Javascript PHP：数据库中的视频插入_Javascript_Php_Mysql

Javascript PHP：数据库中的视频插入

javascript php mysql

Javascript PHP：数据库中的视频插入,javascript,php,mysql,Javascript,Php,Mysql,这是我的表格。它正在将文件插入数据库，我需要在我的系统中检索并播放该视频。。有人能给我更多的代码吗？如何使用和/或如何将视频播放器添加到我的代码中 <?php include "config.php"; $videoname = $_REQUEST["vname"]; $category = $_REQUEST["cat"]; $keyword = $_REQUEST["kwd"]; $target = "C:/xampp/htdocs/upload/"; $target = $ta

这是我的表格。它正在将文件插入数据库，我需要在我的系统中检索并播放该视频。。有人能给我更多的代码吗？如何使用和/或如何将视频播放器添加到我的代码中

<?php 
include "config.php";
$videoname = $_REQUEST["vname"];
$category = $_REQUEST["cat"];
$keyword = $_REQUEST["kwd"];
$target = "C:/xampp/htdocs/upload/"; 
$target = $target . basename( $_FILES['uploaded']['name']) ; 
$ok=1; 

if ($_FILES["uploaded"]["error"] > 0)
{
    echo "Apologies, an error has occurred.";
    echo "Error Code: " . $_FILES["uploaded"]["error"];
}
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) 
{
    echo "The file ". basename( $_FILES['uploaded']['name']). " has been uploaded";
} 
else 
{
    echo "Sorry, there was a problem uploading your file.";
}
?>
<?php
$path = str_replace("C:/xampp/htdocs","http://localhost",$target);
$query = "INSERT INTO videos(videoname, category, keywords, path)";
$query .= "VALUE ('$videoname', '$category', '$keyword', '$path')";
$result = mysql_query($query);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"                    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Untitled Document</title>
</head>
<body>
<form enctype="multipart/form-data" action="upload_file.php" method="POST">
    Video Name :<input type="text" name="vname"/><br/>
    Category : <input type="text" name="cat"/><br/>
    Keywords : <input type="text" name="kwd"/><br/>
    Please choose a file: <input name="uploaded" type="file" /><br />
    <input type="submit" value="Upload" />
</form> 
</body>
</html>

在不知道视频类型的情况下，请尝试此处：-您可能还想查看此页面上的警告消息：始终向问题寻求帮助…而不是代码…如何使用@user3422452从数据库播放该视频路径的视频：您尝试了什么？您需要将值更改为值