Question

我能够检索名称，但为什么我不能在嵌入位置显示整个视频？

这是我正在处理的整个PHP文件。

第一个框主要是关于将视频上传到目录和数据库。

我实际上只是因为视频没有出现而对第二个框有问题。

<?php

    include 'connect.php'; //include the php file into this php file

    ?>

<div id="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']['name'];
            $random_name = rand();
            $tmp = $_FILES['video']['tmp_name'];

            if($type != 'mp4' && $type != 'mp4' && $type != 'wmv'){
                $message = "Video Format Not Supported!";
            }

            else {
                move_uploaded_file($tmp, 'videos/'.$random_name.'.'.$type); 

                mysqli_query($db, "INSERT INTO videos (id, name, url) 
                VALUE ('', '$name', '$random_name.$type')");
                $message = "Successfully Uploaded!";

            }

            echo "$message"; 

           } 
        ?>


        Select Video : <br/>
        <input type='file' name="video" />
        <br/><br/>
        <input type="submit" value="Upload" />
    </form>


</div>


    <div id="box">
        <?php

        $query = mysqli_query($db, "SELECT `id`, `name`, `url` FROM videos");
        while($run = mysqli_fetch_array($query)){
            $video_id = $run['id'];
            $video_name = $run['name'];
            $video_url = $run['url'];

        ?>

        <?php 
              echo $video_name; 


        ?>

        <?php

        $video = $_GET['video'];
        echo "<embed src=`$video` width='560' height='315'></embed>" ;

        ?>


        <?php

        }
        ?>


    </div>

Answer 1

HTML中有反引号而不是引号。将它们更改为单引号或双引号，例如：

    echo "<embed src='$video' width='560' height='315'></embed>" ;
                     ^      ^

Answer 2

你应该

if (isset($_GET['video'])) {
   $video = $_GET['video'];
   echo '<embed src="'.$video.'" width="560" height="315"></embed>';
}

因此，如果没有输入视频ID，人们就无法访问该页面

Answer 3

尝试这种方式，它将要求您更改数据库。但应该工作。只需阅读我使用它的注释说明。

<div id="box">
  <?php
   if (isset($_GET['video'])) {
    $video = $_GET['video'];
    $query = mysql_query("SELECT * FROM `videos` WHERE `id`='$video'");
     $count = mysql_num_rows($query);
     if ($count!=0) {
        $row = mysql_fetch_assoc($query);
        $video_id   = $row['id'];
        $video_name = $row['name'];
        $video_url  = $row['url'];
        echo $video_name;
        echo '<embed src="'.$video_url.'" height="315px" width="560px">';
        //echp <video height="315px" width="560px" controls><source src="movie.mp4" type="video/mp4"></video> 
        //Use the above code if you want a html5 video player.
     } else {
        echo 'Video does not exist!';
     }
   } else {
      echo 'Please enter a video id!';
   }
   /*
    How to use it:
    Create a database called "videos"
    Insert the following three columns:
     id   = varchar(225) as `primary_key`
     name = varchar(225);
     url  = varchar(225);
    To insert a video, you will need to create
    a random id for the video and insert it
    into the mysql database. Example
      $rand = rand(111111111,999999999);
      $id   = md5($rand);
      //example $id = 3174143713413051830531
      $name = "Random video name";
      $url  = "http:/localhost/websitename/videos/nameofvideo.mp4";
      $query = mysql_query("INSERT INTO `videos` VALUES ('$id','$name','$url')");
    Then to select the video you will go to a page
    video.php?video=[video id here];
    video.php?video=3174143713413051830531
    then the php will select the url for the video
    id = 3174143713413051830531
    and it will play that video. 
    example: video_url would be http://localhost/websitename/videos/nameofvideo.mp4
    If you have any further questions,
    feel free to ask me. Thanks
   */
  ?>
</div>

希望这有效。谢谢！

PHP Get方法不起作用

3 个答案: