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py4j.protocol.Py4JNetworkError : An error occurred while trying to connect to the Java server

Tags:

java

python

py4j

I have the following simple example from the py4j document:

from py4j.java_gateway import JavaGateway

def main():

    print("Hello")
    gateway = JavaGateway()                   # connect to the JVM
    random = gateway.jvm.java.util.Random()   # create a java.util.Random instance
    number1 = random.nextInt(10)              # call the Random.nextInt method
    number2 = random.nextInt(10)
    print(number1,number2)

if __name__ == '__main__':
    main()

I tried to run it, but get the following error:

Traceback (most recent call last):
  File "/Users/edamameQ/github/myTest/main.py", line 16, in <module>
    main()
  File "/Users/edamameQ/github/myTest/main.py", line 7, in main
    random = gateway.jvm.java.util.Random()   # create a java.util.Random instance
  File "/Library/Python/2.7/site-packages/py4j/java_gateway.py", line 1188, in __getattr__
    "\n" + proto.END_COMMAND_PART)
  File "/Library/Python/2.7/site-packages/py4j/java_gateway.py", line 624, in send_command
    connection = self._get_connection()
  File "/Library/Python/2.7/site-packages/py4j/java_gateway.py", line 579, in _get_connection
    connection = self._create_connection()
  File "/Library/Python/2.7/site-packages/py4j/java_gateway.py", line 585, in _create_connection
    connection.start()
  File "/Library/Python/2.7/site-packages/py4j/java_gateway.py", line 697, in start
    raise Py4JNetworkError(msg, e)
py4j.protocol.Py4JNetworkError: An error occurred while trying to connect to the Java server

Is there anything I should do in order to connect to a Java server? Thanks!

like image 918
Edamame Avatar asked Oct 19 '22 14:10

Edamame


1 Answers

Actually I need to start the JavaGateway separately. i.e. run the following Java application first, then the python would work.

import py4j.GatewayServer;

public class myTest {

    public static void main(String[] args) {
        myTest app = new myTest();
        // app is now the gateway.entry_point
        GatewayServer server = new GatewayServer(app);
        server.start();
    }
}
like image 103
Edamame Avatar answered Oct 21 '22 04:10

Edamame