PHP MYSQL compare database value to POST value - php

Question

This is a really simple one, I just can't get my head around it sorry. I have this PHP code which picks up my form value, then compares it with the value stored in the database. That works fine.
However I am not sure how to write this logic in terms of this query:
If posted value = database value {
// do something } else { // do something else }
if (empty($_POST['order_id']) === false) {
// prepare data for inserting
$order_id = htmlentities(trim($_POST['order_id']));
$order_id = preg_replace("/[^0-9]/","", $order_id);
$result = mysqli_query($con,"SELECT * FROM listings WHERE order_id = $order_id");
$row = mysqli_fetch_assoc($result);
echo $row['order_id'];
}
SOLVED:
Solved the question, was a silly one I know! Just needed this at the end of the code:
if($order_id === $row['order_id']) {
echo 'found';
} else {
echo 'not found';
}

Answer 1

Try
If ($result->num_rows === 1) { do something } else { do something else }

Answer 2

Since you did the business logic in your query you can just use
if( ! is_null($row)) {
// do
} else {
// nothing
}

Answer 3

Did I read too much into "If posted value = database value "? Are you just referring to the order_id?
if ($row['listingName'] == $_POST['frm_listingName']) {
// something
}
else {
//something else
}

Answer 4

Check this code:
if (empty($_POST['order_id']) === false) {
// prepare data for inserting
$order_id = htmlentities(trim($_POST['order_id']));
$order_id = preg_replace("/[^0-9]/","", $order_id);
$result = mysqli_query($con,"SELECT * FROM listings WHERE order_id = $order_id");
if(mysqli_num_rows($result)>0)
{
//Match found. do something.
}
else
{
//No match found. do something.
}
}
N.B. In place of mysqli_num_rows($result) you can also use $result->num_rows