PHP - Prepared Statement Query doesn't work the second time - php

Question

So, I'm working with PHP and prepared statements sent to a MySQL database. I've ran into a problem that I can't quite debug. Here is my code:
// Check if the input username is in the database
$stmtQuery = "SELECT * FROM updatedplayers WHERE Player=?;";
$preparedStmt = $dbc->prepare($stmtQuery);
$preparedStmt->bind_param("s", $setUsername);
$preparedStmt->execute();
$preparedStmt->bind_result($resultUUID, $resultUsername);
$preparedStmt->fetch();
// If it's not, kill the page.
if ($resultUUID == null) {
incorrect();
}
$stmtQuery = "SELECT Password, Salt FROM logins WHERE UUID=?;";
echo 'flag1 ';
$preparedStmt = $dbc->prepare($stmtQuery);
echo 'flag2 ';
$preparedStmt->bind_param("s", $resultUUID);
echo 'flag3 ';
The fist prepared statement works fine, it's at the line $preparedStmt->bind_param("s", $resultUUID);. There are also a couple other prepared statements before these, so I know I'm doing this correctly, but I'm not too sure about the last statement.
The code just seems to stop running after echo 'flag2 ';, which I put there to find the specific line. I don't get any error messages, it just doesn't print out flag3.
I've tried replacing $resultUUID with a static string, yet I get the same outcome. Also, I know my SQL statement is correctly formatted, I've tested within the console manually.
That's pretty much it, I'd love to hear some criticism, as I am new to PHP. Also, is there any way to get a better idea about the errors I get, instead of trying to pinpoint the error myself? Thanks!

Answer 1

So, adding ini_set('display_error', 1);, suggested by #user2182349, gave me a little more insight, I got "Fatal error: Call to a member function bind_param() on boolean".
After some research, I tried adding mysqli_report(MYSQLI_REPORT_ALL);, which ended up throwing "No index used in query/prepared statement".
I did some research on that to realize that it wasn't a problem, just MySQLI reporting unnecessary errors (which is what I asked it to do lol). In order to get a better, more insightful stack trace, I used mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);.
This threw "Commands out of sync; you can't run this command now". Again, more research taught me to use $preparedStmt->store_result();, in order to allow for another prepared statement to run.
Big thanks to all y'alls' help, hope this can help someone too.

Answer 2

You should be able to use a single select statement similar to this:
SELECT u.UUID, u.Username, l.Password, l.Salt
FROM updatedplayers AS u
JOIN logins AS l ON (u.UUID = l.UUID)
WHERE u.Player = ?
Check the case of the field names to be sure they match the database.
At the top of the file, add ini_set('display_errors',1);. If you have any PHP errors, they will be displayed. Also check the return values from the database calls and use the error display functions.

Answer 3

I think you need to close the prepared statement before you use the variable for another query:
$preparedStmt->close();
Or use another variable name like $preparedStmt2 for the second query.

Answer 4

I would suggest you should start using PDO... I have issues encountered with mysqli prepared statement years ago. Since then, PDO gives me no headaches when it comes to multiple queries at a time.
You should try PDO.. :-) it's more efficient.
http://php.net/manual/en/intro.pdo.php
http://php.net/manual/en/class.pdostatement.php
Or you can do the following "if you want alternative solution"..
//Close connection
$preparedStmt->close();
//AND OPEN YOUR CONNECTION AGAIN TO PREPARE NEW QUERIES..
$stmtQuery = "SELECT Password, Salt FROM logins WHERE UUID=?;";
echo 'flag1 ';
$preparedStmt = $dbc->prepare($stmtQuery);
echo 'flag2 ';
$preparedStmt->bind_param("s", $resultUUID);
echo 'flag3 ';