Handle multiple Tasks to a single Server

[Norbert Forgacs]

Hello guys, I am new to RPYC and I have the following problem and I don't know how to implement:

I have a server:

class MyService(rpyc.Service):

    def exposed_DoSomethingThatTakesTime(self):

          ... code that takes time to execute...

          ...

          ...

          return 1   # return value so the client will know that the service didn't finish

And I have the same client, running on multiple computers:

proxy = rpyc.connect('TMD4701G', 18862, config={'allow_public_attrs': True})

if proxy.root.DoSomethingThatTakesTime == 1:

      print(Service finished running for this computer)

Now my problem is, and what I wan't to achieve is that the if a Client already called the exposed Service, and the Service is running, the second Client which also calls the Service, should wait for the first one to finish --> Something like a task handling for the services exposed in the Server:

example:

Client 1 calls DoSomethingThatTakesTime()                                    Server executes for Client1

Client 2 calls DoSomethingThatTakesTime()                                    Waits for Server to finish executing to Client1

Client 3 calls DoSomethingThatTakesTime()                                    Waits for Server to finish executing to Client1 and Client2