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*
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