Question

im trying to display the picture saved in the database using Php, but what i get is Url link which was saved in the database and not the picture, i think the syntax is wrong.

<?php
echo "<table border=\"1\" align=\"center\">"; 
echo "<tr> 
    <th>Name</th> 
    <th>Description</th> 
    <th>Price</th>
    <th>Manufacturer</th>
    <th>Image</th> 
  </tr>"; 
while($row = mysql_fetch_array($result))
{
    echo "<tr>";
    echo "<td>" .$row['Name']."</td>";
    echo "<td>" .$row['Description'] ."</td>";
    echo "<td>" .$row['Price'] ."</td>";
    echo "<td>" .$row['Manufacturer'] ."</td>";
    echo "<td>" .$row['ImageURL'] ."</td>";
    echo "</tr>";
}
echo "</table>"; 
?>
</p>
<?php

Answer 1

You need to construct an <img> element, not just output the image URL.

Try replacing this: $row['ImageURL']

With this: "<img src='".$row['ImageURL']."' />