所以我上传的视频文件成功上传到文件夹但不会上传到我为我的video_upload系统创建的phpmyadmin表中。我找不到这个问题的解决方案,所以任何帮助都是适用的。附:我对数据库还不熟悉
如果需要提供更多代码,请告诉我,这是我的整个index.php文件 我的程序没有收到任何错误
<html>
<head>
<title>Video Upload System</title>
<link rel='stylesheet' href='style.css' type='text/css' />
</head>
<body>
<?php
include 'connect.php';
?>
<div class='box'>
<form method='post' enctype='multipart/form-data'>
<?php
if(isset($_FILES['video'])){
$name = $_FILES['video']['name'];
$type = explode('.', $name);
$type = end($type);
$size = $_FILES['video']['size'];
$random_name = rand();
$tmp = $_FILES['video']['tmp_name'];
if($type != 'mp4' && $type != 'MP4' && $type != 'flv') {
$message = "Video Format Not Supported !";
} else {
move_uploaded_file($tmp, 'videos/'.$random_name.'.'.$type);
mysqli_query($connection, "INSERT INTO videos VALUES('', '$name', 'videos/$random_name.$type')");
$message = "Successfully Uploaded! ";
}
echo "$message <br/><br/>";
}
?>
Select Video: <br/>
<input type='file' name='video' />
<br/><br/>
<input type='submit' value='Upload' />
</form>
</div>
<div class='box'>
<?php
$run = "";
$video_id = "";
$video_url = "";
$video_name= "";
$query = mysqli_query($connection, "SELECT `id`, `name`, `url` FROM videos");
while($run = mysqli_fetch_array($query)){
$video_id = $run['id'];
$video_name = $run['name'];
$video_url = $run['url'];
}
?>
<a href='view.php?video= <?php echo $video_url; ?>'>
<div id='url'>
<?php echo $video_name; ?>
</div>
</body>
</html>
答案 0 :(得分:0)
试试这个
$sql = "INSERT INTO `videos` (`id`, `name`, `url`)
VALUES ('', '$name', 'videos/$random_name.$type')";
mysqli_query($connection, $sql);