Give argument php function to show database results in another file - php

Question

I have this php code to connect database it works fine.
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "cv";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
mysqli_set_charset($conn,"utf8");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM services";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["service_name"]. " " . $row["service_desc"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
My question is: how to create function getServices() and give argument to show this results in another php file using foreach or while, like this:
<?php foreach ($results as $key=>$result) : ?>
<?php echo $result['service_name']; ?>
<?php endforeach; ?>

Answer 1

Okay. So you have to do the following things:
db.php :
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "cv";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
mysqli_set_charset($conn,"utf8");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
function yourFunctionName() {
$sql = "SELECT * FROM services";
$result = $conn->query($sql);
return $result;
}
$conn->close();
otherfile.php :
<?php
include 'db.php';
$data = yourFunctionName();
if ($data->num_rows > 0) {
// output data of each row
while($row = $data->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["service_name"]. " " . $row["service_desc"]. "<br>";
}
} else {
echo "0 results";
}
?>
Hope this works, haven't tested yet.