Question

我试图找出在用户使用html表单选择视频后如何从我的数据库中检索video_link。根据用户连接，将2个视频上传到我的数据库，原始视频和压缩视频。相应的视频将嵌入viewvideo.php中的html 5视频标签中，但因为ajax期待 $（＆＃34; #speed＆＃34;）。val（html）; 我的回复我想我可以在viewvideo.php中嵌入视频

所以我的问题是我现在应该怎么做？我应该在哪里检索video_link并嵌入视频？

这是我的编码。

我有一个html表单，用于检索用户希望观看的用户连接速度和video_id，并使用ajax发布表单数据

Html表单

<form action="viewvideo.php" method="post" >
                <br/>
                Please select the video
                <br/>

                <select name="video_id">
                    <?php
                        while($row = mysqli_fetch_array($result))
                        {
                    ?>       
                                      <option value="<?php echo $row['video_id']?>">
                    <?php echo $row['videoname']?>
                                      </option>         
                    <?php
                        }
                    ?>
                </select>

                <br />  
                <input type="text" id="speed" name="speed" value="">
                <input type="Submit" id="Submit" value="Submit" />
                </form>

的Ajax

  $.ajax({
      method: "POST",
      url: "viewvideo.php",
      data: {speedMbps: speedMbps,
      video_id: $('[name="video_id"').val()},
      cache: false
    }).done(function( html ) {
        $( "#speed" ).val( html );
});

viewvideo.php

  if(isset($_POST['video_id']) && isset($_POST['speedMbps'] )){
                        $id = trim($_POST['video_id']);
                        $speed = $_POST['speedMbps'];
                        echo $id;

                        $result = mysqli_query($dbc , "SELECT `video_id`, `video_link` FROM `video480p` WHERE `video_id`='".$id."'");
                        $count = mysqli_num_rows($result);

                        if (($speed < 100) && ($count>0)) {     //if user speed is less than 100 retrieve 480p quailtiy video   

                            //does it exist?
                            //if($count>0){
                                //exists, so fetch it in an associative array
                                $video_480p = mysqli_fetch_assoc($result);
                                //this way you can use the column names to call out its values. 
                                //If you want the link to the video to embed it;
                                echo $video_480p['video_link'];                     
                                }

                            else{
                                //does not exist
                            }

        ?>

                        <video id="video" width="640" height="480" controls autoplay>
                        <source src="<?php echo $video_480p['video_link']; ?>" type="video/mp4">
                        Your browser does not support the video tag.
                        </video>
                        <br />

                        <?php

                        $result2 = mysqli_query($dbc , "SELECT `video_id`, `video_link` FROM `viewvideo` WHERE `video_id`='".$video_id."'");
                        $count2 = mysqli_num_rows($result2);

                        // retrieve original video
                         if (($speed >= 100) && ($count2 >0)) { 
                                //does it exist?
                                    //if($count2>0){
                                        //exists, so fetch it in an associative array
                                        $video_arr = mysqli_fetch_assoc($result2);
                                        //this way you can use the column names to call out its values. 
                                        //If you want the link to the video to embed it;
                                        echo $video_arr['video_link'];                      
                                        }
    else{
                                    //does not exist

                                }
        ?>

                        <video id="video" width="640" height="480" controls autoplay>
                        <source src="<?php echo $video_arr['video_link']; ?>" type="video/mp4">
                        Your browser does not support the video tag.
                        </video>
                        <br />

        <?php

                    mysqli_close($dbc);
        ?>

Answer 1

如果您希望用户可以访问视频，则必须将实际视频（MP4）文件放在浏览器可以下载的公共目录中。因此video_link必须包含用户可访问的实际URL。如果视频不在公共目录中，则用户无法直接访问它们，或者您可以使用数据在URL中嵌入视频二进制数据：protocol

 data:[<MIME-type>][;charset=<encoding>][;base64],<video data>

例如

<?php
    function getVideoURLString($file, $type) { 
         return 'data:video/' . $type . ';base64,' .
                base64_encode(file_get_contents($file)); 
    }
?>

然后在HTML中

<video ...>        
     <source type="video/mp4" src="<?php echo getVideoURLString($filename, "mp4");"> 
</video>

在URL中嵌入视频数据的缺点是加载HTML可能需要很长时间。

使用php从数据库中检索视频

1 个答案: