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Parse error: syntax error, unexpected end of file in ... on line 103


williamh69

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Hi guys i have this error,

Parse error: syntax error, unexpected end of file in C:\wamp64\www\nigthclub\videos.php on line 103

 

but i know is probably easy to solve but for some reason i dont see the error. please help me

<?php

session_start();

?>

<?php error_reporting (E_ALL ^ E_NOTICE); ?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />

<link href="style.css" rel="stylesheet" type="text/css" />

<link href="tablefiestas.css" rel="stylesheet" type="text/css" />

<title>La Taverna de Juan</title>

</head>

<body>

<div id="container">

    <div id="header">

    <!--MENU BAR-->

  <?php include("includes/db.php"); ?>

    </div>

    <!--VIDEOS-->

<?php
$query = ("SELECT * from videos");
$result = mysqli_query($connection,$query);

$num_per_page =05;


?>

<table border="1" width="100%">
<tr>
  <td>Video Id</td>
  <td>Titulo</td>
  <td>Video</td>
</tr>
<tr>
  <?php
 
  while ($row=mysqli_fetch_assoc($result))
 
      {
      $videoid = $row['videoid'];
      $titulo = $row['titulo'];
      $url_video = $row['url_video'];
   ?>
   
       <td><?php echo $videoid ?></td>
       <td><?php echo $titulo ?></td>
       <td><?php echo $url_video ?></td>
   </tr>
   
   
   <?php}?>


</table>
 <?php
          $query =  "SELECT * from  videos";
          $pr_result = mysqli_query($connection,$query);
          $totalrecord =  mysqli_num_rows($pr_result);
          echo $totalrecord;

 ?>

    <br>

    <div id="footer">

 <!--FOOTER-->

    <?php include("includes/footer.inc.php"); ?>

    </div>    

</div>

</body>

</html>

 

Edited by Psycho
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Try changing this:
 

$query = ("SELECT * from videos");

to this:

$query = 'SELECT * from videos';

 

Also, you should really avoid escaping in and out "<?php" and "?>" of PHP like this. Instead, keep your HTML in separate template files, and load them via require. This is also known as "views", but I prefer to call it "templates", since that makes more logical sense to me.

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See if this makes sense to you.

<?php
/* ******************************
	script name
	description of what this does
	end of comments
********************************* */
session_start();
error_reporting (E_ALL ^ E_NOTICE); 
include "includes/db.php";
$query = "SELECT videoid, titulo, url_video from videos";
$result = mysqli_query($connection, $query);
$num_per_page =05;
$table_data = "
	<table border='1' width='100%'>
	<tr>
	<th>Video Id</th>
	<th>Titulo</th>
	<th>Video</th>
	</tr>
	";
while (list($videoid, $titulo, $url_video) = mysqli_fetch_row($result))
{
	$table_data .= "
	<tr>
	<td>$videoid</td>
	<td>$titulo</td>
	<td>$url_video</td>
	</tr>";
}
$table_data .= "
	</table>
	<br>
	";
$query =  "SELECT * from  videos";
$pr_result = mysqli_query($connection, $query);
$table_data .= "Results found: " . mysqli_num_rows($pr_result);
//
//  Now output the html data including the generated data above.
//
echo $html=<<<heredocs
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<link href="style.css" rel="stylesheet" type="text/css" />
<link href="tablefiestas.css" rel="stylesheet" type="text/css" />
<title>La Taverna de Juan</title>
</head>
<body>
<div id="container">
<div id="header">
<!--MENU BAR-->
</div>
<!--VIDEOS-->
$table_data
<br>
<div id="footer">
<!--FOOTER-->
heredocs;
//
echo $html;
//
include "includes/footer.inc.php";
echo "
	</div>    
	</div>
	</body>
	</html>
	";
exit();

I may have not inserted the table data in the correct place in your html block but I'm sure you can place it in the correct spot if not.

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