Adding value to path parameter in Java REST?

Question

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,

Answer 1

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());
    }
}

Answer 2

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!

Answer 3

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);

Answer 4

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();