Question

我正在尝试创建人们可以输入视频网址的位置，它将保存到我的数据库中。问题是，在我将视频网址保存到我的数据库后，如何使用PHP在页面上显示它？

例如，使用此嵌入式vimeo视频，如何在从数据库中调用它后显示它。

如果我想从数据库中显示如下：

$result = mysqli_query($connect,"SELECT * FROM web where video <> '' && video IS NOT NULL");

    if ($row = mysqli_fetch_array($result))
    {

如何显示？

<iframe src="https://player.vimeo.com/video/90312869" width="500" height="281" 
        frameborder="0" webkitallowfullscreen mozallowfullscreen allowfullscreen>
            </iframe>
            <p><a href="https://vimeo.com/90312869">360&deg; 
        Video using 6 GoPro Cameras - spherical panorama timelapse</a> 
        from <a href="https://vimeo.com/j0n4s">j0n4s</a> on <a href="https://vimeo.com">Vimeo</a>.</p>

Answer 1

像这样：

$result = mysqli_query($connect,"SELECT * FROM web where video <> '' && video IS NOT NULL");

if ($row = mysqli_fetch_array($result))
{
   $url = $row['url'];
} 

<iframe src="<?php echo $url; ?>" width="500" height="281" 
    frameborder="0" webkitallowfullscreen mozallowfullscreen allowfullscreen></iframe>

        <p><a href="<?php echo $url; ?>">360&deg; 
    Video using 6 GoPro Cameras - spherical panorama timelapse</a> 
    from <a href="https://vimeo.com/j0n4s">j0n4s</a> on <a href="https://vimeo.com">Vimeo</a>.</p>

Answer 2

在html 5中我会像这样写

<?php
 // get the url 
 $sourceURL = value of the extracted column from your table which is a url;
 //Extract the file name
 $fileName = new SplFileInfo::getFilename ( $sourceURL );

 // Get the file extention
 $fileExtention = new SplFileInfo::getExtension ( $fileName );

 $sourceHTML = '<source src='.$sourceURL.' type="video/'.$fileExtention.'">';


?>

<video width="400" controls>
  <?php echo $sourceHTML;  ?>
  Your browser does not support HTML5 video.
</video>

有没有办法在我的网站上从数据库中的嵌入式URL显示vimeo视频？

2 个答案: