dpoxk64080 2015-11-04 23:12
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尝试使用Php-MySQL从数据库中检索视频URL并将其显示在浏览器上

I'm stuck on a php issue where I'm trying to access a url I stored in a local database (stored on a MAMP server) and display it. The URL is a video file on a server and the idea is that the video is retrieved from the url on the database and shown on a browser (preferably on auto-play). For some reason, the php script won't show the table contents at all. I'm kind of new to sql. Any help would be appreciated. Thanks This is the db table: media

This is my code:

<?php
$servername = "localhost";
$username = "media";
$password = "mysqluser";
$dbname = "media";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

// sql to display a record
$sql = "SELECT * FROM media";


if ($conn->query($sql) === TRUE) {
    echo "Record dislayed successfully";
} else {
    echo "Error showing record" . $conn->error;
}

$conn->close();

?>

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1条回答 默认 最新

  • dongque6377 2015-11-05 01:20
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    Since you are new to mysql, I think you should echo the url out and put it maybe a HTML5 video player.

    e.g

    <?php
    $servername = "localhost";
    $username = "media";
    $password = "mysqluser";
    $dbname = "media";
    
    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 
    
    // sql to display a record
    $sql = "SELECT * FROM media";
    $result = $conn->query($sql);
    $data = $result->fetch_assoc();
    $video = $data['video']; // if video is the column name
    
    echo"<video width='320' height='240' controls>
    <source src='".$video."' type='video/mp4'>
    </video>
    ";
    ?>
    

    N/B: The example if for mp4 videos

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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