如何在scala中将“ python.exe”作为子进程启动？

Question

My requirement is to form an integration between scala and python. How do I start "python.exe" and executes python commands from scala.

I have tried the code snippet from following link, but it doesn't seems to be working. Could you please help me this ?

How does the “scala.sys.process” from Scala 2.9 work?

Thanks in Advance !

Answer 1

The process API is quite contrived and absolutely horribly (= not) documented. I often go back to the Java API because it makes more sense to me.

Here is an example. Overwriting the os variable to "store" the output-stream seems like an idiotic approach (in Java you can just query the output-stream from the process). Perhaps the designers of the API can enlighten us what they were thinking. Probably there is a more elegant solution:

import sys.process._

var os: java.io.OutputStream = _
val python = Process(Seq("python","-i")).run(BasicIO.standard(os = _) /* WTF? */)

def pushLine(s: String): Unit = {
  os.write(s"$s\n".getBytes("UTF-8"))
  os.flush()
}

pushLine("1+1")
pushLine("exit()")

Answer 2

I recently had to undertake a very similar project. Rather than just executing a python script and parsing the output, i needed to start up some python code and send commands to it to execute and respond back. After some time trying to do this communication over STDIN/STDOUT, I opted for using ZeroMQ as my interop communications channel

By doing this, the execution of code became as simple as

val process = s"python commandRunner.py $port".run()

where $port is the port on which my code is listening for the python code to connect back to me via ZeroMQ, while the commandRunner script runs a simple ZeroMQ event-loop