I am trying to append the code to display the results of the database by including a link, I got the link to work, but it will not display the results when clicked. Below is the code, If you can help me that would be great! Thanks
<?php
//if we got something through $_POST
if (isset($_POST['search'])) {
// here you would normally include some database connection
include('db.php');
$db = new db();
// never trust what user wrote! We must ALWAYS sanitize user input
$word = mysql_real_escape_string($_POST['search']);
// build your search query to the database
$sql = "SELECT Customer_Name, ID FROM Client_Computer_Info WHERE Customer_Name LIKE '%" . $word . "%' ORDER BY ID LIMIT 10";
// get results
$row = $db->select_list($sql);
if(count($row)) {
$end_result = '';
foreach($row as $r) {
$result = "\n\t<li><a href=\"search.php?ID={$r['ID']}\">{$r['Customer_Name']}</a></li>";
// we will use this to bold the search word in result
$bold = '<span class="found">' . $word . '</span>';
$end_result .= '<li>' . str_ireplace($word, $bold, $result) . '</li>';
}
echo $end_result;
} else {
echo "<p>Sorry no results for $word</p>";
}
}elseif(isset($_GET['ID'])){
$id = intval($_GET['ID']);
$query = "SELECT Customer_Name, Computer_Type, Computer_Make, Computer_Model, Serial_Number, Product_Number, OS, `Key` FROM Client_Computer_Info WHERE ID = $id LIMIT 1";
$n = mysql_query($query);
echo mysql_error();
if(mysql_num_rows($n) == 1){
$d = mysql_fetch_array($n);
$result .= "
\n<p>Name: {$n['Customer_Name']}</p>
\n<p>Computer Type: {$n['Computer_Type']}</p>
\n<p>Computer Make: {$n['Computer_Make']}</p>
\n<p>Computer Model: {$n['Computer_Model']}</p>
\n<p>Serial Number: {$n['Serial_Number']}</p>
\n<p>Product Number: {$n['Product_Number']}</p>
\n<p>OS: {$n['OS']}</p>
\n<p>Key: {$n['Key']}</p>";
}else{
$result = "<p>Sorry the ID supplied for the customer does not exist.</p>";
}
}
?>
