PHP MYSQL SELECT Returns NULL - php

Question

I executing the following code which creates a company_id as a UUID_SHORT in a temporary table.
This company_id will then be used to insert records in multiple tables with the UUID as the primary key. My issue is when I try retrieve the company_id that is $company_id in my code it is null. However if I json_encode ($tempResult) the company_id value is there. What am I doing wrong?
Any help is much appreciated, thank you!
try {
$conn = new PDO("mysql:host=localhost;dbname=$dbname", $db->id, $db->pass); //connect to db
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); //error modes
$temp = $conn->prepare('CREATE TEMPORARY TABLE tempId (user_id VARCHAR(17) PRIMARY KEY, company_id VARCHAR(17))');
$temp->execute();
$temp = $conn->prepare('INSERT INTO tempId(user_id, company_id) VALUES(:user_id, UUID_SHORT())');
$temp->bindParam(':user_id', $_SESSION['username'], PDO::PARAM_INT);
$temp->execute();
$temp = $conn->prepare('SELECT company_id FROM tempId WHERE user_id = :user_id ');
$temp->bindParam(':user_id', $_SESSION['username'], PDO::PARAM_INT);
$temp->execute();
$tempResult= $temp->fetchAll(PDO::FETCH_ASSOC);
$company_id = $tempResult->company_id;
// $result[1] =$_SESSION('username');
} catch(PDOException $e) {
$result = $e->getMessage();
}
print json_encode($company_id);

Answer 1

Here:
$tempResult= $temp->fetchAll(PDO::FETCH_ASSOC);
If the fetchAll is successful, then $tempResult will be an array. For debugging, we can verify this using the convenient var_dump, e.g.
var_dump($tempResult);
If $tempResult is an array, I'm wondering about this expression:
$tempResult->company_id
What does that return? What do you expect that to return? Why?
EDIT: I know better than to answer a question with a question, or three questions.
However, I can't (in good conscience) bring myself to giving an "answer" to the problem with OP code...
at least not without (figuratively) scratching my head wondering about the actual SQL being used in the code.
What is the purpose of the TEMPORARY TABLE? Why is there an INSERT to it? Why is the UNSIGNED BIGINT datatype (returned by UUID_SHORT() function) being cast to a VARCHAR(17)? Is there some reason we want to lop off 1 or 2 digits when the function returns 18 or 19 decimal digits?
If the intent of this block of code is to return a value from MySQL UUID_SHORT() function, I'm not understanding why we need more than one statement. Obviously, there's something I'm missing, why this wouldn't suffice:
try {
$conn = new PDO("mysql:host=localhost;dbname=$dbname", $db->id, $db->pass);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sth = $conn->prepare('SELECT UUID_SHORT() AS company_id');
$sth->execute();
$company_id = $sth->fetchColumn();
} catch(PDOException $e) {
//var_dump($e->getMessage);
} finally {
if(isset($sth)){ $sth->close(); }
if(isset($conn)){ $conn->close(); }
}
(An application wouldn't churn database connections like this; there would either be a connection pool, or the connection would be passed in to this routine.)

Answer 2

Not sure, but as soon as fetchAll returns array, your code:
$company_id = $tempResult->company_id;
is invalid, you should:
$company_id = $tempResult[0]['company_id'];
or
$tempResult= $temp->fetch(PDO::FETCH_ASSOC);
$company_id = $tempResult['company_id'];