-1

I don't have source code of application, I just downloaded an application from play store.

Is there any way to do load testing of users.

My scenario is 100,000 users using the App simultaneously.

  • 2
    What do you mean load testing the app? The local app on the phone is only going to have ~1 user at a time, do you mean load testing of the services (if any) that the app communicates with? – jonrsharpe Jul 16 at 8:55
  • I mean I don't have a source code of app, means I don't know that what's the url of APIs and server. In that case, Is there any tool which is providing an environment to test app with multiple users? I need to test the app that 100,000 users logged in at a time and app still working fine. if yes than increase the no. of logged in users to check when app will starting to crash. – Sameed Ahmed Siddiqui Jul 17 at 7:03
  • Why don't you ask where those services are? Or if you're just randomly testing the app without any connection to the people who made it, then... why? – jonrsharpe Jul 17 at 7:05
  • team has left and not providing anything and app is live now. just confirming that is there any way to test the app. – Sameed Ahmed Siddiqui Jul 17 at 7:16
  • In that case is it a good idea to be doing this at all? Think this through; if you knock it over, what are you going to do about it? This is less of a test and more of a potential DDoS. – jonrsharpe Jul 17 at 7:18
0

If your goal is to load test the backend of the mobile application you need to simulate the application's network traffic, to wit:

  1. Capture the network calls which are being made by the application
  2. Use JMeter Samplers or Plugins to mimic it from JMeter
  3. Add 100 000 virtual users in the Thread Group (at this stage most probably you will have to go for Distributed Testing)

If the application uses HTTP protocol for communicating with the backend you can even record its network activity using HTTP(S) Test Script Recorder, you will need to configure your device to use JMeter as the proxy

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.