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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();

Upload file on server using Rest API

I want to commit text file "demo2.txt" to bitbucket server using rest API. I can upload the same file using Postman but it's not working with Java code. As shown in the below code I want to send string object "str" as the body. Can someone help me here to upload the file on the bitbucket server? Also Please let me know if there is any other way to do this.
URL url = new URL("https://api.bitbucket.org/2.0/repositories/{team name}/{repository name}/src");
HttpURLConnection httpCon = (HttpURLConnection) url.openConnection();
httpCon.setRequestProperty("X-Requested-with", "Curl");
httpCon.setDoOutput(true);
httpCon.setDoInput(true);
httpCon.setRequestProperty("Connection", "Keep-Alive");
httpCon.setRequestProperty("Content-Type", "multipart/form-data; boundary="+boundary);
httpCon.setRequestProperty("Accept", "application/x-www-form-urlencoded");
httpCon.setRequestProperty("Authorization", basicauth);
httpCon.setRequestMethod("POST");
String str =
"{"
+ "\"-F\":\"File3=#/D:/log/demo2.txt\" "
+ "}";
try {
OutputStream output = httpCon.getOutputStream();
output.write(str.getBytes());
output.close();
} catch(Exception e){
System.out.println(e.getMessage());
}
int responseCode = httpCon.getResponseCode();
String inputLine;
StringBuffer response = new StringBuffer();
if (responseCode == HttpURLConnection.HTTP_OK || responseCode == HttpURLConnection.HTTP_CREATED){
BufferedReader in = new BufferedReader(new .
InputStreamReader(httpCon.getInputStream()));
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();
List<String> message = new ArrayList<>();
message.add(response.toString());
}

If this is all of your code, then your problem may be as simple as the fact that you're not making any sort of call to finalize the request...to tell HttpURLConnection that you're done forming the request and want it to complete. There are two things you can do to help this:
close the output stream when you're done writing to it. You're generally supposed to do this. Here, you can call output.close(). Better still, since you have a try/catch block already anyway, use a "try with resources" construct to make sure that the stream is closed no matter what happens (assuming you're using a newer version of Java that supports this).
make some sort of call to query the response to the request. It may
be that the request is not being fully sent until you do this. Try
calling httpCon.getResponseCode() at the bottom of your code.
Given that you have provided no information as to what "it's not working with Java code" means, this may be useful information but not the ultimate solution to your problem. Your code does look good other than exhibiting these omissions.

Java GET request with HttpURLConnection not working as expected

I am using core java to retrieve single web page content as String using proxy.
private static HttpURLConnection getConnection(String urlStr) throws MalformedURLException, IOException {
URL url = new URL(urlStr);
Proxy proxy = new Proxy(Proxy.Type.HTTP, new InetSocketAddress("216.56.48.118", 9000));
HttpURLConnection uc = (HttpURLConnection) url.openConnection(proxy);
return uc;
}
public static void printHomePageContent() throws Exception {
String queryParam = URLEncoder.encode(
"YBuD64pgMuP3hiqNmR4hB6H8xdAebeBPfdEPbdNUq3ptkbhSYGkdhuKaPCXE+KXT3unjfaI3tRJzQno10f/FiC7IzNAdhbrPK9d4smyxHpE=",
"UTF-8");
HttpURLConnection conn = getConnection(
"https://www.example.com?params=" + queryParam);
conn.setRequestMethod("GET");
conn.setReadTimeout(60 * 1000);
conn.setRequestProperty("Accept-Charset", "UTF-8");
conn.connect();
Reader in = new BufferedReader(new InputStreamReader(conn.getInputStream(), "UTF-8"));
for (int c; (c = in.read()) >= 0;) {
System.out.print((char) c);
}
}
NOTE: www.example.com is replaced with the actual url as I can't share it in public, all other things are exactly same as my original code.
PROBLEM: When I call printHomePageContent it's printing wrong page(same page if I don't send query param) content that means it's not considering params query parameter value as expected. While if If I hit the same URL on browser or POSTMAN(Rest Client), it's displaying right page. I am using proxy from browser as well using chrome extension.
I know you can't replicate the issue your own as I have replaced the URL, but the description I wrote is exactly what is happening. If anybody can suggest some hints based on their past experienced would be helpful.
Thanks in advance.

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.

Connect to web that requires user/password

I'm a bit new to Java and more to connections stuff with it. I'm trying to create a program to connect to a website ("www.buybackprofesional.com") where I would like to download pictures and get some text from cars (after the login you have to enter a plate number to access a car's file).
This is what I have right now, but it always says that the session has expired, I need a way to login using the username and password of the mainpage, am I right? can someone give me some advice? Thanks
Note: I want to do it in Java, maybe I was not clear in the question.
//URL web = new URL("http://www.buybackprofesional.com/DetallePeri.asp?mat=9073FCV&fec=27/07/2010&tipo=C&modelo=4582&Foto=0");
URL web = new URL("http://www.buybackprofesional.com/");
HttpURLConnection con = (HttpURLConnection) web.openConnection();
con.setRequestMethod("GET");
con.setRequestProperty("User-Agent", "Mozilla/4.0 (compatible; JVM)");
con.setRequestProperty("Pragma", "no-cache");
con.connect();
BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream()));
String line = null;
while ((line = reader.readLine()) != null) {
System.out.println(line);
}
A colleage helped me with this so I'll post the code that works:
public static URLConnection login(String _url, String _username, String _password) throws IOException, MalformedURLException {
String data = URLEncoder.encode("Usuario", "UTF-8") + "=" + URLEncoder.encode(_username, "UTF-8");
data += "&" + URLEncoder.encode("Contrase", "UTF-8") + "=" + URLEncoder.encode(_password, "UTF-8");
// Send data
URL url = new URL(_url);
URLConnection conn = url.openConnection();
conn.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(data);
wr.flush();
wr.close();
return conn;
}
This will submit the form info on the page I need and after that, using cookies I can stay connected!

To connect to a website using java consider using httpunit or httpcore (offered by apache). They handle sessions much better then you (or I) could do on your own.
Edit: Fixed the location of the link. Thanks for the correction!