从特定链接读取流(Read stream from specific link)

从特定链接读取流(Read stream from specific link)

我需要使用OpenCV库从mjpeg读取流。 更多细节，我需要阅读http://194.126.108.66:8887/ 。 但是当我尝试用它时

VideoCapture ipCam; ipCam.open("http://194.126.108.66:8887/")

我收到错误icvOpenAvi_XINE(): Unable to initialize video driver 。

我已经使用另一个链接mjpeg测试了这段代码 - http://c-cam.uchicago.edu/mjpg/video.mjpg它运行正常。 这两个链接有什么区别？ 以及如何阅读http://194.126.108.66:8887/ ？

I need to read stream from mjpeg with OpenCV library. In more details, I need to read http://194.126.108.66:8887/. But when I try to do it with

VideoCapture ipCam; ipCam.open("http://194.126.108.66:8887/")

I get the error icvOpenAvi_XINE(): Unable to initialize video driver.

I have tested this code with another link to mjpeg - http://c-cam.uchicago.edu/mjpg/video.mjpg It works fine. What is the difference between these 2 links? And how to read http://194.126.108.66:8887/?

最满意答案

OpenCV期望其VideoCapture参数的文件扩展名，即使并非总是需要（如你的情况）。

您可以通过传入以mjpg扩展名结尾的伪参数来“欺骗”它：

ipCam.open("http://194.126.108.66:8887/?dummy=param.mjpg")

这适用于我类似的OpenCV Python案例，祝你好运！

OpenCV expects a filename extension for its VideoCapture argument, even though one isn't always necessary (like in your case).

You can "trick" it by passing in a dummy parameter which ends in the mjpg extension:

ipCam.open("http://194.126.108.66:8887/?dummy=param.mjpg")

This worked in my similar OpenCV Python case, so good luck!