415 error while calling post API from jersey client - java

I have below API which returns back the access_token.
POST https://idcs-xxxxxxxxxbf08128c3d93a19c.identity.c9dev2.oc9qadev.com/oauth2/v1/token
in header content-type is application/x-www-form-urlencoded. also in body it contains below parameter.
I send user name and password and it is secured with basic authentication. It provides access_token when I call from postman. also it provides output when I consume using HttpUrlConnection
url = new URL(tokenURL);
connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setRequestProperty("Authorization", auth);
connection.setRequestProperty("Accept", "application/json");
OutputStream os = connection.getOutputStream();
OutputStreamWriter osw = new OutputStreamWriter(os, "UTF-8");
osw.write("grant_type=client_credentials&scope=" + scope);
The above code is working properly. But when I use jersey it gives 415 error. I am using below code.
String user="idcs-oda-zzzxxxxxf93560b94eb8a2e2a4c9aac9a3ff-t0_APPID";
String password="xxxxxxx-6f71-4af2-b5cc-9110890d1456";
String scope = "https://idcs-oda-xxxxxxxxxxxxxxxxe2a4c9aac9a3ff-t0.data.digitalassistant.oci.oc-test.com/api/v1";
String tokenURL = "https://idcs-xxxxxxxxxxxxxxxx28c3d93a19c.identity.c9dev2.oc9qadev.com/oauth2/v1/token";
HttpAuthenticationFeature feature= HttpAuthenticationFeature
.basicBuilder()
.nonPreemptive()
.credentials(user,password)
.build();
ClientConfig clientConfig = new ClientConfig();
clientConfig.register(feature);
Client client = ClientBuilder.newClient(clientConfig);
WebTarget webTarget= client.target(tokenURL);
PostDetails post= new PostDetails("client_credentials",scope); //Bean class to assign body parameter
Response response= webTarget.request()
.header("Content-Type", "application/x-www-form-urlencoded")
.post(Entity.json(post));
System.out.println(response);
Can somebody tell me what mistake I am doing in Response line.

You need to set your Accept on the request method:
Response response= webTarget.request(MediaType.APPLICATION_JSON)
.header("Content-Type", "application/x-www-form-urlencoded")
.post(Entity.json(post));
You also need to ensure that if your API accepts application/x-www-form-urlencoded content, that is what you are sending.
Currently, you are sending application/json content based on your usage of Entity.json(post).
I don't know what type is assigned to post, but you need to figure out how to convert it either to a Form or a MultiValuedMap<String,String>, and then use the form method on Entity to submit your content.
Response response= webTarget.request(MediaType.APPLICATION_JSON)
.header("Content-Type", "application/x-www-form-urlencoded")
.post(Entity.form(postForm)); //assuming postForm typed as Form or MultiValuedMap<String,String>
Taking a guess regarding post, creating postForm as a MultiValuedMap<String,String> may be as simple as the following (which you would place prior to your request, of course).
MultiValuedMap<String,String> postForm = new MultiValuedHashMap<>();
postForm.add("client_credentials",scope);

What you need is:
Response response= webTarget.request()
.accept("application/json") // Accept field from header of request
.header("Content-Type", "application/x-www-form-urlencoded") //manually set content-tyoe
.post(Entity.entity(input, MediaType.TEXT_PLAIN)); // request body
The best way to see what is Jersey actually is sending is to register logger, and log network. For example:
clientConfig.register(
new LoggingFeature(
new Slf4jLogger(this.getClass().getName(), null)));
where Slf4jLogger is from org.apache.cxf:cxf-core.

Related

java 11 HttpClient send header then body

I want to use Java 11 HttpClient and send header first, check response and if response is OK then send the body.
How can I send header only?
this is my current code:
HttpClient httpClient = HttpClient.newBuilder()
.version(HttpClient.Version.HTTP_2)
.followRedirects(HttpClient.Redirect.NORMAL)
.connectTimeout(Duration.ofSeconds(10))
.authenticator(Authenticator.getDefault())
.build();
HttpRequest httpRequest = HttpRequest.newBuilder("someEndpoint)
.header(CONTENT_TYPE, MediaType.APPLICATION_JSON_VALUE)
.header(AUTHORIZATION, "someApiKey)
.build();
HttpResponse<String> httpResponse = httpClient.send(httpRequest, HttpResponse.BodyHandlers.ofString());
However with such httpResponse I understand I send the body.
By default, the header comes first in requests.
What you asked is, The first request with header and then with a body are two different requests. A single request can't be broken this way.
If you are talking about, Http HEAD method usage, then
The HEAD method asks for a response identical to that of a GET request, but without the response body.
The HTTP HEAD method requests the headers that would be returned if the HEAD request's URL was instead requested with the HTTP GET method. For example, if a URL might produce a large download, a HEAD request could read its Content-Length header to check the file size without actually downloading the file.
an example to use HEAD method:-
var httpClient: HttpClient = HttpClient.newBuilder()
.connectTimeout(Duration.ofSeconds(10))
.build();
var requestHead = HttpRequest.newBuilder()
.method("HEAD", HttpRequest.BodyPublishers.noBody())
.uri(URI.create("https://www.test.com"))
.build();
val httpResponse = httpClient.send(requestHead, BodyHandlers.discarding());
HttpHeaders headers = response.headers();
headers.map().forEach((key, values) -> {
System.out.printf("%s: %s%n", key, values);
});

How change state Openhab item using Jersey?

How change state Openhab item using Jersey?
Source-code:
ClientConfig clientConfig = new ClientConfig();
Client client = ClientBuilder.newClient(clientConfig);
WebTarget webTarget = client.target("http://demo.openhab.org:8080/rest");
WebTarget resourceWebTarget = webTarget.path("things");
Form form =new Form();
if (item.getState().equals("ON"))
form.param("state", "OFF");
else
form.param("state", "ON");
response =
webTarget.path("items").path("Light_GF_Corridor_Ceiling").request()
.header("Content-Type", "text/plain")
.header("Accept", "application/json")
.post(Entity.entity(form,MediaType.APPLICATION_JSON),Response.class);
System.out.println("Status Info Response " + response.getStatusInfo());
Output:
Status Info Response Unsupported Media Type
Form is not meant to be used with application/json (it's meant to be used with application/x-www-form-urlencoded). With JSON, you are either supposed to used POJOs, collections of POJOS, or Strings. If you want to send collections of POJOs, you need to wrap it in GenericEntity.

Post Request to WebService in Android

I have a problem with a WebService on Android. I am getting a 400 error but there is no information on the ErrorStream.
What I am trying to do is a POST request on a WCF Webservice using JSON.
I must add that I have includeExceptionDetailInFaults Enabled on my Service. The last time I got a 400 error, it was because I hadn't defined the RequestProperty. Now I don't get any error in the stream.
Here is the code:
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
try {
// In my last error I had not included these lines. Maybe they are still wrong?
urlConnection.setRequestProperty("Content-Type", "application/json");
urlConnection.setRequestProperty("Accept", "application/json");
urlConnection.setRequestMethod("POST");
urlConnection.setDoOutput(true);
urlConnection.setChunkedStreamingMode(0);
OutputStream out = new BufferedOutputStream(urlConnection.getOutputStream());
OutputStreamWriter outputStreamWriter = new OutputStreamWriter(out);
outputStreamWriter.write(jsonObject.toString(), 0, jsonObject.length());
outputStreamWriter.flush();
//outputStreamWriter.close();
int code = urlConnection.getResponseCode();
System.out.println(code);
if(code == 400) {
BufferedInputStream errorStream = new BufferedInputStream(urlConnection.getErrorStream());
InputStreamReader errorStreamReader = new InputStreamReader(errorStream);
BufferedReader bufferedReader = new BufferedReader(errorStreamReader);
StringBuilder builder = new StringBuilder();
String aux = "";
while ((aux = bufferedReader.readLine()) != null) {
builder.append(aux);
}
String output = builder.toString(); // The output is empty.
System.out.print(output);
}
Check Retrofit library from Square it's more easy and thin for GET/POST request and especially for REST. I suggest you to try it. It will make your life easy.
You can use different JSON parsers, error handlers, etc. Very flexible.
POST request definition using retrofit it's simple like this:
An object can be specified for use as an HTTP request body with the #Body annotation.
#POST("/users/new")
void createUser(#Body User user, Callback<User> cb);
Methods can also be declared to send form-encoded and multipart data.
Form-encoded data is sent when #FormUrlEncoded is present on the method. Each key-value pair is annotated with #Field containing the name and the object providing the value.
#FormUrlEncoded
#POST("/user/edit")
User updateUser(#Field("first_name") String first, #Field("last_name") String last);
After you define method inside your Java interface like shown above instantiate it:
RestAdapter restAdapter = new RestAdapter.Builder()
.setEndpoint("https://api.soundcloud.com")
.build();
MyInterface service = restAdapter.create(MyInterface.class);
And then you can call your method synchronously or asynchronously (in case you pass Callback instance).
service.myapi(requestBody);
See Retrofit documentation (http://square.github.io/retrofit/javadoc/index.html) and samples on GitHub for more details.
A 400 error might be occuring (and usually occurs in my case) because of incorrect URL or bad JSON format in post. please check those two
Using an HttpPost object will make your job a lot easier in my opinion
HttpPost post = new HttpPost(url);
if(payload != null){
try {
StringEntity entity = new StringEntity(payload,HTTP.UTF_8);
entity.setContentType(contentType);
post.setEntity(entity);
} catch (UnsupportedEncodingException e) {
LOG.d(TAG, "post err url : " + url);
LOG.e(TAG, "post err url" , e);
throw new Exception(1, e);
}
}
HttpResponse response=executeRequest(owner, post);

Get response Uri parameters from apache HttpResponse

I'm sending an http get/head request using Apache HttpClient 4.x. I'm sending a request with a url like "http://example.com/getAccessToken". I'm expecting the response to be a redirect url with parameters in the returned url like "http://redirecturl.com/?code=accessTokenStuff". I want to be able to parse the response redirect url parameters, i.e. I want to get "accessTokenStuff". How can I do that?
HttpClient client = new DefaultHttpClient();
HttpHead request = new HttpHead(authUrl);
HttpResponse response = client.execute(request);
System.out.println(response.getStatusLine().getStatusCode());//returns 200
request.releaseConnection();
In a nutshell: what I want is executing an original url and then getting the result which is another url that has a parameter called "code". Then I want to get the value of that parameter.
EDIT:
I also tried this but it returns the same original URL
DefaultHttpClient client = new DefaultHttpClient();
HttpParams params = client.getParams();
HttpClientParams.setRedirecting(params, false);
HttpGet request = new HttpGet(authUrl);
HttpResponse response = client.execute(request);
String location = response.getLastHeader("Location").getValue();//returns same original url
System.out.println(location);
request.releaseConnection();
Setting HttpClientParams.setRedirecting(params, true); return null
Http response do not take a form of a "redirect url". Redirect and response are both different (but related) concepts. Redirect usually means "get your response from this address instead of original one".
Having said this, you can prevent HttpClient from following a redirect, see this answer: How to prevent apache http client from following a redirect
When your HttpClient is not following the redirect, you can inspect the 'Location:' header of its response, eg:
HeaderIterator iterator = httpResponse.headerIterator("Location");
while(iterator.hasNext()) {
Header header = iterator.nextHeader();
String redirectUrl = header.getValue();
}

Http POST in Java (with file upload)

What I want to do is submit a web form from a java application. The form I need to fill out is located here: http://cando-dna-origami.org/
When the form is submitted, the server sends a confirmation email to the email address given, which for now I'm just checking by hand. I've tried filling out the form manually, and the emails get sent fine. (It should also be noted that when the form is filled out incorrectly, the page just refreshes and doesn't give any feedback).
I've never done anything with http before, but I looked around for a while, and came up with the following code, which is supposed to send a POST request to the server:
String data = "name=M+V&affiliation=Company&email="
+ URLEncoder.encode("m.v#gmail.com", "UTF-8")
+ "&axialRise=0.34&helixDiameter=2.25&axialStiffness=1100&bendingStiffness=230" +
"&torsionalStiffness=460&nickStiffness=0.01&resolution=course&jsonUpload="
+ URLEncoder.encode("C:/Users/Marjie/Downloads/twisted_DNA_bundles/monotwist.L1.v1.json",
"UTF-8") + "&type=square";
URL page = new URL("http://cando-dna-origami.org/");
HttpURLConnection con = (HttpURLConnection) page.openConnection();
con.setDoOutput(true);
con.setRequestMethod("POST");
con.connect();
OutputStreamWriter out = new OutputStreamWriter(con.getOutputStream());
out.write(data);
out.flush();
System.out.println(con.getResponseCode());
System.out.println(con.getResponseMessage());
out.close();
con.disconnect();
However, when it runs it doesn't appear to do anything - that is, I don't get any emails, although the program does print "200 OK" to System.out, which seems to indicate that something got received from the server, although I'm not sure what it means exactly. I think the problem might be in the file uploading, since I wasn't sure whether that data type required a different format.
Is this a correct way to send a POST request using Java? Do I need to do something different for the file uploading? Thanks!
After reading Adam's post, I used Apache HttpClient and wrote the following code:
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("type", "square"));
//... add more parameters
UrlEncodedFormEntity entity = new UrlEncodedFormEntity(params, HTTP.UTF_8);
HttpPost post = new HttpPost("http://cando-dna-origami.org/");
post.setEntity(entity);
HttpResponse response = new DefaultHttpClient().execute(post);
post = new HttpPost("http://cando-dna-origami.org/");
post.setEntity(new FileEntity(new File("C:/Users/Marjie/Downloads/twisted_DNA_bundles/monotwist.L1.v1.json"), "text/plain; charset=\"UTF-8\""));
HttpResponse responseTwo = new DefaultHttpClient().execute(post);
However, it still doesn't seem to be working; again, I wasn't sure how the uploaded file fit into the form, so I tried just sending two separate POST requests, one with the form and one with the other data. I am still looking for a way to combine these into one request; does anybody know something about this?
You would probably be better off using something like Apache HttpClient, with which you can build up a POST request programatically.
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://.../whatever");
List <NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("param1", "value1"));
params.add(new BasicNameValuePair("param2", "value2"));
...
httpost.setEntity(new UrlEncodedFormEntity(params, HTTP.UTF_8));
HttpResponse response = httpclient.execute(httppost);
If you need to upload a file along with your form, you will need to use a MultipartEntity instead:
MultipartEntity reqEntity = new MultipartEntity();
reqEntity.addPart("someParam", "someValue");
reqEntity.addPart("someFile", new FileBody("/some/file"));
....
httpost.setEntity(reqEntity);
There are some sample programs over on their site. The "Form based logon" and "Multipart encoded request entity" are good examples to start from.
It may also be worthwhile testing out your connections and taking a look at the underlying network data to see what is happening. Something like Firebug will let you see exactly what is happening in your browser, and you can turn up the HttpClient logging to see all of the data exchanged in your program. Alternatively, you can use something like Wireshark or Fiddler to watch your network traffic in real-time. This may give you a better idea of exactly what your browser is doing, versus what your program is doing.
As most of the suggested Java HTTP POST request code out there is not operational, I decided to give you my fully operational code that I'm sure you'll find helpful to create any Java-based POST request in the future.
This POST request is of multipart type to allow sending/uploading a file to the server.
Multipart request consist of a main header and a separator string called boundary to tell each part from the other (this separator will come in the stream with "--" (two dashes) string before it, and each part has its own small header to tell its type and some more meta-data.
My task was to create a PDF file using some online services but all the multipart POST examples just didn't do the trick...
I needed to pack an HTML document along with its pics, JS and CSS files in a ZIP/TAR file, upload it to an online html2pdf conversion service and get the result as a PDF document back to me as the response (stream) from the service.
The current service I've checked the using following code is: Htmlpdfapi.com but I'm sure that with minor adjustments you'll be able to use it with any other service.
The method call (for that service) looks something like:
[class instance name].sendPOSTRequest("http://htmlpdfapi.com/api/v1/pdf", "Token 6hr4-AmqZDrFVjAcJGykjYyXfwG1wER4", "/home/user/project/srv/files/example.zip", "result.pdf");
Here is my code that was checked and 100% works:
public void sendPOSTRequest(String url, String authData, String attachmentFilePath, String outputFilePathName)
{
String charset = "UTF-8";
File binaryFile = new File(attachmentFilePath);
String boundary = "------------------------" + Long.toHexString(System.currentTimeMillis()); // Just generate some unique random value.
String CRLF = "\r\n"; // Line separator required by multipart/form-data.
int responseCode = 0;
try
{
//Set POST general headers along with the boundary string (the seperator string of each part)
URLConnection connection = new URL(url).openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
connection.addRequestProperty("User-Agent", "CheckpaySrv/1.0.0");
connection.addRequestProperty("Accept", "*/*");
connection.addRequestProperty("Authentication", authData);
OutputStream output = connection.getOutputStream();
PrintWriter writer = new PrintWriter(new OutputStreamWriter(output, charset), true);
// Send binary file - part
// Part header
writer.append("--" + boundary).append(CRLF);
writer.append("Content-Disposition: form-data; name=\"file\"; filename=\"" + binaryFile.getName() + "\"").append(CRLF);
writer.append("Content-Type: application/octet-stream").append(CRLF);// + URLConnection.guessContentTypeFromName(binaryFile.getName())).append(CRLF);
writer.append(CRLF).flush();
// File data
Files.copy(binaryFile.toPath(), output);
output.flush();
// End of multipart/form-data.
writer.append(CRLF).append("--" + boundary + "--").flush();
responseCode = ((HttpURLConnection) connection).getResponseCode();
if(responseCode !=200) //We operate only on HTTP code 200
return;
InputStream Instream = ((HttpURLConnection) connection).getInputStream();
// Write PDF file
BufferedInputStream BISin = new BufferedInputStream(Instream);
FileOutputStream FOSfile = new FileOutputStream(outputFilePathName);
BufferedOutputStream out = new BufferedOutputStream(FOSfile);
int i;
while ((i = BISin.read()) != -1) {
out.write(i);
}
// Cleanup
out.flush();
out.close();
}
catch(Exception e)
{
e.printStackTrace();
}
}
I am currently writing a small web server, and I tested your request client. My server is receiving the following request:
User-Agent: Java/1.6.0_20
Host: localhost:1700
Accept: text/html, image/gif, image/jpeg, *; q=.2, */*; q=.2
Connection: keep-alive
Content-type: application/x-www-form-urlencoded
Content-Length: 287
name=M+V&affiliation=Company&email=m.v%40gmail.com&axialRise=0.34&helixDiameter=2.25&axialStiffness=1100&bendingStiffness=230&torsionalStiffness=460&nickStiffness=0.01&resolution=course&jsonUpload=C%3A%2FUsers%2FMarjie%2FDownloads%2Ftwisted_DNA_bundles%2Fmonotwist.L1.v1.json&type=square
You should check the format of the POST data you are sending, most probably it is not processed by the server as you would expect.
You should definitively use apaches HTTPClient for that job! It makes life much easier. Here is an example how to upload a file with apaches HttpClient.
byte[] data = outStream.toByteArray()
HttpClient client = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://localhost:8080/YourResource");
ByteArrayBody byteArrayBody = new ByteArrayBody(data, "application/json", "some.json");
MultipartEntity multipartEntity = new MultipartEntity();
multipartEntity.addPart("upload", byteArrayBody);
httpPost.setEntity( multipartEntity );
HttpResponse response = client.execute(httpPost);
Reader reader = new InputStreamReader(response.getEntity().getContent());
Let me know if you have further questions.
Here's an example I got working that uses apache httpclient. Also, don't forget to add these dependencies:
<dependency>
<groupId>org.apache.httpcomponents</groupId>
<artifactId>httpclient</artifactId>
<version>4.4.1</version>
</dependency>
<dependency>
<groupId>org.apache.httpcomponents</groupId>
<artifactId>httpmime</artifactId>
<version>4.4.1</version>
</dependency>
The code:
HttpClient httpclient = HttpClientBuilder.create().build();
HttpPost httppost = new HttpPost(DataSources.TORRENT_UPLOAD_URL);
MultipartEntityBuilder builder = MultipartEntityBuilder.create();
builder.addPart("a_field_name", new FileBody(torrentFile));
HttpEntity entity = builder.build();
httppost.setEntity(entity);
HttpResponse response = httpclient.execute(httppost);

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