视频文件移至文件夹但未上传到数据库

时间:2017-02-07 20:35:47

标签: php mysql

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>

1 个答案:

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);