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JSON in java by POST getting None return

I request POST by this code
URL url = new URL("adress/discordnotifi");
HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
httpURLConnection.setReadTimeout(5000);
httpURLConnection.setConnectTimeout(5000);
httpURLConnection.setRequestMethod("POST");
httpURLConnection.connect();
JSONObject object = new JSONObject();
myPWord = ((EditText) (findViewById(R.id.edit_Id))).getText().toString();
object.put("token", tokens);
object.put("discordid", mydiscord);
OutputStream outputStream = httpURLConnection.getOutputStream();
outputStream.write(object.toString().getBytes("UTF-8"));
Log.d("debug",object.toString());
outputStream.flush();
outputStream.close();
and my tokens and mydiscord is the information that I want to send as JSON format like
{"token":"tokens","discordid":"mydiscord"}
and from python flask
#app.route('/discordnotifi', methods=['POST'])
def post():
content = request.json
print(content)
Discordid = int(content["discordid"])
token = content['token']
"""Discordid = request.form.get("discordid")
token = request.form.get("token")"""
print(Discordid, token)
return ("Thx")
at print(content) I get None I really don't know whts wrong here. I was planning to send Json with information but I getting None from Json.

The very least you're missing is enabling output on the connection. Add:
httpURLConnection.setDoOutput(true);
The server may also require that you set some headers, for example content-type to tell it that you are sending JSON.
httpURLConnection.setRequestProperty("Content-Type", "application/json; utf-8");
If you can use Java 11, consider using the new HttpClient class instead of HttpUrlConnection. It simplifies creating correct requests.

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();

Protocol message contained an invalid tag 0

My goal is to create a realtime service alert feed and send it over to a server that I made in Java with an HTTP post request. The first step I did was to create a copy of the example alert feed posted here and it seems I was successfully able to do that as I was able to print it out the message. https://developers.google.com/transit/gtfs-realtime/examples/alerts
The next step that I did is to create an HTTP connection and send the feed over with the POST request. This is what I have in my client code and example here is the feed name.
String url = "https://localhost:8080";
URL obj = new URL(url);
HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();
//add reuqest header
con.setRequestMethod("POST");
con.setRequestProperty("User-Agent", "Mozilla/5.0");
con.setRequestProperty("Accept-Language", "en-US,en;q=0.5");
con.setRequestProperty("Content-Type", "application/x-protobuf");
con.setDoOutput(true);
DataOutputStream wr = new DataOutputStream(con.getOutputStream());
example.build().writeDelimitedTo(wr);
wr.flush();
wr.close();
My server code is simply this so far.
ServerSocket server = new ServerSocket(8080);
System.out.println("Listening for connection on port 8080 ....");
while (true) {
try (Socket socket = server.accept()) {
FeedMessage feed = FeedMessage.parseDelimitedFrom(socket.getInputStream());
Date today = new Date();
String httpResponse = "HTTP/1.1 200 OK\r\n\r\n" + today;
socket.getOutputStream().write(httpResponse.getBytes("UTF-8"));
}
}
The question here is that I get the Protocol message contained an invalid tag 0 on the server side. I would like some help on trying to resolve this issue. Maybe I am not parsing it correctly.
Update #2
I have tried to parse the HTTP headers to get to the payload like comments have said. But my code hangs and the output to print the headers on the terminal looks serialized.
DataInputStream isr = new DataInputStream(socket.getInputStream());
Reader reader = new InputStreamReader(isr);
BufferedReader reader2 = new BufferedReader(reader);
String line = reader2.readLine();
System.out.println("get lines");
while (!line.isEmpty()) {
System.out.println(line);
line = reader2.readLine();
}

You are dealing with raw sockets at the server, but the payload is encoded in an http request body. You are going to need to parse the http request through to the payload, and when you have just the body : send that to protobuf. Right now, you're sending the http headers to protobuf, which doesn't make sense. That could mean parsing through to \r\n\r\n, but it would help if you could make use of the content-length header, and even better if you can use a pre-built http library.

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.