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Adding value to path parameter in Java REST?

NOTICE UPDATE!!
The problem got solved and i added my own answer in the thread
In short, I have attempted to add the parameter "scan_id" value but since it is a POST i can't add the value directly in the url path.
using the code i already have, how would i go about modifying or adding so that the url is correct, that is, so that it accepts my POST?.
somehow i have been unable to find any examples that have helped me in figuring out how i would go about doing this..
I know how to do a POST with a payload, a GET with params. but a post with Params is very confusing to me.
Appreciate any help. (i'd like to continue using HttpUrlConnection unless an other example is provided that also tells me how to send the request and not only configuring the path.
I've tried adding it to the payload.
I've tried UriBuilder but found it confusing and in contrast with the rest of my code, so wanted to ask for help with HttpUrlConnection.
URL url = new URL("http://localhost/scans/{scan_id}/launch");
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setRequestMethod("POST");
con.setRequestProperty("tmp_value_dont_mind_this", "432432");
con.setRequestProperty("X-Cookie", "token=" + "43432");
con.setRequestProperty("X-ApiKeys", "accessKey="+"43234;" + " secretKey="+"43234;");
con.setDoInput(true);
con.setDoOutput(true); //NOT NEEDED FOR GETS
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
//First example of writing (works when writing a payload)
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
//second attemp at writing, doens't work (wanted to replace {scan_id} in the url)
DataOutputStream writer = new DataOutputStream(con.getOutputStream());
writer.writeChars("scan_id=42324"); //tried writing directly
//writer.write(payload);
writer.close();
Exception:
Exception in thread "main" java.io.IOException: Server returned HTTP response code: 400 for URL: http://localhost/scans/launch
I'd like one of the three response codes because then i know the Url is correct:
200 Returned if the scan was successfully launched.
403 Returned if the scan is disabled.
404 Returned if the scan does not exist.
I've tried several urls
localhost/scans/launch,
localhost/scans//launch,
localhost/scans/?/launch,
localhost/scans/{scan_id}/launch,

So with the help of a friend and everyone here i solved my problem.
The below code is all the code in an entire class explained bit by bit. at the bottom you have the full class with all its syntax etc, that takes parameters and returns a string.
in a HTTP request there are certain sections.
Such sections include in my case, Request headers, parameters in the Url and a Payload.
depending on the API certain variables required by the API need to go into their respective category.
My ORIGINAL URL looked like this: "http://host:port/scans/{scan_id}/export?{history_id}"
I CHANGED to: "https://host:port/scans/" + scan_Id + "/export?history_id=" + ID;
and the API i am calling required an argument in the payload called "format" with a value.
String payload = "{\"format\" : \"csv\"}";
So with my new URL i opened a connection and set the request headers i needed to set.
HttpsURLConnection con = (HttpsURLConnection) url.openConnection();
The setDoOutput should be commented out when making a GET request.
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
con.setRequestProperty("X-Cookie", "token=" + token);
con.setRequestProperty("X-ApiKeys", "accessKey="+"23243;" +"secretKey="+"45543;");
Here i write to the payload.
//WRITING THE PAYLOAD to the http call
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
After i've written the payload i read whatever response i get back (this depends on the call, when i do a file download (GET Request) i don't have a response to read as i've already read the response through another piece of code).
I hope this helps anyone who might encounter this thread.
public String requestScan(int scan_Id, String token, String ID) throws MalformedInputException, ProtocolException, IOException {
try {
String endpoint = "https://host:port/scans/" + scan_Id + "/export?history_id=" ID;
URL url = new URL(endpoint);
String payload= "{\"format\" : \"csv\"}";
HttpsURLConnection con = (HttpsURLConnection) url.openConnection();
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
con.setRequestProperty("X-Cookie", "token=" + token);
con.setRequestProperty("X-ApiKeys", "accessKey="+"324324;" +
"secretKey="+"43242;");
//WRITING THE PAYLOAD to the http call
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
//READING RESPONSE
BufferedReader br = new BufferedReader(new InputStreamReader(con.getInputStream()));
StringBuffer jsonString = new StringBuffer();
String line;
while ((line = br.readLine()) != null) {
jsonString.append(line);
}
br.close();
con.disconnect();
return jsonString.toString();
} catch (Exception e) {
throw new RuntimeException(e.getMessage());
}
}

As discussed here the solution would be to change the content type to application/x-www-form-urlencoded, but since you are already using application/json; charset=UTF-8 (which I am assuming is a requirement of your project) you have no choise to redesign the whole thing. I suggest you one of the following:
Add another GET service;
Add another POST service with content type application/x-www-form-urlencoded;
Replace this service with one of the above.
Do not specify the content type at all so the client will accept anything. (Don't know if possible in java)
If there are another solutions I'm not aware of, I don't know how much they would be compliant to HTTP protocol.
(More info)
Hope I helped!

Why you are not using like this. Since you need to do a POST with HttpURLConnection, you need to write the parameters to the connection after you have opened the connection.
String urlParameters = "scan_id=42324";
byte[] postData = urlParameters.getBytes(StandardCharsets.UTF_8);
DataOutputStream dataOutputStream = new DataOutputStream(conn.getOutputStream());
dataOutputStream.write(postData);
Or if you have launch in the end, just change the above code to the following,
String urlParameters = "42324/launch";
byte[] postData = urlParameters.getBytes(StandardCharsets.UTF_8);
DataOutputStream dataOutputStream = new DataOutputStream(conn.getOutputStream());
dataOutputStream.write(postData);

URL url = new URL("http://localhost/scans/{scan_id}/launch");
That line looks odd to me; it seems you are trying to use a URL where you are intending the behavior of a URI Template.
The exact syntax will depend on which template implementation you choose; an implementation using the Spring libraries might look like:
import org.springframework.web.util.UriTemplate;
import java.net.url;
// Warning - UNTESTED code ahead
UriTemplate template = new UriTemplate("http://localhost/scans/{scan_id}/launch");
Map<String,String> uriVariables = Collections.singletonMap("scan_id", "42324");
URI uri = template.expand(uriVariables);
URL url = uri.toURL();

Why HttpURLConnection does not send data unless I try to receive something

I cannot comprehend why doesn't the following code does not put a packet onto wire (confirmed via wireshark). It is a fairly standard method of sending an HTTP POST request, as I believe. I don't intend to read anything just POST.
private void sendRequest() throws IOException {
String params = "param=value";
URL url = new URL(otherUrl.toString());
HttpURLConnection con = (HttpURLConnection)url.openConnection();
con.setDoOutput(true);
con.setDoInput(true); //setting this to `false` does not help
con.setRequestMethod("POST");
con.setRequestProperty("Content-Type", "text/plain");
con.setRequestProperty("Content-Length", "" + Integer.toString(params.getBytes().length));
con.setRequestProperty("Accept", "text/plain");
con.setUseCaches(false);
con.connect();
DataOutputStream wr = new DataOutputStream(con.getOutputStream());
wr.writeBytes(params);
wr.flush();
wr.close();
//Logger.getLogger("log").info("URL: "+url+", response: "+con.getResponseCode());
con.disconnect();
}
What happens is... actually nothing, unless I try to read anything. For example by uncommenting the above log line which reads the response code. Trying to read a response via con.getInputStream(); also works. There is no movement of packets. When I uncomment the getResponseCode, I can see that http POST is sent, and then 200 OK is sent back. The order is proper. I.e. I don't get some wild response before sending POST. Everything else looks exactly the same (I can attach wireshark screenshots if needed.). In the debugger the code executes (i.e. does not block anywhere).
I don't understand under what circumstances this can be happening. I belive it should be possible, to send a POST request with con.setDoInput(false);. Currently it doesn't send anything or fails (when trying to execute con.getResponseCode()) with an exception because I obviously promised I won't read anything.
It might be relevant, that before sendRequest I do request some data from the same site, but I trust I close everything properly. I.e:
public static String getData(String urlAddress) throws MalformedURLException, IOException {
URL url = new URL(urlAddress);
HttpURLConnection con = (HttpURLConnection)url.openConnection();
con.setDoOutput(false);
InputStream in = con.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
StringBuilder data = new StringBuilder();
String line;
while((line = reader.readLine()) != null) {
data.append(line);
}
reader.close();
in.close();
con.getResponseCode();
con.disconnect();
return data.toString();
}
The server for url in both cases is the same, port also, so I believe it is possible to use the same socket for communication. The above code works and retrieves the data properly.
I am not sure, maybe I don't clean something, and it gets cached, so with out an explicit read the POST gets delayed. There is no other traffic on the socket.

Unless you're using fixed-length or chunked transfer mode, HttpURLConnection will buffer all your output until you call getInputStream() or getResponseCode(), so that it can send a correct Content-length header.
If you call getResponseCode() you should have a look at its value.

Simulate URL entering on java

So I have a problem where if I type this link on the browser and hit enter, an activation happens. I just want to do the same through Java. I don't need any kind of response from the URL. It should just do the same as entering the URL on a browser. Currently my code doesn't throw an error, but I don't think its working because the activation is not happening. My code:
public static void enableMachine(String dns){
try {
String req= "http://"+dns+"/username?username=sputtasw";
URL url = new URL(req);
URLConnection connection = url.openConnection();
connection.connect();
/*BufferedReader br = new BufferedReader(new InputStreamReader(url.openStream()));
String strTemp = "";
while (null != (strTemp = br.readLine())) {
System.out.println(strTemp);
}*/
} catch (Exception ex) {
ex.printStackTrace();
}
}
What's the problem?

If you want to do that with an URLConnection, it isn't sufficient to just open the connection with connect, you also have to send e.g. an HTTP request etc.
That said, i think it would be easier, if you use an HTTP client like the one from Apache HttpComponents (http://hc.apache.org/). Just do a GET request with the HTTP client, this would be the same as visiting the page with a browser (those clients usually also supports redirection etc.).

You may use HttpUrlConnectionClass to do the job:
URL url = new URL("http://my.url.com");
HttpURLConnection httpCon = (HttpURLConnection) url.openConnection();
httpCon.setRequestProperty("Content-Type", "application/json");
httpCon.setDoOutput(true);
httpCon.setRequestMethod("POST");
String params = "foo=42&bar=buzz";
DataOutputStream wr = new DataOutputStream(httpCon.getOutputStream());
wr.writeBytes(params);
wr.flush();
wr.close();
httpCon.connect();
int responseCode = httpCon.getResponseCode();
You may as well use "GET" request method and just append parameters to the url.

How to post a JSON object to an URL?

I'm trying to post a JSON to a web service so that I can get a JSON in response as return, I have searched in Google but most of the response I found is for Android but not for core Java, this question is for a Swing application I'll give the code the code I used below.
Connection class
public class Connection extends Thread {
private String url1;
private JSONObject data;
String line;
//Constuctor to initialize the variables.
public Connection(String url1, JSONObject data) {
this.url1 = url1;
this.data = data;
start();
}
public void run() {
ConnectionReaderWriter();
}
//To fetch the data from the input stream
public String getResult() {
return line;
}
public String ConnectionReaderWriter() {
URL url;
HttpURLConnection connection = null;
ObjectOutputStream out;
try {
/*URL url = new URL(Url.server_url + url1); //Creating the URL.
URLConnection conn = url.openConnection(); //Opening the connection.
conn.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(data); //Posting the data to the ouput stream.
wr.flush();
BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
line=rd.readLine(); //Reading the data from the input stream.
wr.close();
rd.close();*/
url = new URL(Url.server_url + url1); //Creating the URL.
connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "application/json");
connection.setRequestProperty("Accept", "application/json");
connection.setRequestProperty("api_key", "123456");
connection.setUseCaches(false);
connection.setDoInput(true);
connection.setDoOutput(true);
//Send request
out = new ObjectOutputStream(connection.getOutputStream());
out.writeObject(data);
out.flush();
out.close();
} catch (MalformedURLException ex) {
Logger.getLogger(Connection.class.getName()).log(Level.SEVERE, null, ex);
String nonet = "No Network Connection";
line = nonet;
} catch (IOException ex) {
Logger.getLogger(Connection.class.getName()).log(Level.SEVERE, null, ex);
String nonet = "No Server Connection";
line = nonet;
}
return line; //Return te stream recived from the input stream.
}
}
The commented code is the one I used before when I was passing as text encoded to the URL. The function call is given below
JSONObject json = new JSONObject();
json.put("username", username);
json.put("password", passwordenc);
Connection conn = new Connection(Url.login, json);
conn.join();
On execution I get the exception shown below
Jan 20, 2014 1:18:32 PM SupportingClass.Connection ConnectionReaderWriter
SEVERE: null
java.io.NotSerializableException: org.json.JSONObject
at java.io.ObjectOutputStream.writeObject0(ObjectOutputStream.java:1180)
at java.io.ObjectOutputStream.writeObject(ObjectOutputStream.java:346)
at SupportingClass.Connection.ConnectionReaderWriter(Connection.java:74)
at SupportingClass.Connection.run(Connection.java:40)
Please tell me the problem in this code or an alternative to this method.

I'll give the answer:
replace out.writeObject(data); with out.write(data.toString().getBytes());
You were trying to write the JSONObject object, and the writeObject() method will attempt to serialize the object, but it will fail since the JSONObject class does not implement Serializable.

JSon is text, so you cannot use an ObjectOutputStream. The POST method uses the first line of the content as the arguments, so you would need a blank line before the actual content:
OutputStream stream = connection.getOutputStream();
stream.write('\r');
stream.write('\n');
out = new OutputStreamWriter(stream, "UTF-8");
out.write(data.toString());
out.flush();
out.close();
EDIT: actually we need CR LF, so println() may not work.

If you find it awkward to use HttpURLConnection natively, you might use an abstraction library. There are many powerful out there. One of them is DavidWebb. You can find a long list of alternatives at the end of that page.
With this library your code would be shorter and more readable:
JSONObject nameAndPassword = new JSONObject();
// set name and password
Webb webb = Webb.create();
JSONObject result = webb.post("your_serverUrl")
.header("api_key", "123456")
.useCaches(false)
.body(nameAndPassword)
.ensureSuccess()
.asJsonObject()
.getBody();
The code shows only the part that would run in your run() method of the Thread. It is not necessary to set the Content-Type header, because this is done automatically by detecting the type of object you set as the body. The same is true for the Accept header.
I assumed that you receive a JSON Object from your REST service, since you set the "Accept" header to "application/json". For receiving a plain String one could write String result = ... .asString().getBody().
BTW the library has been developed with Android in mind, but it can be used with Swing or server-side as well. Maybe you choose another library (e.g. Jersey Client), because you have no restriction about the size. DavidWebb weighs about 20 kilobytes while most other libraries add several hundred kilobytes to your deployment artifacts.
Another thing: you use JSONObject. For small web services, this is not a problem, but if you have to marshal many and/or big objects, you could think about using JAXB + Jackson or Gson. Less code, less errors, more fun!

FileNotFoundException while getting the InputStream object from HttpURLConnection

I am trying to send a post request to a url using HttpURLConnection (for using cUrl in java).
The content of the request is xml and at the end point, the application processes the xml and stores a record to the database and then sends back a response in form of xml string. The app is hosted on apache-tomcat locally.
When I execute this code from the terminal, a row gets added to the db as expected. But an exception is thrown as follows while getting the InputStream from the connection
java.io.FileNotFoundException: http://localhost:8080/myapp/service/generate
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1401)
at org.kodeplay.helloworld.HttpCurl.main(HttpCurl.java:30)
Here is the code
public class HttpCurl {
public static void main(String [] args) {
HttpURLConnection con;
try {
con = (HttpURLConnection) new URL("http://localhost:8080/myapp/service/generate").openConnection();
con.setRequestMethod("POST");
con.setDoOutput(true);
con.setDoInput(true);
File xmlFile = new File("test.xml");
String xml = ReadWriteTextFile.getContents(xmlFile);
con.getOutputStream().write(xml.getBytes("UTF-8"));
InputStream response = con.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(response));
for (String line ; (line = reader.readLine()) != null;) {
System.out.println(line);
}
reader.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}
Its confusing because the exception is traced to the line InputStream response = con.getInputStream(); and there doesn't seem to be any file involved for a FileNotFoundException.
When I try to open a connection to an xml file directly, it doesn't throw this exception.
The service app uses spring framework and Jaxb2Marshaller to create the response xml.
The class ReadWriteTextFile is taken from here
Thanks.
Edit:
Well it saves the data in the DB and sends back a 404 response status code at the same time.
I also tried doing a curl using php and print out the CURLINFO_HTTP_CODE which turns out to be 200.
Any ideas on how do I go about debugging this ? Both service and client are on the local server.
Resolved:
I could solve the problem after referring to an answer on SO itself.
It seems HttpURLConnection always returns 404 response when connecting to a url with a non standard port.
Adding these lines solved it
con.setRequestProperty("User-Agent","Mozilla/5.0 ( compatible ) ");
con.setRequestProperty("Accept","*/*");

I don't know about your Spring/JAXB combination, but the average REST webservice won't return a response body on POST/PUT, just a response status. You'd like to determine it instead of the body.
Replace
InputStream response = con.getInputStream();
by
int status = con.getResponseCode();
All available status codes and their meaning are available in the HTTP spec, as linked before. The webservice itself should also come along with some documentation which overviews all status codes supported by the webservice and their special meaning, if any.
If the status starts with 4nn or 5nn, you'd like to use getErrorStream() instead to read the response body which may contain the error details.
InputStream error = con.getErrorStream();

FileNotFound is just an unfortunate exception used to indicate that the web server returned a 404.

To anyone with this problem in the future, the reason is because the status code was a 404 (or in my case was a 500). It appears the InpuStream function will throw an error when the status code is not 200.
In my case I control my own server and was returning a 500 status code to indicate an error occurred. Despite me also sending a body with a string message detailing the error, the inputstream threw an error regardless of the body being completely readable.
If you control your server I suppose this can be handled by sending yourself a 200 status code and then handling whatever the string error response was.

For anybody else stumbling over this, the same happened to me while trying to send a SOAP request header to a SOAP service. The issue was a wrong order in the code, I requested the input stream first before sending the XML body. In the code snipped below, the line InputStream in = conn.getInputStream(); came immediately after ByteArrayOutputStream out = new ByteArrayOutputStream(); which is the incorrect order of things.
ByteArrayOutputStream out = new ByteArrayOutputStream();
// send SOAP request as part of HTTP body
byte[] data = request.getHttpBody().getBytes("UTF-8");
conn.getOutputStream().write(data);
if (conn.getResponseCode() != HttpURLConnection.HTTP_OK) {
Log.d(TAG, "http response code is " + conn.getResponseCode());
return null;
}
InputStream in = conn.getInputStream();
FileNotFound in this case was an unfortunate way to encode HTTP response code 400.

FileNotFound in this case means you got a 404 from your server - could it be that the server does not like "POST" requests?

FileNotFound in this case means you got a 404 from your server
You Have to Set the Request Content-Type Header Parameter
Set “content-type” request header to “application/json” to send the request content in JSON form.
This parameter has to be set to send the request body in JSON format.
Failing to do so, the server returns HTTP status code “400-bad request”.
con.setRequestProperty("Content-Type", "application/json; utf-8");
Full Script ->
public class SendDeviceDetails extends AsyncTask<String, Void, String> {
#Override
protected String doInBackground(String... params) {
String data = "";
String url = "";
HttpURLConnection con = null;
try {
// From the above URL object,
// we can invoke the openConnection method to get the HttpURLConnection object.
// We can't instantiate HttpURLConnection directly, as it's an abstract class:
con = (HttpURLConnection)new URL(url).openConnection();
//To send a POST request, we'll have to set the request method property to POST:
con.setRequestMethod("POST");
// Set the Request Content-Type Header Parameter
// Set “content-type” request header to “application/json” to send the request content in JSON form.
// This parameter has to be set to send the request body in JSON format.
//Failing to do so, the server returns HTTP status code “400-bad request”.
con.setRequestProperty("Content-Type", "application/json; utf-8");
//Set Response Format Type
//Set the “Accept” request header to “application/json” to read the response in the desired format:
con.setRequestProperty("Accept", "application/json");
//To send request content, let's enable the URLConnection object's doOutput property to true.
//Otherwise, we'll not be able to write content to the connection output stream:
con.setDoOutput(true);
//JSON String need to be constructed for the specific resource.
//We may construct complex JSON using any third-party JSON libraries such as jackson or org.json
String jsonInputString = params[0];
try(OutputStream os = con.getOutputStream()){
byte[] input = jsonInputString.getBytes("utf-8");
os.write(input, 0, input.length);
}
int code = con.getResponseCode();
System.out.println(code);
//Get the input stream to read the response content.
// Remember to use try-with-resources to close the response stream automatically.
try(BufferedReader br = new BufferedReader(new InputStreamReader(con.getInputStream(), "utf-8"))){
StringBuilder response = new StringBuilder();
String responseLine = null;
while ((responseLine = br.readLine()) != null) {
response.append(responseLine.trim());
}
System.out.println(response.toString());
}
} catch (Exception e) {
e.printStackTrace();
} finally {
if (con != null) {
con.disconnect();
}
}
return data;
}
#Override
protected void onPostExecute(String result) {
super.onPostExecute(result);
Log.e("TAG", result); // this is expecting a response code to be sent from your server upon receiving the POST data
}
and call it
new SendDeviceDetails().execute("");
you can find more details in this tutorial
https://www.baeldung.com/httpurlconnection-post

The solution:
just change localhost for the IP of your PC
if you want to know this: Windows+r > cmd > ipconfig
example: http://192.168.0.107/directory/service/program.php?action=sendSomething
just replace 192.168.0.107 for your own IP (don't try 127.0.0.1 because it's same as localhost)

Please change
con = (HttpURLConnection) new URL("http://localhost:8080/myapp/service/generate").openConnection();
To
con = (HttpURLConnection) new URL("http://YOUR_IP:8080/myapp/service/generate").openConnection();