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

hello gays have php code connect database 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 * 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 create function getservices() , give argument show results in php file using foreach or while, this:

   <?php foreach ($results $key=>$result) : ?>      <?php echo $result['service_name']; ?>    <?php endforeach; ?> 

okay. have 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 * 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 works, haven't tested yet.