ros2 pub/sub 自定义消息通过 ros2-web-bridge 到客户端应用程序

Question

我想通过 ros2-web-bridge 发布 ros2 自定义消息来响应应用程序。在反应中，我订阅了已发布的自定义消息。 ros2-web-bridge 在端口 ws://localhost:9090 上运行。我已经在ros2上创建了自定义消息接口。

   // this is client side web app subscribe method
   var example = new ROSLIB.Topic({
  ros: ros,
  name: "/sample",
  messageType: `tutorial_interfaces/msg/Num`,
});

// Subscribe a Topic
example.subscribe(function (message) {
  console.log("Subscribe data", message);
});


// This is ros2 publish python code
import rclpy
from rclpy.node import Node
from tutorial_interfaces.msg import Num    # CHANGE
class MinimalPublisher(Node):

def __init__(self):
    super().__init__('minimal_publisher')
    self.publisher_ = self.create_publisher(Num, 'sample', 10)     # CHANGE
    timer_period = 0.5
    self.timer = self.create_timer(timer_period, self.timer_callback)
    self.i = 0

// This is ros2 published class and method
def timer_callback(self):
    msg = Num()                                           # CHANGE
    msg.num = self.i                                      # CHANGE
    self.publisher_.publish(msg)
    self.get_logger().info('Publishing: "%d"' % msg.num)  # CHANGE
    self.i += 1

Answer 1

如果您已经在 React 端创建并订阅了一条自定义消息，您只需在此处包含消息标头，它将像 std_msg 一样工作。例如：如果您有包含自定义消息类型 my_msg 的包 custom_interface，您的代码将如下所示：

// This is ros2 publish python code
import rclpy
from rclpy.node import Node
from custom_interface.msg import my_msg
class MinimalPublisher(Node):

def __init__(self):
    super().__init__('minimal_publisher')
    self.publisher_ = self.create_publisher(my_msg, 'sample', 10)
    timer_period = 0.5
    self.timer = self.create_timer(timer_period, self.timer_callback)
    self.i = 0

// This is ros2 published class and method
def timer_callback(self):
    msg = my_msg()
    my_msg.custom_field = self.i
    self.publisher_.publish(my_msg)
    self.get_logger().info('Publishing: "%d"' % my_msg.custom_field)
    self.i += 1

如果网络桥已经像你说的那样设置好了，那一切都会起作用。