我想从php文件中获取视频url。但是当我尝试运行html文件时，视频播放错误显示
以下是我的html代码：
头
<head>
<link href="https://vjs.zencdn.net/7.11.4/video-js.css" rel="stylesheet" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script src="https://vjs.zencdn.net/7.11.4/video.min.js"></script>
<script src="https://vjs.zencdn.net/ie8/1.1.2/videojs-ie8.min.js"></script></head>身体
<div style="display: block;" id="resultMsg"></div>
<div style="display: block;" id="resultMsgs"></div>
<dev id="vPlayer"></dev>php代码示例
<?Php
$url = ("video.url");
echo = $url;
?>javascript
<Script>
var vUrl = $("#resultMsg").load("https://funbees.cf/test.php");
setTimeout(function (){
var v1Url = vUrl;
var x = document.createElement("video");
x.setAttribute("class", "video-js");
x.setAttribute("id", "my_player_1");
x.setAttribute("controls", "controls");
x.setAttribute("height", "100%");
x.setAttribute("weight", "100%");
document.getElementById("vPlayer").innerHTML = x.outerHTML
//IMPORTANT//
var myPlayer = videojs("my_player_1");
//IMPORTANT//
var url = v1Url;
// var url = "https://web.funbees.cf/movie.mp4";
myPlayer.src({type: "video/mp4", src: url});
myPlayer.ready(function() {
console.log(myPlayer.isFullscreen());
myPlayer.isFullscreen(true);
console.log(myPlayer.isFullscreen());
});
videojs("my_player_1", {
autoplay: false,
preload: "auto"
});
}, 5000);
<Script>如何解决此问题。并成功获取视频url以播放视频。
暂无答案！
目前还没有任何答案，快来回答吧！