How to parse mjpeg http stream from ip camera?

Question

Given below is the code written for getting live stream from an IP Camera.

from cv2 import *
from cv2 import cv
import urllib
import numpy as np
k=0
capture=         


        

Answer 1

import cv2
import urllib 
import numpy as np

stream = urllib.urlopen('http://localhost:8080/frame.mjpg')
bytes = ''
while True:
    bytes += stream.read(1024)
    a = bytes.find('\xff\xd8')
    b = bytes.find('\xff\xd9')
    if a != -1 and b != -1:
        jpg = bytes[a:b+2]
        bytes = bytes[b+2:]
        i = cv2.imdecode(np.fromstring(jpg, dtype=np.uint8), cv2.CV_LOAD_IMAGE_COLOR)
        cv2.imshow('i', i)
        if cv2.waitKey(1) == 27:
            exit(0)   

edit (explanation)

I just saw that you mention that you have c++ code that is working, if that is the case your camera may work in python as well. The code above manually parses the mjpeg stream without relying on opencv, since in some of my projects the url will not be opened by opencv no matter what I did(c++,python).

Mjpeg over http is multipart/x-mixed-replace with boundary frame info and jpeg data is just sent in binary. So you don't really need to care about http protocol headers. All jpeg frames start with marker 0xff 0xd8 and end with 0xff 0xd9. So the code above extracts such frames from the http stream and decodes them one by one. like below.

...(http)
0xff 0xd8      --|
[jpeg data]      |--this part is extracted and decoded
0xff 0xd9      --|
...(http)
0xff 0xd8      --|
[jpeg data]      |--this part is extracted and decoded
0xff 0xd9      --|
...(http)

edit 2 (reading from mjpg file)

Regarding your question of saving the file, yes the file can be directly saved and reopened using the same method with very small modification. For example you would do curl http://IPCAM > output.mjpg and then change the line stream=urllib.urlopen('http://localhost:8080/frame.mjpg')so that the code becomes this

import cv2
import urllib 
import numpy as np

stream = open('output.mjpg', 'rb')
bytes = ''
while True:
    bytes += stream.read(1024)
    a = bytes.find('\xff\xd8')
    b = bytes.find('\xff\xd9')
    if a != -1 and b != -1:
        jpg = bytes[a:b+2]
        bytes = bytes[b+2:]
        i = cv2.imdecode(np.fromstring(jpg, dtype=np.uint8), cv2.CV_LOAD_IMAGE_COLOR)
        cv2.imshow('i', i)
        if cv2.waitKey(1) == 27:
            exit(0)   

Of course you are saving a lot of redundant http headers, which you might want to strip away. Or if you have extra cpu power, maybe just encode to h264 first. But if the camera is adding some meta data to http header frames such as channel, timestamp, etc. Then it may be useful to keep them.

edit 3 (tkinter interfacing)

import cv2
import urllib 
import numpy as np
import Tkinter
from PIL import Image, ImageTk
import threading

root = Tkinter.Tk()
image_label = Tkinter.Label(root)  
image_label.pack()

def cvloop():    
    stream=open('output.mjpg', 'rb')
    bytes = ''
    while True:
        bytes += stream.read(1024)
        a = bytes.find('\xff\xd8')
        b = bytes.find('\xff\xd9')
        if a != -1 and b != -1:
            jpg = bytes[a:b+2]
            bytes = bytes[b+2:]
            i = cv2.imdecode(np.fromstring(jpg, dtype=np.uint8), cv2.CV_LOAD_IMAGE_COLOR)            
            tki = ImageTk.PhotoImage(Image.fromarray(cv2.cvtColor(i, cv2.COLOR_BGR2RGB)))
            image_label.configure(image=tki)                
            image_label._backbuffer_ = tki  #avoid flicker caused by premature gc
            cv2.imshow('i', i)
        if cv2.waitKey(1) == 27:
            exit(0)  

thread = threading.Thread(target=cvloop)
thread.start()
root.mainloop()

Answer 2

I don't think the first anwser is fine with other format image data, eg png. So I write the following code, which can handle other type of images

"""
MJPEG format

Content-Type: multipart/x-mixed-replace; boundary=--BoundaryString
--BoundaryString
Content-type: image/jpg
Content-Length: 12390

... image-data here ...


--BoundaryString
Content-type: image/jpg
Content-Length: 12390

... image-data here ...
"""
import io
import requests
import cv2
import numpy as np


class MjpegReader():
    def __init__(self, url: str):
        self._url = url

    def iter_content(self):
        """
        Raises:
            RuntimeError
        """
        r = requests.get(self._url, stream=True)

        # parse boundary
        content_type = r.headers['content-type']
        index = content_type.rfind("boundary=")
        assert index != 1
        boundary = content_type[index+len("boundary="):] + "\r\n"
        boundary = boundary.encode('utf-8')

        rd = io.BufferedReader(r.raw)
        while True:
            self._skip_to_boundary(rd, boundary)
            length = self._parse_length(rd)
            yield rd.read(length)

    def _parse_length(self, rd) -> int:
        length = 0
        while True:
            line = rd.readline()
            if line == b'\r\n':
                return length
            if line.startswith(b"Content-Length"):
                length = int(line.decode('utf-8').split(": ")[1])
                assert length > 0


    def _skip_to_boundary(self, rd, boundary: bytes):
        for _ in range(10):
            if boundary in rd.readline():
                break
        else:
            raise RuntimeError("Boundary not detected:", boundary)

mr = MjpegReader("http://127.0.0.1/mjpeg.cgi")
for content in mr.iter_content():
    i = cv2.imdecode(np.frombuffer(content, dtype=np.uint8), cv2.IMREAD_COLOR)
    cv2.imshow('i', i)
    if cv2.waitKey(1) == 27:
        break

Answer 3

First of all, please be aware that you should first try simply using OpenCV's video capture functions directly, e.g. cv2.VideoCapture('http://localhost:8080/frame.mjpg')!

This works just fine for me:

import cv2
cap = cv2.VideoCapture('http://localhost:8080/frame.mjpg')

while True:
  ret, frame = cap.read()
  cv2.imshow('Video', frame)

  if cv2.waitKey(1) == 27:
    exit(0)

Anyways, here is Zaw Lin's solution ported to OpenCV 3 (only change is cv2.CV_LOAD_IMAGE_COLOR to cv2.IMREAD_COLOR and Python 3 (string vs byte handling changed plus urllib):

import cv2
import urllib.request
import numpy as np

stream = urllib.request.urlopen('http://localhost:8080/frame.mjpg')
bytes = 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:]
        i = cv2.imdecode(np.fromstring(jpg, dtype=np.uint8), cv2.IMREAD_COLOR)
        cv2.imshow('i', i)
        if cv2.waitKey(1) == 27:
            exit(0)

Answer 4

I had the same problem. The solution without requests or urllib: just add the user and password in the cam address, using VideoCapture, like this:

E.g.

cv2.VideoCapture('http://user:password@XXX.XXX.XXX.XXX/video')

using IPWebcam for android.

Answer 5

Here is an answer using the Python 3 requests module instead of urllib.

The reason for not using urllib is that it cannot correctly interpret a URL like http://user:pass@ipaddress:port

Adding authentication parameters is more complex in urllib than the requests module.

Here is a nice, concise solution using the requests module:

import cv2
import requests
import numpy as np

r = requests.get('http://192.168.1.xx/mjpeg.cgi', auth=('user', 'password'), stream=True)
if(r.status_code == 200):
    bytes = bytes()
    for chunk in r.iter_content(chunk_size=1024):
        bytes += chunk
        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:]
            i = cv2.imdecode(np.fromstring(jpg, dtype=np.uint8), cv2.IMREAD_COLOR)
            cv2.imshow('i', i)
            if cv2.waitKey(1) == 27:
                exit(0)
else:
    print("Received unexpected status code {}".format(r.status_code))