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Hi, I am current using flask to run a python web application that currently just allows...

Hi, I am current using flask to run a python web application that currently just allows users to access a video camera on a localhost. As of now, I need to type python main.py to actually run the project and turn on the localhost. What I want to do is, be able to click a button in HTML, through google chrome, and just run the webcam like that. To sum it up, I want to click a simple button and have the main.py file run and open up a new window that starts the webcam.

CODE:

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='localhost', debug=True)

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()

index.html

<html>
<head>
<title>Video Streaming Demonstration</title>
</head>
<body>
<h1>Video Streaming Demonstration</h1>
<img id="bg" src="{{ url_for('video_feed') }}">
</body>
</html>

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Homework Answers

Answer #1

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='localhost', debug=True)

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()

index.html

<html>
<head>
<title>Video Streaming Demonstration</title>
</head>
<body>
<h1>Video Streaming Demonstration</h1>
<img id="bg" src="{{ url_for('video_feed') }}">

<form action="main.py"

<input type="button" value="web cam">
</form>
</body>
</html>

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