Undefined variable, unsure why - php

Question

<?php
$tid = $_GET['tid'];
$id = $_SESSION['userid'];
$sql1 = "SELECT * FROM topics WHERE id='$tid' LIMIT 1";
$res1 = mysqli_query($connect, $sql1) or die(mysqli_error($connect));
while ($row = mysqli_fetch_array($res1, MYSQLI_ASSOC)) {
$title = $row['topic_title'];
$creator = $row['topic_creator'];
}
$sql = "SELECT * FROM users WHERE id='$creator' LIMIT 1";
$user_query = mysqli_query($connect, $sql) or die(mysqli_error($connect));
while ($row = mysqli_fetch_array($user_query, MYSQLI_ASSOC)) {
$name = $row["first"].$row["last"];
}
echo $name;
?>
I'm a little new to PHP, but I've done things exactly like this, but this time I'm getting an error. Everything here works except for $name. I checked my SQL tables and made sure users exist and that there's first and a last area. I don't see what else could be wrong.
Notice: Undefined variable: name in * on line **
Thank you.

Answer 1

Try this code on for size:
<?php
$tid = $_GET['tid'];
$id = $_SESSION['userid'];
$tid = mysqli_escape_string($connect, $tid);
$sql1 = "SELECT * FROM topics WHERE id='{$tid}' LIMIT 1";
$res1 = mysqli_query($connect, $sql1) or die(mysqli_error($connect));
// Check for rows first.
if($res1 and mysqli_num_rows($res1)){
// Use if as while is pointless on LIMIT 1
if($row = mysqli_fetch_array($res1, MYSQLI_ASSOC)) {
$title = $row['topic_title'];
$creator = $row['topic_creator'];
$creator = mysqli_escape_string($connect, $creator);
$sql = "SELECT * FROM users WHERE id='{$creator}' LIMIT 1";
$user_query = mysqli_query($connect, $sql) or die(mysqli_error($connect));
// Check for rows first.
if($user_query and mysqli_num_rows($user_query)){
// Use if as while is pointless on LIMIT 1
if ($row = mysqli_fetch_array($user_query, MYSQLI_ASSOC)) {
$name = $row["first"].$row["last"]; // NO HIT!
}
echo $name;
}else{
echo 'no rows found (query 2).';
}
}
}else{
echo 'no rows found (query 1).';
}
?>
Variable $name is undefined because the $name = ...; line is not reached. So make sure you $sql query actually returns results. It has to in order to define $name.