ROS2 サービス通信でサーバの処理が終わるまで待機したい

やりたいこと

サービス通信で，クライアントがトリガーを投げたら，それを受け取ったサーバが処理を実行し終わるまで待ちたい

方法

spin_until_future_complete　を使う
サービスタイプはTrigerで実装した

service_server.py

import rclpy
from rclpy.node import Node
from std_srvs.srv import Trigger #service type
 
import time
 
 
class ServiceServer(Node):
 
    def __init__(self):
        super().__init__('service_server')
        self.srv = self.create_service(Trigger, 'service_name', self.handle_request)
        self.get_logger().info('server is ready')
 
    def handle_request(self, request, response):
        self.get_logger().info('executing hundle_request')
        
        '''
        実行したい処理
        例
        time.sleep(10)
        '''
 
        response.success = True
        response.message = "finished"
        self.get_logger().info('finished hundle_request')
 
        return response
 
 
def main(args=None):
    rclpy.init(args=args)
    server = ServiceServer()
    rclpy.spin(server)
    server.destroy_node()
    rclpy.shutdown()
    
 
 
if __name__ == '__main__':
    main()

service_client.py

from std_srvs.srv import Trigger #service type
import rclpy
from rclpy.node import Node
 
 
class ServiceClient(Node):
 
    def __init__(self):
        super().__init__('service_client')
        self.cli = self.create_client(Trigger, 'service_name')
        while not self.cli.wait_for_service(timeout_sec=1.0):
            self.get_logger().info('service not available, waiting again...')
        self.req = Trigger.Request()
 
    def send_request(self):
        self.get_logger().info('send a request')
        self.future = self.cli.call_async(self.req)
        rclpy.spin_until_future_complete(self, self.future)
        return self.future.result()
 
 
 
def main(args=None):
    rclpy.init(args=args)
 
    client = ServiceClient()
    response = client.send_request()
    #サーバの処理が終わるまで待機
    client.get_logger().info(response.message)
 
    client.destroy_node()
    rclpy.shutdown()
 
 
if __name__ == '__main__':
    main()