I am using the video tag to play videos.
我正在使用视频标签来播放视频。
I using php files for the playback like this:
我使用php文件进行回放,如下所示:
<video id="playvideo" preload="auto" width="845" height="395"
poster="http://video-js.zencoder.com/oceans-clip.png">
<source src="../getvideo_webm.php" type='video/webm' />
<source src="../getvideo_mp4.php" type='video/mp4'/>
<source src="../getvideo_ogv.php" type='video/ogg' />
</video>
All .php files are playing fine when i check them directly in the brower. But the above setup with all .php as source files will not play. If i give a straight .mp4 source it will play fine.
当我直接在浏览器中检查它们时,所有的.php文件都运行良好。但是上面所有.php作为源文件的设置不能运行。如果我给出一个直的。mp4源代码,它将会运行良好。
The getvideo_mp4.php looks like this:
getvideo_mp4。php是这样的:
$path = 'oceans-clip.mp4';
if (file_exists($path))
{
$size=filesize($path);
$fm=@fopen($path,'rb');
if(!$fm) {
// You can also redirect here
header ("HTTP/1.0 404 Not Found");
die();
}
$begin=0;
$end=$size;
if(isset($_SERVER['HTTP_RANGE'])) {
if(preg_match('/bytes=\h*(\d+)-(\d*)[\D.*]?/i',
$_SERVER['HTTP_RANGE'],$matches)){
$begin=intval($matches[0]);
if(!empty($matches[1])) {
$end=intval($matches[1]);
}
}
}
if($begin>0||$end<$size)
header('HTTP/1.0 206 Partial Content');
else
header('HTTP/1.0 200 OK');
header("Content-Type: video/mp4");
header('Accept-Ranges: bytes');
header('Content-Length:'.($end-$begin));
header("Content-Disposition: inline;");
header("Content-Range: bytes $begin-$end/$size");
header("Content-Transfer-Encoding: binary\n");
header('Connection: close');
$cur=$begin;
fseek($fm,$begin,0);
while(!feof($fm)&&$cur<$end&&(connection_status()==0))
{ print fread($fm,min(1024*16,$end-$cur));
$cur+=1024*16;
usleep(1000);
}
die();
}
So what am i doing wrong ?
我做错了什么?
3 个解决方案
#1
3
The above code is working. After i changed the src url for .php files, it did actually work. Now it plays in moz, ie, chrome with only php files as sources in the video tag.
上面的代码正在工作。在我为。php文件修改了src url之后,它确实工作了。现在它在moz(即chrome)中运行,只有php文件作为视频标签的源文件。
#2
0
You will have to echo the path after retrieving it and pass it to the 'src' attribute of the Video Tag of HTML5. Your Current Strategy won't work well, i hope...
获取路径后,必须对路径进行回送,并将其传递给HTML5视频标记的“src”属性。我希望你目前的策略不会奏效。
For example,
例如,
<source src="<?php echo getMp4VideoUrl(); ?>" type='video/mp4'/>
#3
0
The browser identifies the video content from the header sent to it with the request. Just manipulate the header and keep the PHP extension. It will work perfect
浏览器从发送给它的请求头中识别视频内容。只需要操作header并保留PHP扩展名。它将工作完美
#1
3
The above code is working. After i changed the src url for .php files, it did actually work. Now it plays in moz, ie, chrome with only php files as sources in the video tag.
上面的代码正在工作。在我为。php文件修改了src url之后,它确实工作了。现在它在moz(即chrome)中运行,只有php文件作为视频标签的源文件。
#2
0
You will have to echo the path after retrieving it and pass it to the 'src' attribute of the Video Tag of HTML5. Your Current Strategy won't work well, i hope...
获取路径后,必须对路径进行回送,并将其传递给HTML5视频标记的“src”属性。我希望你目前的策略不会奏效。
For example,
例如,
<source src="<?php echo getMp4VideoUrl(); ?>" type='video/mp4'/>
#3
0
The browser identifies the video content from the header sent to it with the request. Just manipulate the header and keep the PHP extension. It will work perfect
浏览器从发送给它的请求头中识别视频内容。只需要操作header并保留PHP扩展名。它将工作完美