Access IP camera with OpenCV

Question

Can't access the video stream. Can any one please help me to get the video stream. I have searched in google for the solution and post another question in stack overflow but unfortunately nothing can't solve the problem.

import cv2
cap = cv2.VideoCapture()
cap.open('http://192.168.4.133:80/videostream.cgi?user=admin&pwd=admin')
while(cap.isOpened()):
    ret, frame = cap.read()
    cv2.imshow('frame', frame)
    if cv2.waitKey(1) & 0xFF == ord('q'):
        break
cap.release()
cv2.destroyAllWindows()

Answer 1

You can use urllib to read frames from video stream.

import cv2
import urllib
import numpy as np

stream = urllib.urlopen('http://192.168.100.128:5000/video_feed')
bytes = ''
while True:
    bytes += stream.read(1024)
    a = bytes.find(b'\xff\xd8')
    b = bytes.find(b'\xff\xd9')
    if a != -1 and b != -1:
        jpg = bytes[a:b+2]
        bytes = bytes[b+2:]
        img = cv2.imdecode(np.fromstring(jpg, dtype=np.uint8), cv2.IMREAD_COLOR)
        cv2.imshow('Video', img)
        if cv2.waitKey(1) == 27:
            exit(0)

Check this out if you want to stream video from webcam of your pc. https://github.com/shehzi-khan/video-streaming

Answer 2

Thank You. May be, now urlopen is not under utllib. It is under urllib.request.urlopen.I use this code:

import cv2
from urllib.request import urlopen
import numpy as np

stream = urlopen('http://192.168.4.133:80/video_feed')
bytes = ''
while True:
    bytes += stream.read(1024)
    a = bytes.find(b'\xff\xd8')
    b = bytes.find(b'\xff\xd9')
    if a != -1 and b != -1:
        jpg = bytes[a:b+2]
        bytes = bytes[b+2:]
        img = cv2.imdecode(np.fromstring(jpg, dtype=np.uint8), cv2.IMREAD_COLOR)
        cv2.imshow('Video', img)
        if cv2.waitKey(1) == 27:
            exit(0)

Answer 3

You can use this code to get live video feeds in browser.

for accessing camera other than your laptop's webcam, you can use RTSP link like this

rtsp://admin:12345@192.168.1.1:554/h264/ch1/main/av_stream"

where

  username:admin password:12345 your camera ip address and port ch1 is first camera on that DVR 

replace cv2.VideoCamera(0) with this link like this for your camera and it will work

camera.py

import cv2

class VideoCamera(object):
    def __init__(self):
        # Using OpenCV to capture from device 0. If you have trouble capturing
        # from a webcam, comment the line below out and use a video file
        # instead.
        self.video = cv2.VideoCapture(0)
        # If you decide to use video.mp4, you must have this file in the folder
        # as the main.py.
        # self.video = cv2.VideoCapture('video.mp4')

    def __del__(self):
        self.video.release()

    def get_frame(self):
        success, image = self.video.read()
        # We are using Motion JPEG, but OpenCV defaults to capture raw images,
        # so we must encode it into JPEG in order to correctly display the
        # video stream.
        ret, jpeg = cv2.imencode('.jpg', image)
        return jpeg.tobytes()

main.py

from flask import Flask, render_template, Response
from camera import VideoCamera

app = Flask(__name__)

@app.route('/')
def index():
    return render_template('index.html')

def gen(camera):
    while True:
        frame = camera.get_frame()
        yield (b'--frame\r\n'
               b'Content-Type: image/jpeg\r\n\r\n' + frame + b'\r\n\r\n')

@app.route('/video_feed')
def video_feed():
    return Response(gen(VideoCamera()),
                    mimetype='multipart/x-mixed-replace; boundary=frame')

if __name__ == '__main__':
    app.run(host='0.0.0.0', debug=True)

then you can follow this blog to increase your FPS of video stream

Answer 4

You can Use RTSP instead of direct video feed.

Every IP Camera have RTSP to Stream Live Video.

So you can use RTSP Link instead of videofeed

Answer 5

Use code below to access ipcam directly through opencv. Replace the url in VideoCapture with your particular camera rtsp url. The one given generally works for most cameras I've used.

import cv2

cap = cv2.VideoCapture("rtsp://[username]:[pass]@[ip address]/media/video1")

while True:
    ret, image = cap.read()
    cv2.imshow("Test", image)
    if cv2.waitKey(1) & 0xFF == ord('q'):
        break
cv2.destroyAllWindows()

Answer 6

If using python 3, you will probably need to use a bytearray instead of a string. (modifying the current top answer)

with urllib.request.urlopen('http://192.168.100.128:5000/video_feed') as stream:

    bytes = bytearray()

    while True:
        bytes += stream.read(1024)
        a = bytes.find(b'\xff\xd8')
        b = bytes.find(b'\xff\xd9')
        if a != -1 and b != -1:
            jpg = bytes[a:b+2]
            bytes = bytes[b+2:]
            img = cv2.imdecode(np.frombuffer(jpg, dtype=np.uint8), cv2.IMREAD_COLOR)
            cv2.imshow('Video', img)
            if cv2.waitKey(1) & 0xFF == ord('q'):
                break

cv2.destroyAllWindows()