如何通过ROSTOPIC发布包含python或xml文件等文件的文件夹？

# 在ROS工作空间中创建一个新的消息类型
cd ~/catkin_ws/src
catkin_create_pkg file_publisher roscpp std_msgs
cd file_publisher
mkdir msg
echo "string python_file_content" > msg/FileContent.msg
echo "string xml_file_content" >> msg/FileContent.msg

cd ~/catkin_ws
catkin_make
source devel/setup.bash

#!/usr/bin/env python

import rospy
from file_publisher.msg import FileContent

def publish_file_content():
    pub = rospy.Publisher('file_content_topic', FileContent, queue_size=10)
    rospy.init_node('file_content_publisher', anonymous=True)
    rate = rospy.Rate(10)  # 10 Hz

    while not rospy.is_shutdown():
        # 读取文件内容
        with open('path_to_your_python_file.py', 'r') as f:
            python_content = f.read()
        with open('path_to_your_xml_file.xml', 'r') as f:
            xml_content = f.read()

        # 创建消息并发布
        msg = FileContent()
        msg.python_file_content = python_content
        msg.xml_file_content = xml_content
        pub.publish(msg)

        rate.sleep()

if __name__ == '__main__':
    try:
        publish_file_content()
    except rospy.ROSInterruptException:
        pass

rosrun file_publisher file_content_publisher.py

#!/usr/bin/env python

import rospy
from file_publisher.msg import FileContent

def callback(msg):
    rospy.loginfo("Received Python file content:")
    rospy.loginfo(msg.python_file_content)
    rospy.loginfo("Received XML file content:")
    rospy.loginfo(msg.xml_file_content)

def subscribe_file_content():
    rospy.init_node('file_content_subscriber', anonymous=True)
    rospy.Subscriber('file_content_topic', FileContent, callback)
    rospy.spin()

if __name__ == '__main__':
    try:
        subscribe_file_content()
    except rospy.ROSInterruptException:
        pass

rosrun file_publisher file_content_subscriber.py