PHP video playback

Question

I am trying to get a video play to play from a database. I have a form with the following code:

<form action="abs3xvideos.php" method="POST" enctype="multipart/form-data">

<input type="file" name="id" />
<input type="submit" name="submit" value="UPLOAD!" />

<form action="abs3xvideos.php">
  Search ABS3X:
  <input type="search" name="googlesearch">
  <input type="submit">
</form>

I then have another page the form is linked to with the following code:

<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);

define('DB_Name', 'gaufensr_abs3x');
define('DB_User', 'gaufensr_owner');
define('DB_Password', 'password');
define('DB_Host', 'localhost');

$link = mysql_connect(DB_Host, DB_User, DB_Password);

if (!$link) {
      die('could not connect:' . mysql_error());
}

$db_selected = mysql_select_db(DB_Name, $link);

if (!@db_selected) {
    die('can\t use' . DB_Name. ': ' . mysql_error());
}

echo 'CONNECTED SUCCESSFULLY';
$id = $_POST['id'];
$value = $_POST['id'];

$sql = "INSERT INTO videos (video_name) VALUES ('$value')";

if (!mysql_query($sql)) {
     die(`ERROR: ` .mysql_error());
}

if (isset($_POST['id'])){
      $id = $_POST['id'];
      $query = mysql_query("SELECT * FROM `videos` WHERE id='$id'");
       while($row = mysql_fetch_assoc($query))
       {
             $id = $row['id'];
         $video_name = $row['video_name'];      
       }

    echo "You are watching " .$id. "<br />";
    echo "<embed src=`$id` width='560' height='315'></embed>";
}
else
{
   echo "Error!";
}  
mysql_close();
?>

I get the following error message when I try to upload a video using the form page that I created:

CONNECTED SUCCESSFULLY
Notice: Undefined index: id in /home1/gaufensr/public_html/abs3xvideos.php on line 39

Notice: Undefined index: id in /home1/gaufensr/public_html/abs3xvideos.php on line 40 Error!

I am at a loss. I spoke with someone on stackflow earlier and they suggested that something might be wrong with my while loop but I am not to sure what the mistake could be. Should I separate the PHP code into different pages maybe?

Why do you have two <form> tags? One of them isn't closed either. — Machavity
– Machavity ♦, Commented Apr 14, 2015 at 17:48
And does the upload not work? These errors SHOULD show, when you navigate to the page, without submitting the form, because $_POST is not yet set. Do the errors also show after submitting a file? — user169600
– user169600, Commented Apr 14, 2015 at 18:03
Btw, I just realized ... "upload"? There's no code to handle uploading a file. — user169600
– user169600, Commented Apr 14, 2015 at 18:07

Misunderstood · Accepted Answer · 2015-04-14 17:57:20Z

1

Did you urlencode the name?

$video_name = urlencode($row['video_name']);

Or rawurldecode may work better.

$video_name = rawurldecode($row['video_name']);

answered Apr 14, 2015 at 17:57

Misunderstood

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Comments

Saty · Accepted Answer · 2015-04-14 18:45:40Z

0

you forget to close form

Use

<form action="abs3xvideos.php" method="post" enctype="multipart/form-data">
<input type="file" name="id" />
<input type="submit" name="submit" value="UPLOAD!" />
</form>//from close here

<form action="abs3xvideos.php">
  Search ABS3X:
  <input type="search" name="googlesearch">
  <input type="submit">
</form>

edited Apr 14, 2015 at 18:45

answered Apr 14, 2015 at 17:50

Saty

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2 Comments

Misunderstood Over a year ago

"POST" will work the same as "post". "POST" will generate and HTML error with an XHTML DocType, but will still work.

ed-ta Over a year ago

I don't think it matters

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