As PeteBlackerThe3rd suggested you're publishing the image in your __init__ method right after the publisher is created. Here is an example class that works for your case:
import rospy
from sensor_msgs.msg import Image
import cv2
from cv_bridge import CvBridge, CvBridgeError
import time
class ImagePublisher(object):
def __init__(self):
self.node_rate = 1
self.image_pub = rospy.Publisher("image_topic", Image)
cv_image = cv2.imread('/Pictures/image.png',0)
self.bridge = CvBridge()
try:
self.image_message = self.bridge.cv2_to_imgmsg(cv_image, "passthrough")
except CvBridgeError as e:
print(e)
time.sleep(5)
self.image_pub.publish(self.image_message)
def doSmth(self):
rospy.loginfo('its working!')
# self.image_pub.publish(self.image_message)
def run(self):
loop = rospy.Rate(self.node_rate)
while not rospy.is_shutdown():
self.doSmth()
loop.sleep()
If you want to publish the image only once then use the time.sleep() in your __init__ . Otherwise if you want to publish on some rate then uncomment the publish line in doSmth().
EDIT:
WIth image_view I am encountering conversion errors due to encoding/color formats. In my case it is trying to convert image from '8UC3' to 'bgr8' so I had to change this line:
self.image_message = self.bridge.cv2_to_imgmsg(cv_image, encoding="bgr8")
Now I can see the image using:
rosrun image_view image_view image:=/image_topic
Anyway the initial problem you had was still related to publishing the image in your __init__ method right after the publisher is create.
Also note that if you are publishing it only once as in your example, make sure to run image_view before you run your node.
