2015-09-07 11:05
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I am trying to prepare a webcam streaming page. I created a vlc streaming and here is what I wrote in the command line to run my webcam.

cvlc -vvv v4l2:///dev/video0 --sout '#transcode{vcodec=mp2v,vb=800,acodec=none}:rtp{dst=,port=5004,mux=ts}'

After typing this code i can see my webcam by typing


to the browser. Its okay up to here.

I prepared a php streaming file and it opens static video files with


command and it works properly. But when I pass "rtp://"/ in

 fopen( 'rtp://', "rb" )

command, I get en error 502 gateway that probably means it has not opened rtp file.

What should I do ? Thanks

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我正在尝试准备网络摄像头流媒体页面。 我创建了一个 vlc 流媒体,这是我在命令行中编写的用于运行我的网络摄像头的内容。

  cvlc -vvv v4l2:/// dev  / video0 --sout'#transcode {vcodec = mp2v,vb = 800,acodec = none}:rtp {dst =,port = 5004,mux = ts}'
   \  n 


  rtp:// / 

到浏览器。 它还可以到这里。


  fopen('localhost / sample  .mp4','rb')

命令,它可以正常工作。 但是当我传递“rtp://”/ in

  fopen('rtp:// /',“rb”)  


我应该怎么做 ? 谢谢

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1条回答 默认 最新

  • drc4925
    drc4925 2015-09-07 11:19

    PHP can only open resources using some protocols.

    file:// — Accessing local filesystem 
    http:// — Accessing HTTP(s) URLs  
    ftp:// — Accessing FTP(s) URLs
    php:// — Accessing various I/O streams
    zlib:// — Compression Streams
    data:// — Data (RFC 2397) 
    glob:// —    Find pathnames matching pattern
    phar:// — PHP Archive
    ssh2:// —    Secure Shell 2
    rar:// — RAR
    ogg:// — Audio streams
    expect:// —    Process Interaction Streams

    As you see rtp is not one of theese. you need to find/write rtp wrapper to reed this resource.

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