all you need to do is change line 28 to the following:

$sql = "SELECT * FROM customers WHERE lastName = '$name'";

and move line 32 onto line 27.... So the code would look like this:

<?php include '../view/header.php'; ?>
<?php require('../model/database.php');
require('../model/customer_db.php'); ?>
<div id="main">
 
<h1>Customer Search</h1>
 
<div id="content">
<!-- display a table of products -->
 
<form method="post" action="search.php?go" id ="searchform">
<label>Last Name:</label>
<input type="text" name="lastName">
<input type="submit" name ="submit" value="Search">
</form>
 
<?php
if(isset($_POST['submit'])) {
if(isset($_GET['go'])) {
if(preg_match("/^[A-za-z]+/", $_POST['lastName'])){
//connect
$db = mysql_connect("localhost", "username", "password") or die ('I cannot connect to the database because: ' . mysql_error());
 
//select
$mydb=mysql_select_db("dbname");

$name=$_POST['lastName'];
//
$sql = "SELECT * FROM customers WHERE lastName = '$name'"; //my tablename "customers"
 
//run
$result = mysql_query($sql);

 
//Create a while loop
while ($row = mysql_fetch_array($result)){
$lastName = $row['lastName'];
$firstName = $row['firstName'];
$email = $row['email'];
$city = $row['city'];
 
//display
echo "<tr> <th>Last Name,</th> <th>First Name</th> <th>Email</th> <th>City</th> <th></th></tr>";
 
echo "<u1>\n";
echo "<li>" . "<a href=\"search.php?id=$row\">" .$lastName . " " . $firstName . " " . $email . " " . $city . "</li>\n";
echo "</ul>";
 
}
 
// close table>
echo "</table>";
 
}
}
}
else
{
echo "<p>Please input a valid name</p>";
 
}
?>
}
 
</div>
</div>
<?php include '../view/footer.php'; ?>

Your issue was you never gave mysql a condition so it would always pull all customers.

EDIT: also, look up mysql_real_escape_string() to help sanitize your data before using against a database

New issue now:

Warning: mysqli_stmt::bind_param(): Undefined fieldtype 0 (parameter 2) in /srv/www/htdocs/framework/system/library/db_mysqli.php on line 103 Warning: mysqli_stmt::bind_param(): Undefined fieldtype 1 (parameter 2) in /srv/www/htdocs/framework/system/library/db_mysqli.php on line 103 Notice: Trying to get property of non-object in /srv/www/htdocs/framework/system/application/controllers/blog_controller.php on line 42 Notice: Trying to get property of non-object in /srv/www/htdocs/framework/system/application/controllers/blog_controller.php on line 43 Notice: Trying to get property of non-object in /srv/www/htdocs/framework/system/application/controllers/blog_controller.php on line 44 Notice: Trying to get property of non-object in /srv/www/htdocs/framework/system/application/controllers/blog_controller.php on line 45 

EDIT: I also noticed I was querying against title not entry_title but changing it made no different

and when I do so I get the following (without using your suggestion yet codeprada)

string(159) "SELECT b.entry_title, b.date_time, u.username, b.poster, b.entry FROM blog_post b INNER JOIN user u ON u.id = b.poster WHERE b.title = 'this is a test'

I var_dump'd it, sorry I forgot to mention, when echo'd the SQL Query was exactly how it should have been.

string(175) "SELECT b.entry_title, b.date_time, u.username, b.poster, b.entry FROM blog_post b INNER JOIN user u ON u.id = b.poster WHERE b.entry_title = 'this is a test' AND published = 1"

it is in blog_post, i did try it as b.title but the same issue is present