How to use query results as conditional in PHP? - php

Question

I have the following SELECT query that looks to if there are records that meet the given criteria:
$stmt = $conn->prepare(' SELECT (`Active`)
FROM `Table1`
WHERE `Name` = ":name" AND Active <> Yes ');
$stmt->execute([
'name' => $_POST['name']
]);
If results of Active could either be a No or just no results. If the results are a No then the following query should run:
$stmt = $conn->prepare('UPDATE `Table1`
SET `active` = `Yes`
WHERE `Name` = `:name`');
$stmt->execute([
'name' => $_POST['name']
]);
If the results of the SELECT query does not find anything then the following query should run:
$stmt = $conn->prepare(' INSERT INTO `Table1` (`Name`,`Active`)
VALUES (:name, :active ) ');
$stmt->execute([
'name' => $_POST['name'],
'active' => $_POST['active']
]);
How can this conditional be written?
I specifically do not want to set a unique constraint to the name column and create stored procedures I need it to be done as a conditional described above.

Answer 1

This should work..
$stmt = $conn->prepare(' SELECT (`Active`)
FROM `Table1`
WHERE `Name` = ":name" AND Active <> Yes ');
$stmt->execute([
'name' => $_POST['name']
]);
if ($stmt->rowCount() === 0) {
$stmt = $conn->prepare(' INSERT INTO `Table1` (`Name`,`Active`)
VALUES (:name, :active ) ');
$stmt->execute([
'name' => $_POST['name'],
'active' => $_POST['active']
]);
} else {
$stmt = $conn->prepare('UPDATE `Table1`
SET `active` = `Yes`
WHERE `Name` = `:name`');
$stmt->execute([
'name' => $_POST['name']
]);
}
I have not tried this one out but it should work. Tell me if it doesnt.