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What should be url value for the paramter in TestRailService class constuctor?


I am trying to integrate testrail in my automation framework.
I have got url , username, password from testrail team.
I am using the following code

String username="";
String password="fdhfksdjdflskfljskfbjckljkf";

String projectName=”CATS PROJECT2”;
URL url= new URL("");

TestRailService testRailService =null;
       testRailService =new TestRailService(url,username,password);

Project project = testRailService.getProjectByName(projectName);  // last line

When I try to run the code
I get this error in last line

Exception in thread “main” java.lang.IllegalArgumentException: Illegal character(s) in message header value: Basic dmt1bWFyMTM5QHNhcGllbnQuY29tOnNFUm9vdEZxNVhxUWlXeEZNV2dN

the url is am giving is the one i got from testrail during the time of registration.

I think i am giving wrong url parameter , can you please suggest me what should be url parameter for testrailservice class ??


Hi Vatsal,

Thank you for your post. I believe you may have submitted a support ticket for this as well. Is this correct?


yes , i have submitted support ticket but I didn’t get the proper answer for that , so i Updated the same here.
Plz help me out


Hi Vatsal,

Thanks for your reply! When specifying your TestRail address, you wouldn’t include the ‘index.php’, and you can just use the base address (e.g. “”). We also have a basic Java binding that you can use to access TestRail’s API, and you can learn more about this on our website here:

Hope this helps!



hey I already tried with “”, but it is not accepting it. giving the same exception , as I mentioned before. Plz provide some other solution to achieve the same.
I found that there is one more way like-
TestRailService testRailService =new TestRailService(clientId, username, password);
But I don’t understand what is clientid,
how can I get the clientId ??

How to get clientId in testrail?

I tried with that too , but that did not work.
There is one other way like
TestRailService testRailService =new TestRailService(clientId, username, password);
How to find the clientid?


Hi Vatsal,

Thank you for the update. Could you please take a screenshot of your project so we can see that you have imported the TestRail JAVA binding correctly? The usage of this is pretty straight forward.

import com.gurock.testrail.APIClient;
import com.gurock.testrail.APIException;
import java.util.Map;
import java.util.HashMap;
import org.json.simple.JSONObject;
public class Program
public static void main(String[] args) throws Exception
APIClient client = new APIClient(“http:///testrail/”);

This is the example we provide along with the binding. Keep in mind that there is a dependency on JSON.simple that needs to be added as well: