Realice la modificacion en list_records.php y lo deje asi de esta manera:
Código PHP:
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="charter"; // Database name
$tbl_name="clientes"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name WHERE id_cliente='$id_clientes";
$result=mysql_query($sql);
?>
<table width="400" border="0" cellspacing="1" cellpadding="0">
<tr>
<td>
<table width="400" border="1" cellspacing="0" cellpadding="3">
<tr>
<td colspan="4"><strong>Listar clientes </strong> </td>
</tr>
<tr>
<td align="center"><strong>Apellido</strong></td>
<td align="center"><strong>Nombre</strong></td>
<td align="center"><strong>Direccion</strong></td>
<td align="center"><strong>Update</strong></td>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td><? echo $rows['$Apellido']; ?></td>
<td><? echo $rows['$Nombre']; ?></td>
<td><? echo $rows['$Direccion']; ?></td>
<td align="center"><a href="update.php?id=<? echo $rows['id_cliente']; ?>">Update</a></td>
</tr>
<?php
}
?>
</table>
</td>
</tr>
</table>
<?php
mysql_close();
?>
Cuando lo ejecuto me tira el siguiente error:
Notice: Undefined variable: id_clientes in C:\Program Files\EasyPHP 2.0b1\www\Charter\list_records.php on line 12
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\EasyPHP 2.0b1\www\Charter\list_records.php on line 30