Multiple mysql query resulting in null return - php

Question

While doing a PHP function using two mysql query, the second one returns null, which shouldn't have, since both of them works in phpmyadmin SQL query submition.
Here is the function:
public static function getStates($countryId) {
try {
$query = "SELECT * FROM `Apostas` WHERE `idConfronto` = ".$countryId;
$result = dbconfig::run($query);
if(!$result) {
throw new exception("Confronto invalido.");
}
$res = array();
while($resultSet = mysqli_fetch_assoc($result)) {
$sql2 = "SELECT * FROM `Mercados` WHERE `idMercado` = ".$resultSet["idMercado"];
$result2 = dbconfig::run($sql2);
if(!$result2) {
throw new exception("Não há mercados.");
}
while ($row2 = mysqli_fetch_assoc($result2)) {
$res[$resultSet['idMercado']] = $row2['nomeMercado'];
}
}
$data = array('status'=>'success', 'tp'=>1, 'msg'=>"States fetched successfully.", 'result'=>$res);
} catch (Exception $e) {
$data = array('status'=>'error', 'tp'=>0, 'msg'=>$e->getMessage());
} finally {
return $data;
}
}
The first query works everytime, but the second one goes towards the exception "Não há mercados".
After debugging the function the variable $resultSet["idMercado"] is working and the sql query also works, but the code results in the exception.
What I'm doing wrong? Perhaps something with the syntax?
--EDIT1:
As requested, the code for dbconfig::run:
public static function run($query) {
try {
if(empty($query) && !isset($query)) {
throw new exception("Query string is not set.");
}
$result = mysqli_query(self::$con, $query);
self::close();
return $result;
} catch (Exception $e) {
echo "Error: ".$e->getMessage();
}
}
--EDIT2:
As Cavid suggested, do all querys before closing the connection, here is how the function turned out:
public static function getStates($countryId) {
try {
$query = "SELECT * FROM `Apostas` WHERE `idConfronto` = ".$countryId;
$result = $conn->query($query);
if(!$result) {
throw new exception("Confronto invalido.");
}
$res = array();
if ($result->num_rows > 0) {
while ($resultSet = $result->fetch_assoc()) {
$sql2 = "SELECT * FROM `Mercados` WHERE `idMercado` = ".$resultSet['idMercado'];
$result2 = $conn->query($sql2);
if ($result2->num_rows > 0) {
while ($row2 = $result2->fetch_assoc()) {
$res[$resultSet['idMercado']] = $row2['nomeMercado'];
}
}
}
}
$data = array('status'=>'success', 'tp'=>1, 'msg'=>"States fetched successfully.", 'result'=>$res);
} catch (Exception $e) {
$data = array('status'=>'error', 'tp'=>0, 'msg'=>$e->getMessage());
} finally {
return $data;
}
$conn->close();
}
Its giving me now a error code 500 "Internal server error"

Answer 1

Instead of going in two different tables that has a Reference id in both, i suggest you to JOIN them and only use one while loop:
public static function getStates($countryId) {
try {
// Inner join both tables
$query = "SELECT a.idMercado, a.idConfronto, b.nomeMercado FROM Apostas AS a ";
$query .= "INNER JOIN ";
$query .= "Mercados AS b ";
$query .= "ON a.idMercado = b.idMercado ";
$query .= "WHERE a.idConfronto = " . $countryId;
$result = dbconfig::run($query);
if(!$result) {
throw new exception("Confronto invalido.");
}
// Results
$res = array();
while ($row = mysqli_fetch_assoc($result)) {
$res[$row['idMercado']] = $row['nomeMercado'];
}
$data = array('status'=>'success', 'tp'=>1, 'msg'=>"States fetched successfully.", 'result'=>$res);
} catch (Exception $e) {
$data = array('status'=>'error', 'tp'=>0, 'msg'=>$e->getMessage());
} finally {
return $data;
}
}