Is this code correct?

Hello friends,

The logic is to retrieve all the videos from the database and play them in the browser.

<?php

$con=mysql_connect("localhost","root","");

if(!$con)
{
mysql_error();
}

mysql_select_db("my_upload",$con);

$result = mysql_query("SELECT * FROM upload");

 $assoc = mysql_fetch_assoc( $result );

$videofolder = './upload/';
while($row = mysql_fetch_array($result))
  {
  $vid1= $row['vid_name'];
  if (!preg_match('/\.(mpg|wmv|avi)$/i', $vid1) OR !file_exists($videofolder . $vid1))
{
     exit();
}
  echo "<img dynsrc=$videofolder . $vid1 WIDTH=400 HEIGHT=200 type=application/mpg/>";

  echo "<br/>";

  }
mysql_close($con);
?>

Code (markup):

I need help as quickly as possible,

 

It looks fine to me.

I would add start="FILEOPEN" and fix a few things to make the html valid.

echo '<img dynsrc="'.$videofolder.$vid1.'" width="400" height="200" start="FILEOPEN" type="application/mpg/">';

PHP:

 

Add exit if can't connect to database server.

if(!$con)
{
     mysql_error();
     exit();
}

PHP: