Cant get my user database to sort. - php

Question

I have a user database which I want the admin to be able view users. I want the admin to be able to sort the users by different fields. (E.g first name or surname).
My problem:
Is that it wont sort. Im trying to do it using a drop down box and a submit button. I have have tried the Sql in phpMyAdmin to make sure the statement is correct which it is. I have also added a or die to make sure its not a mysql error. Im getting no errors its just not sorting. I have also tried echoing out the drop down to make sure the right value is being added to the variable.
code:
HTML drop down
<form action="View_users.php" enctype="multipart/form-data" name="sort" id="sort" method="post" align="right">
<label>
<select name="sortdropdown" id="sortdropdown">
<option value="<?php echo $sortby; ?>"><?php echo $sortby; ?></option>
<option value="name">First Name</option>
<option value="surname">Surname</option>
<option value="email">Email</option>
<option value="signupdate">Date</option>
</select>
</label>
<label>
<input type="submit" name="button" id="button" value="Sort" />
</label>
</form>
logic:
$sortby = "";
if (!isset($_POST['sortdropdown'])) {
$sortby = "surnname";
}else{
$sortby = mysql_real_escape_string($_POST['sortdropdown']);
}
// This block grabs the whole list for viewing
$product_list = "";
$counter = 0;
$sql = mysql_query("SELECT * FROM users ORDER BY '$sortby' ASC") or die(mysql_error());
$productCount = mysql_num_rows($sql); // count the output amount
if ($productCount > 0) {
while($row = mysql_fetch_array($sql)){
$counter++;
$id = $row["id"];
$email = $row["email"];
$name = $row["name"];
$surname = $row["surname"];
$lastlogin = $row["lastlogin"];
$signupdate = $row["signupdate"];
if (is_float($counter/2)) {$class = "#CCCCCC"; }
else {$class = "white";}
$product_list .= '<tr bgcolor="'.$class.'">
<td>'.$surname.'</td>
<td>'.$name.'</td>
<td>'.$email .'</td>
<td>View More</td></tr>';
}
} else {
$product_list = "There are no users in the system yet";
}
Thanks

Answer 1

The comments are correct, you should be using back ticks `$sortby` instead of '$sortby'
The reason is - single quotes are used to wrap values in MySQL, which is not what you want. You're passing the name of a table column, so you use back ticks because they are used to wrap table column names.