php - 使用 MySql 和 php 获取视频

Question

我正在开发一个网站，该网站获取用户上传的视频并放置名称、大小、类型、路径 和 tmp_name 到 MySQL 数据库中。 upload.php 文件如下，

<?php 

$is_form_submitted =  (isset($_POST['submit']))?true:false;

    if($is_form_submitted)
    {
        //defines variables
        $name=$_FILES['file']['name'];
        $type=$_FILES['file']['type'];
        $size=$_FILES['file']['size'];
        $tmp_name=$_FILES['file']['tmp_name'];
        $target_path="videos/";


        $allowedTypes   = array("video/wmv","video/avi",
         "video/mpeg","video/mpg","video/mp4");
        $is_valid_type  = (in_array($_FILES['file']['type'],                           $allowedTypes))?true:false;
       if ( $is_valid_type&& ($_FILES["file"]["size"] < 20000000000))
      {
        if ($_FILES["file"]["error"] > 0)
        {
        echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
        }
        else
        {
        echo "Upload:&nbsp; " . $_FILES["file"]["name"] . "<br>";
        echo "Type:&nbsp; " . $_FILES["file"]["type"] . "<br>";
        echo "Size:&nbsp; " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
        echo "Temp file:&nbsp; " . $_FILES["file"]["tmp_name"] . "<br>";
        mysql_query("INSERT INTO vids(name, type, size, tmp_name, target_path)
        VALUES('$name', '$type', '$size', '$tmp_name', '$target_path')");

        if (file_exists("images/" . $_FILES["file"]["name"]))
          {
          echo $_FILES["file"]["name"] . " already exists. ";
          }
        else
          {
          move_uploaded_file($_FILES["file"]["tmp_name"],
          "videos/" . $_FILES["file"]["name"]);
          echo "Stored in: " . "videos/" . $_FILES["file"]["name"];
          }
        }
      }
    else
    {
      echo "Invalid file";
    }

    }
?> 

一切都正确上传，但是当我尝试获取视频的名称和路径时，它说视频播放器找不到受支持的视频，那么我该如何获取从数据库中随机视频并将其作为视频播放器的来源？ (视频文件可以在播放器中播放，所以不是视频的错)

这是文档代码中的php代码

<?php

$vid_url = "videos/";
    $result = mysql_query("SELECT * FROM `vids` WHERE 1");
    while($row = mysql_fetch_assoc($result))
    {
        echo 

        <div name="video">
        <video width="100%" height="100%" controls>
        <source src=".$vid_url.$row."type="video/mp4">
     Error: Video Not working
        </object>
        </video> 
        </div>'; 

    } 
?>

我存储视频的位置是一个名为视频的目录

Answer 1

我想通了以防万一有人想引用这个

<?php    
    //Connects to database
    $host="################"; // Host name
    $username="##########"; // Mysql username
    $password="###########"; // Mysql password
    $db_name="#######"; // Database name
    $tbl_name="####"; // Table name

    // Connect to server and select databse.
    mysql_connect("$host", "$username", "$password")or die("cannot connect");
    mysql_select_db("$db_name")or die("cannot select DB");
    //Tells variable video where to get the name from
    $vid_url = "videos/";
        $result = mysql_query("SELECT * FROM vids ORDER BY RAND() LIMIT 1");
    $row = mysql_fetch_assoc($result);
    $video = $vid_url.$row["name"];
<?