PHP/Mysqli insert : Cannot pass parameter 2 by reference - php

Question

Thanks for helping me out.
I'm trying to do something simple : just 2 inserts in the database.
The first one works well:
$stmt = $db->prepare("INSERT INTO sessions(date, lieu, trainer) VALUES (?,?,?)");
$stmt->bind_param("sss", $date, $location, $trainer);
$stmt->execute();
$session = $stmt->insert_id;
$stmt->close();
Then, I try to do another insert: I have a table 'users' which has 4 columns : an Id, auto-incremented, a username (column called 'fullName'), a sessionId (an int) and a contactInfo (a varchar).
Here's my code for the second insert:
//var_dump($session); returns an int
$stmt = $db->prepare("INSERT INTO users(fullName, sessionId, contactInfo) VALUES (?,?,?)");
$stmt->bind_param("sis", "$username", "$session", "$contactinfo");
$stmt->execute();
$newId = $stmt->insert_id;
$stmt->close();
From what I could read from other posts, it seems that this issue comes when you try to pass a parameter that's not a parameter (e.g.:an int) but in my case it's a variable, I reuse "$session" from the first block ($session = $stmt->insert_id;)
Am I allowed to do that? What did I miss?
Thanks!
EDIT: removed the single quotes and put double quotes, but that doesn't seem to cut it. Tried to put them for both strings but not for session but it doesn't change the result.
EDIT2:following the good advice from serjoscha, I printed the query to have an idea of what it looks like:
echo "INSERT INTO users (fullName, sessionId, contactInfo) VALUES ($username,$session,$contactinfo)";
which gives me something like
INSERT INTO users (fullName, sessionId, contactInfo) VALUES (Paul Honour,56,Location Liège Belgium Email ph#ph.be [phone])
The query only works if I put it like this:
INSERT INTO users (fullName, sessionId, contactInfo) VALUES ('Paul Honour',56,'Location Liège Belgium Email ph#ph.be [phone]')
Any idea what's wrong?

Answer 1

Your issue are the single quoted variables. Just remove the single quotes or use double quotes for the content of double quotes is partitial evaluated / parsed:
$a=3;
echo '$a'; // prints: $a
echo "$a"; // prints: 3
echo $a; // prints: 3 and this is just what you need
You do not need the quotes for variables, so just remove them:
$stmt->bind_param("sis", $username, $session, $contactinfo);

Answer 2

Found out the issue:
it was failing on "Liège", some problem with the accent. I made sure I had the same encoding on both sides, and I can finally insert!
My code now looks like:
$contactinfo = $db->real_escape_string($contactinfo);
$stmt = $db->prepare("INSERT INTO users (fullName, sessionId, contactInfo) VALUES (?,?,?)");
$stmt->bind_param("sis", $username, $session, $contactinfo);
$stmt->execute();
$userId = $stmt->insert_id;
$stmt->close();