How to only show ID's not already in another table - php

Question

I am trying to display records from a categories table that don't exist in another table. This works to a certain extent whereby if there are 3 records in the sp_cats table, the display from $stmt2 will show all records except 1 where it should be all except 3. So, it isn't looping through for some reason.
$stmt = $link->prepare("SELECT `sp_cat_id` FROM `sp_cats` WHERE `sp_id` = ?");
$stmt->bind_param("i", $_GET['id']);
$stmt->execute();
$result = $stmt->get_result();
$numRows = $result->num_rows;
if($numRows > 0) {
while($row = $result->fetch_assoc()){
$spCats = $row['sp_cat_id'];
echo $spCats . "<br/>";
}
}
$stmt->close();
echo "<br>";
$stmt2 = $link->prepare("SELECT `cat_id`FROM `categories` WHERE `cat_id` != ?");
$stmt2->bind_param("i", $spCats);
$stmt2->execute();
$result2 = $stmt2->get_result();
$numRows2 = $result2->num_rows;
if($numRows2 > 0){
while($row2 = $result2->fetch_assoc()){
$cat_id = $row2['cat_id'];
echo $cat_id . "<br/>";
}
}
$stmt2->close();

Answer 1

See this while() loop here,
while($row = $result->fetch_assoc()){
$spCats = $row['sp_cat_id'];
...
}
The problem is, in each iteration of while() loop you're overwriting $spCats value. Besides, you don't have to use two different queries, you can do everything in just one query, like this:
$stmt = $link->prepare("SELECT `cat_id`FROM `categories` WHERE `cat_id` NOT IN (SELECT `sp_cat_id` FROM `sp_cats` WHERE `sp_id` = ?)");
$stmt->bind_param("i", $_GET['id']);
$stmt->execute();
$result = $stmt->get_result();
$numRows = $result->num_rows;
if($numRows > 0){
while($row = $result->fetch_assoc()){
$cat_id = $row['cat_id'];
echo $cat_id . "<br/>";
}
}
$stmt->close();