, which runs not on the SCP / HCP I'm receiving "400-Bad Request". I used the original Olingo libary.
I only used the $metadata Parameter, so there is no filter value or something else.
public void sendGet(String user, String password, String url) throws IOException, URISyntaxException {
// String userPassword = user + ":" + password;
// String encoding = Base64.encodeBase64String(userPassword.getBytes("UTF-8"));
URL obj = new URL(url);
URL urlToEncode = new URL(url);
URI uri = new URI(urlToEncode.getProtocol(), urlToEncode.getUserInfo(), urlToEncode.getHost(), urlToEncode.getPort(), urlToEncode.getPath(), urlToEncode.getQuery(), urlToEncode.getRef());
// open Connection
HttpURLConnection con = (HttpURLConnection) uri.toURL().openConnection();
// Basis Authentifizierung
con.setRequestProperty("Authorization", "Basic " + user);
// optional default is GET
con.setRequestMethod("GET");
// add request header
con.setRequestProperty("Content-Type", "application/xml");
int responseCode = con.getResponseCode();
System.out.println("\nSending 'GET' request to URL : " + url);
System.out.println("Response Code : " + responseCode);
BufferedReader in = new BufferedReader(new InputStreamReader(con.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
response.append("\n");
}
in.close();
// print result
System.out.println(response.toString());
// Schließt eine Vorhandene Verbindung
con.disconnect();
in User is already the encoded value. by manipulating this one, i'm receiving an authorization error, so already tested.
May somebody can help me in that case :)
Thanks in advance.
Tim
So I solved it by myself.
i added the statement con.setRequestProperty("Accept", "application/xml"); and it works fo me.
Maybe it could help somebody else.
I am using the following code to perform POST requests on a REST API. It is all working fine. What I am being unable to do is after POST is successful the API returns response JSON in body with headers, this JSON has information which I require. I am unable to get the JSON response.
I need this response as this response includes the ID generated by DB. I can see the response while using REST Client plugin of firefox. Need to do implement the same in Java.
String json = "{\"name\": \"Test by JSON 1\",\"description\": \"Test by JSON 1\",\"fields\": {\"field\": []},\"typeDefinitionId\": \"23\",\"primaryParentId\": \"26982\"}";
String url = "http://serv23/api/contents";
URL obj = new URL(url);
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
//Setting the Request Method header as POST
con.setRequestMethod("POST");
//Prepairing credentials
String cred= "user123:p#ssw0rd";
byte[] encoded = Base64.encodeBase64(cred.getBytes());
String credentials = new String(encoded);
//Setting the Authorization Header as 'Basic' with the given credentials
con.setRequestProperty ("Authorization", "Basic " + credentials);
//Setting the Content Type Header as application/json
con.setRequestProperty("Content-Type", "application/json");
//Overriding the HTTP method as as mentioned in documentation
con.setRequestProperty("X-HTTP-Method-Override", "POST");
con.setDoOutput(true);
JSONObject jsonObject = (JSONObject)new JSONParser().parse(json);
OutputStream os = con.getOutputStream();
os.write(jsonObject.toJSONString().getBytes());
os.flush();
WriteLine( con.getResponseMessage() );
int responseCode = con.getResponseCode();
Get the input stream and read it.
String json_response = "";
InputStreamReader in = new InputStreamReader(con.getInputStream());
BufferedReader br = new BufferedReader(in);
String text = "";
while ((text = br.readLine()) != null) {
json_response += text;
}
I'm using the 'MultipartUtil' class as described in the first answer of the SO question.
I've changed the constructor, by adding additional lines, as follows:
public MultiPartUtil(String requestURL, String charset) throws IOException {
this.charset = charset;
// creates a unique boundary based on time stamp
boundary = "===" + System.currentTimeMillis() + "===";
URL url = new URL(requestURL);
httpConn = (HttpsURLConnection) url.openConnection();
httpConn.setUseCaches(false);
httpConn.setDoOutput(true); // indicates POST method
httpConn.setDoInput(true);
httpConn.setRequestMethod("POST");
httpConn.setRequestProperty("Connection", "Keep-Alive");
httpConn.setRequestProperty("Cache-Control", "no-cache");
httpConn.setRequestProperty("enctype", "multipart/form-data");
httpConn.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
httpConn.setRequestProperty("User-Agent", "MTApp");
outputStream = httpConn.getOutputStream();
writer = new PrintWriter(new OutputStreamWriter(outputStream, charset), true);
}
And the code to upload a file is:
MultiPartUtil multipart = new MultiPartUtil(url, "UTF-8");
multipart.addFilePart("archive", new File("archivefile.zip"));
multipart.finish();
On the server side, I have:
[HttpPost]
public void upload()
{
if (!Request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(Request.CreateResponse(HttpStatusCode.UnsupportedMediaType, "This request is not properly formatted"));
}
//Process file here...
}
What might I be missing in the Java code that does not allow the file to be accepted?
Can you please try to change the boundary to render like ---1234567--- instead of ===1234567===?
Because tokens are not allowed in http header parameters.
I am trying to send a request to get public transport information. Here's a screenshot of an example below, stating that I must send an XML request to the site, defining the method and the service reference (in the example it's StopMonitoringRequest and 020035811).
So far I have managed to connect to the service, but I have no idea what to do from here. I have so far done this...
String user = "";
String pass = "";
String url = "http://nextbus.mxdata.co.uk/nextbuses/1.0/1";
String authString = user + ":" + pass;
byte[] authEncBytes = Base64.encodeBase64(authString.getBytes());
String authStringEnc = new String(authEncBytes);
HttpURLConnection connection = (HttpURLConnection) new URL(url).openConnection();
connection.setRequestProperty("Authorization", "Basic " + authStringEnc);
connection.setRequestMethod("POST");
connection.setRequestProperty( "Content-Type", "application/x-www-form-urlencoded");
connection.setRequestProperty( "charset", "utf-8");
connection.setUseCaches(false);
connection.setDoOutput(true);
InputStream is = connection.getInputStream();
InputStreamReader isr = new InputStreamReader(is);
int numCharsRead;
char[] charArray = new char[1024];
StringBuffer sb = new StringBuffer();
while ((numCharsRead = isr.read(charArray)) > 0) {
sb.append(charArray, 0, numCharsRead);
}
String result = sb.toString();
System.out.print(result);
...receiving this:
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Siri version="1.0" xmlns="http://www.siri.org.uk/">
<ServiceDelivery>
<ResponseTimestamp>2015-11-08T20:33:03.574Z</ResponseTimestamp>
</ServiceDelivery>
</Siri>
How do I enter the required parameters and method?
So what I had to do was create a HttpPost and set the xml request up as an entity, binding it to the post. Here is the code, in case anyone wants to request information via HTTP POST using XML, outputting the XML as a string:
// basic autthorization security
String url = "http://nextbus.mxdata.co.uk/nextbuses/1.0/1";
String authString = "<username>:<password>";
byte[] authEncBytes = Base64.encodeBase64(authString.getBytes());
String authStringEnc = new String(authEncBytes);
DefaultHttpClient client = new DefaultHttpClient();
HttpPost post = new HttpPost(url);
post.setHeader("Authorization", "Basic " + authStringEnc);
StringEntity input = new StringEntity(request);
input.setContentType("text/xml");
post.setEntity(input);
HttpResponse response = client.execute(post);
HttpEntity entity = response.getEntity();
String unformattedXML = EntityUtils.toString(entity);
I am trying to post xml data to API using HTTP post method with credentials but a getting HTTP/1.1 400 Bad Request error .. Can anyone pl help me out ....
Here is my sample code:
BufferedReader br = new BufferedReader(new FileReader(new File("Data.xml")));
StringBuilder sb = new StringBuilder();
while((line=br.readLine())!= null){
sb.append(line.trim());
}
System.out.println("xml: "+sb);
params=sb.toString();
HttpPost request = new HttpPost("*****************url***************");
String urlaparam=URLEncoder.encode("importFormatCode:1&data:"+params,"UTF-8");
String userCredentials = "****:******";
byte[] auth = Base64.encodeBase64(userCredentials.getBytes());
StringEntity entity=new StringEntity(urlaparam);
request.addHeader("Content-type","application/x-www-form-urlencoded");
request.addHeader("Accept", "application/xml");
request.addHeader("Accept-Language", "en-US,en;q=0.5");
request.addHeader("Authorization", "Basic " + new String(auth));
request.setEntity(entity);
HttpResponse response = httpClient.execute(request);
System.out.println(response.getStatusLine());
System.out.println(request);
}
catch(Exception e)
{
}
First of all, your form parameters are not encoded correctly. You are using colon (:) to separate keys from their values, but instead, the equal sign (=) must be used:
Wrong: "importFormatCode:1&data:" + params
Correct: "importFormatCode=1&data=" + params
(See also W3C.org - Forms in HTML Documents - application/x-www-form-urlencoded)
Apart from that, you must not URL-encode the entire string but only the keys and the values. Otherwise you'll also encode the separator characters = and &!
The easiest way is to use the existing utility class org.apache.http.client.utils.URLEncodedUtils (assuming that you're using Apache HTTP Components):
String xmlData = // your xml data from somewhere
List<NameValuePair> params = Arrays.asList(
new BasicNameValuePair("importFormatCode", "1"),
new BasicNameValuePair("data", xmlData)
);
String body = URLEncodedUtils.format(params, encoding); // use encoding of request
StringEntity entity = new StringEntity(body);
// rest of your code