我是 Django 的新手,现在正在尝试制作一个应用程序,它可以同时通过 rtsp 访问来自多个 IP 摄像机的实时视频流。
首先,我用 12 个 IP 摄像机尝试了以下代码,一切正常。
import cv2
import threading
class camThread(threading.Thread):
def __init__(self, previewName, camID):
threading.Thread.__init__(self)
self.previewName = previewName
self.camID = camID
def run(self):
print("Starting " + self.previewName)
camPreview(self.previewName, self.camID)
def camPreview(previewName, camID):
cv2.namedWindow(previewName)
cam = cv2.VideoCapture(camID)
if cam.isOpened():
rval, frame = cam.read()
else:
rval = False
while rval:
cv2.imshow(previewName, frame)
rval, frame = cam.read()
key = cv2.waitKey(20)
if key == 27:
break
cv2.destroyWindow(previewName)
thread = camThread('Cam', 'rtsp://admin:pass@1.2.3.4/axis-media/media.amp?videocodec=h264&resolution=320x240'')
thread.start()
但是当我尝试像这样在 Django 中使用它时:
模型.py
class VideoCamera(threading.Thread):
def __init__(self, cam_ip):
threading.Thread.__init__(self)
self.cam_ip_str = str(cam_ip)
url = 'rtsp://admin:pass@' + self.cam_ip_str + '/axis-media/media.amp?videocodec=h264&resolution=320x240'
self.video = cv2.VideoCapture(url)
print("Starting " + self.cam_ip_str)
def __del__(self):
self.video.release()
def get_frame(self):
ret, image = self.video.read()
ret, jpeg = cv2.imencode('.jpg', image)
return jpeg.tobytes()
视图.py
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')
def cam_index(request, cam_name):
camera = Cameras.objects.get(camera_name = cam_name)
video_thread = gen(VideoCamera(cam_ip=camera.IP_adress))
stream = StreamingHttpResponse(video_thread, content_type="multipart/x- mixed-replace;boundary=frame")
return stream
然后我只是将我的流粘贴到带有<img src='...'>标签的 HTML 文件中。一切正常,但我不能同时看到超过 6 个流,我至少需要 12 个。
或者这种方法可能不好,我应该使用 ffmpeg 或 gstreamer 之类的东西来达到我的目的?