mysql result into php array - php

Question

I'm trying to convert the result that i'm getting from mysql to a php array
can anyone helps me
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "women";
$conn = new mysqli($servername, $username, $password, $dbname);
$id=$_GET['id'];
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT DAY(ADDDATE(`dateDebutC`, `dureeC`)) AS MONTHS,
DAY(ADDDATE(ADDDATE(`dateDebutC`, `dureeC`),`dureeR`))AS DAYS
FROM normalW
where id = '$id'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
foreach($new_array as $array){
echo $row['DAYS'].'<br />';
echo $row['MONTHS'].'<br />';
}
} else {
echo "0 results";
}
$conn->close();
?>
Problem solved Thank you guys

Answer 1

To answer your question you must first declare the new array
$new_array = array();
Then loop through your query results to populated the array
while ($row = $result->fetch()) {
$new_array[] = $row;
}
But as one of the comments mentioned you really should be using prepared statements to protect yourself from sql injection.
$stmt = $mysqli->prepare("SELECT DAY(ADDDATE(`dateDebutC`, `dureeC`)) AS MONTHS, DAY(ADDDATE(ADDDATE(`dateDebutC`, `dureeC`),`dureeR`)) AS DAYS FROM normalW where id = ?");
/* bind parameters i means integer type */
$stmt->bind_param("i", $id);
$stmt->execute();
$new_array = array();
while($row = $stmt->fetch()) {
$new_array[] = $row;
}