Question

Consider the following code

<?php

 $username = "root";
 $password = "";
 $host = "localhost";
 $database = "binaries";

 @mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

 @mysql_select_db($database) or die("Can not select the database: ".mysql_error());

 $id = 5;

 if(!isset($id) || empty($id)){
 die("Please select your image!");
 }else{

$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['imag'];
header('Content-type: image/jpg');
echo '<table><tr><td height="700" width="700">';// Line X
print $content;

echo '</td></tr></table>';//Line Y

}

?>  

When I comment the lines X and Y the image gets displayed, otherwise not.What could be the possible reason?

EDIT: After following Matt’s advice.

show.php

 echo '<table><tr><td>
  <img src="image.php"/>
  </td></tr></table>';

image.php

$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['imag'];
header('Content-type: image/jpg');

print $content;

Even after doing this I am not getting the expected result.

EDIT:
code of ‘image.php’ :

 <?php
  $username = "root";
  $password = "";
  $host = "localhost";
  $database = "binaries";

  @mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

 @mysql_select_db($database) or die("Can not select the database: ".mysql_error());




 $query = mysql_query("SELECT * FROM tbl_images WHERE id=5");
 $row = mysql_fetch_array($query);
 $content = $row['imag'];
 header('Content-type: image/jpg');
 echo $content;

?>

Answer 1

Because when the browser is told the MIME type is image/jpg, the last thing it is expecting to see is <table ...

When you set the MIME type, you are telling the browser “I am sending you an image.” However, HTML markup is certainly not image data, so the browser doesn’t know how to render it.