<html>
<head>
<link rel="stylesheet" type="text/css" href="video.css">
</head>
<body>
<div id="banner" class="headerContent">
<div id="header">
<div id="headerContent" class="headerContent">
<video autoplay id="awsome_video" >
<source src="http://derrylahan.comze.com/promo.mp4" type="video/mp4">
</video>
<div class="overlay">
<h1 id="rowdy"> ROWDY RONDY ROUSEY </h1>
<a href="test.html" id="enter">ENTER</a>
</div>
</div>
</div>
</body>
</html>
此代码适用于另一个mp4视频但不是这个想法吗?
提前感谢任何帮助工作正常离线但不在服务器即时通讯使用000webhost免费托管
当我在chrome上打开inspect元素并查看网络时,我可以看到加载的内容然后显示“无法写入临时文件:在处理操作时找不到请求的文件或目录”。显示在屏幕底部
也从inspect元素
获取此信息资源解释为文档但使用MIME类型video / mp4传输:“http://derrylahan.comze.com/UFCR.mp4”。 导航到http://derrylahan.comze.com/UFCR.mp4
答案 0 :(得分:0)
请查看
<html>
<head>
<link rel="stylesheet" type="text/css" href="video.css">
</head>
<body>
<div id="banner" class="headerContent">
<div id="header">
<div id="headerContent" class="headerContent">
<video autoplay id="awsome_video" width="320" height="240" controls>
<source src="http://derrylahan.comze.com/promo.mp4" type="video/mp4">
</video>
<div class="overlay">
<h1 id="rowdy"> ROWDY RONDY ROUSEY </h1>
<a href="test.html" id="enter">ENTER</a>
</div>
</div>
</div>
</body>
</html>
答案 1 :(得分:0)
<?php
include '../db/connect6.php';
if(isset($_POST['id'])) {
$id = intval($_POST['id']);
$result = mysqli_query("SELECT id,name,email,company FROM yourtable WHERE yourtable.id = $id") or die("Could not search");
// since we expect only one result we don't need a loop
$row = mysqli_fetch_assoc($result);
// let's return the $row in json format
// first let's prepare the http header
header('Content-Type: application/json');
// and now we return the json payload
echo json_encode($row);
}
现在发生了一些用于测试的视频,但视频在页面加载时结束时却没有 玩