Accessing two tables within same statement? - php

Question

I'm trying to create an upvote system where, after checking to see if you're logged in and after getting your userid, it checks if the table has your userid with the postid already in it and if it does then it means it was already upvoted. I just want to know what's wrong in my code, this is being used for learning, I don't need any complex thing.
Code:
if (isset($_GET['upvote'])) {
if ($_SESSION["loggedin"] == true) {
$upvoteid = $_GET["upvote"];
$servername = "";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id FROM users WHERE username=".$_SESSION["loggedinusername"];
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$userid = $row["id"];
}
$sql2 = "SELECT userid, postid FROM upvotedposts WHERE userid='".$userid."' AND postid='".$upvoteid."'";
$result2 = $conn->query($sql);
if (!$result2->numrows > 0) {
$sql = "UPDATE posts SET upvotes = upvotes + 1 WHERE id = ".$upvoteid;
if ($conn->query($sql) === TRUE) {
echo "Sucessfully upvoted";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
} else {
echo "Failed;
}
$conn->close();
}
}
When I do click the upvote button, it simply does nothing. The issue here is that as far as I know it looks like it would work but I may be forgetting something that I am unaware of or incorrectly using something.

Answer 1

You are missing a closing "
echo "Failed;
should be:
echo "Failed";