php fopen:身份验证错误:访问被拒绝,身份验证协议错误

php fopen:身份验证错误:访问被拒绝,身份验证协议错误,php,authentication,fopen,mjpeg,Php,Authentication,Fopen,Mjpeg,我喜欢用php读取一个mjpeg流,从流中分割出一个jpg图像。该流将从我的网络摄像头创建。网络摄像头有密码保护 这是我的代码: function mjpeg2jpg_convert(){ $camurl="http://admin:secretpass@192.168.1.60/Streaming/Channels/1/httppreview"; $boundary="\n--"; $f = fopen($camurl,"r") ; if(!$f) { //

我喜欢用php读取一个mjpeg流,从流中分割出一个jpg图像。该流将从我的网络摄像头创建。网络摄像头有密码保护

这是我的代码:

function mjpeg2jpg_convert(){
$camurl="http://admin:secretpass@192.168.1.60/Streaming/Channels/1/httppreview";

$boundary="\n--";

$f = fopen($camurl,"r") ;


   if(!$f)
   {
        //**** cannot open
        echo "error";
   }else{
    $cnt = 0;    
    //**** URL OK
    $r = "";
    $l = "";
         while (substr_count($r,"Content-Length") != 2 && $cnt<10000){
            $l.=fread($f,512);
            echo $l;
            $r.=$l;
            $cnt++;
         }

         $start = strpos($r,'ÿ');
         $end   = strpos($r,$boundary,$start)-1;
         $frame = substr("$r",$start,$end - $start);
echo "FRAME";
         //header("Content-type: image/jpeg");
         echo $frame;
   }
fclose($f);
}
函数mjpeg2jpg_convert(){ $camurl=”http://admin:secretpass@192.168.1.60/Streaming/Channels/1/httppreview”; $boundary=“\n--”; $f=fopen($camurl,“r”); 如果(!$f) { //****打不开 回声“错误”; }否则{ $cnt=0; //****URL OK $r=“”; $l=“”;
虽然(substr_count($r,“Content Length”)!=2&&$cntI也有同样的问题,但您找到了解决方案吗?