I'm very new to PHP and web-design, and I'm having trouble sending the value of a variable to another php.
Here is the current code:
<?php
$username = "root";
$password = "";
$database = "banners";
mysql_connect('localhost',$username,$password);
@mysql_select_db($database) or die ("Não foi possível estabelecer uma ligação a Base de Dados");
$data = mysql_query("SELECT * FROM banner") or die(mysql_error());
echo "<table border = 1>";
echo "<tr>";
echo "<th>Cliente</th>";
echo "<th>Status</th>";
echo "<th>Visualizar</th>";
echo "<th>Excluir</th>";
echo "</tr>";
while($info = mysql_fetch_array( $data ))
{
echo "<tr>";
echo "<td value=".$info['cliente'].">".$info['cliente'] . "</td> ";
if($info['activo']='SIM'){echo "<td>Activo</td>"; }else {echo "<td>Desactivo</td> ";}
echo '<td><a href="visualizar.php" target="_blank">
<img src="imagens/search.png" />
</a></td>';
echo "<td>".$info['ficheiro'] . "</td> ";
}
echo "</table>";
?>
The code that is underlined+bolded is supposed to send me to another page, and display the data of the entire information of the table, of the entry the user clicked on.
My problem is letting the other page know that I'm asking it to show "certain" data, or better said, sending the variable form here to another .php file.
I give an image example:
http://i.imgur.com/Og7St.png[
^]
I hope I didn't make it sound so complicated.
Thanks in advance !