Question

用户可以从我的网站流式传输视频。我想添加一个按钮，所以当用户点击按钮时，它将允许浏览器下载文件，

我该怎么做？

检索要流式传输的视频

  if(isset($_POST['video_id'])){
        $id = trim($_POST['video_id']);
        //check if it exists
        $result = mysqli_query($dbc , "SELECT `video_id`, `video_link` FROM `viewvideo` WHERE `video_id`='".$video_id."'");
        $count = mysqli_num_rows($result);
        //does it exist?
        if($count>0){
            //exists, so fetch it in an associative array
            $video_arr = mysqli_fetch_assoc($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 Link: ".$video_arr['video_link'];
            echo $video_arr['video_link'];
            ?>

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

下载按钮

<button id="download" onclick="download()">download</button>

function download() {
                    <?php
                     if(isset($_POST['video_id'])){
                // get the video link...
                $fileName=$video_arr['video_link'];
                $fileName=str_replace("..",".",$fileName); //required. if somebody is trying parent folder files
                $file = "../folder/".$fileName;
                if (file_exists($file)) {
                    $mime = 'application/force-download';

                    header('Content-Type: '.$mime);

                    header('Content-Description: File Transfer');
                    header('Content-Disposition: attachment; filename='.$fileName);
                    header('Content-Transfer-Encoding: binary');
                    header('Expires: 0');
                    header('Cache-Control: must-revalidate');
                    header('Pragma: public');
                    header('Content-Length: ' . filesize($file));
                    ob_clean();
                    flush();
                    readfile($file);
                    exit;
                    ?>
                    }

Answer 1

在某处添加链接标记，链接到您的视频文件：

<a href='your-video.mp4' download>

Answer 2

是的，您只需使用HTML的下载属性，如下所示

<a href="**YOUR FILE LINK**" download>Click here to Download</a>

这可能对你有帮助！

Answer 3

正如我在评论中已经提到的，下载属性在IE浏览器中不起作用（我自己测试了！| Also on can i use）。

相反，我会选择这样的东西：

          if(isset($_POST['video_id'])){
                // get the video link...
                $fileName=$video_arr['video_link'];
                $fileName=str_replace("..",".",$fileName); //required. if somebody is trying parent folder files
                $file = "../destinationOfYourFile/".$fileName;
                if (file_exists($file)) {
                    $mime = 'application/force-download';

                    header('Content-Type: '.$mime);

                    header('Content-Description: File Transfer');
                    header('Content-Disposition: attachment; filename='.$fileName);
                    header('Content-Transfer-Encoding: binary');
                    header('Expires: 0');
                    header('Cache-Control: must-revalidate');
                    header('Pragma: public');
                    header('Content-Length: ' . filesize($file));
                    ob_clean();
                    flush();
                    readfile($file);
                    exit;
                }
            }

修改

HTML：

<button id="download" onclick="download()">download</button>

JS：

function download(){ var fileName = FILENAME; // you need to get your filename, e.g. "video.mp4" window.location.href = 'https://www.YOURSITE.com/LOCATION/TO/download.php?filename=' + fileName; }

的download.php：

if(isset($_GET['filename'])){ $fileName=$_GET['filename']; $fileName=str_replace("..",".",$fileName); //required. if somebody is trying parent folder files $file = "../destinationOfYourFile/".$fileName; // you need to change the destination to your video folder if (file_exists($file)) { $mime = 'application/force-download'; header('Content-Type: '.$mime); header('Content-Description: File Transfer'); header('Content-Disposition: attachment; filename='.$fileName); header('Content-Transfer-Encoding: binary'); header('Expires: 0'); header('Cache-Control: must-revalidate'); header('Pragma: public'); header('Content-Length: ' . filesize($file)); ob_clean(); flush(); readfile($file); exit; } }

所以你只需要更改我在代码中提到的路径并获取你的文件名，一切都应该正常工作。我希望我可以帮助你:)。

Answer 4

我假设您的网络和数据库位于一台服务器中。否则，实施将完全不同，或者可以通过FTP完成。

<强> HTML

如果您使用的是bootstrap，则可以轻松地将锚标记转换为按钮显示。我建议在下载链接的按钮上使用锚标记。

<a href="<?php echo 'http://baseurl/download_video.php?video_id='.$video_arr['video_id']; ?>">Download</a>

php - download_video.php

我推荐使用GET方法而不是POST，因为你只需要获取数据。

// query video info here using $_GET['video_id']
$filepath = $result['video_path'];
$filename = base_name($filepath);

// enable php_fileinfo.dll/.so in php.ini as necessary
$finfo = finfo_open(FILEINFO_MIME_TYPE);
$mime = finfo_file($finfo, $filepath);
finfo_close($finfo);

header("Pragma: public");
header("Expires: 0");
header("Cache-Control: must-revalidate, post-check=0, pre-check=0");
header("Content-Type: ".$mime);
header("Content-Transfer-Encoding: Binary");
header("Content-disposition: attachment; filename=\"".$fname."\"");

ob_clean();
flush();

readfile($filepath);                
exit;

Answer 5

$file_url = $_GET['link'];           // http://www.somedomain.com/test.mp4
header('Content-Type: video/mp4'); 
header("Content-disposition: attachment; filename=\"" . 
basename($file_url) . "\""); 
readfile($file_url);

PHP从服务器下载视频文件

5 个答案: