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zmq version for video file

master
Lennart Heimbs 2 years ago
parent
commit
41ef2c4cb6

+ 15
- 6
camera/video_stream/zmq_video_client.py View File

@@ -1,11 +1,20 @@
# run this program on each RPi to send a labelled image stream
import socket
import time
from imutils.video import VideoStream
import imagezmq

sender = imagezmq.ImageSender(connect_to='tcp://jeff-macbook:5555')

import argparse
import socket
import time
# construct the argument parser and parse the arguments
ap = argparse.ArgumentParser()
ap.add_argument("-s", "--server-ip", required=True,
help="ip address of the server to which the client will connect")
args = vars(ap.parse_args())
# initialize the ImageSender object with the socket address of the
# server
sender = imagezmq.ImageSender(connect_to="tcp://{}:5555".format(
args["server_ip"]))
rpi_name = socket.gethostname() # send RPi hostname with each image
picam = VideoStream(usePiCamera=True).start()
time.sleep(2.0) # allow camera sensor to warm up

+ 23
- 0
camera/video_stream/zmq_video_client_file.py View File

@@ -0,0 +1,23 @@
from imutils.video import FileVideoStream
import imagezmq
import argparse
import socket
import time
# construct the argument parser and parse the arguments
ap = argparse.ArgumentParser()
ap.add_argument("-s", "--server-ip", required=True,
help="ip address of the server to which the client will connect")
args = vars(ap.parse_args())
# initialize the ImageSender object with the socket address of the
# server
sender = imagezmq.ImageSender(connect_to="tcp://{}:5555".format(
args["server_ip"]))

rpi_name = socket.gethostname() # send RPi hostname with each image
video_file = FileVideoStream("../run.mp4").start()
time.sleep(2.0) # allow camera sensor to warm up
while video_file.more():
image = video_file.read()
sender.send_image(rpi_name, image)

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