jQuery code
$.ajax({
    url: "far-area_m.php",
    type: "POST",
    dataType: 'json',
    data: values ,
    success: function (response) {
        alert("sdfds");
        // you will get response from your php page (what you echo or print)                        
    },
    error: function(jqXHR, textStatus, errorThrown) {
        alert(textStatus);
        console.log(textStatus, errorThrown);
    }               
});
far-area_m.php
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname='lateral';
$conn = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$len = $_REQUEST["len"];
$bre = $_REQUEST["bre"];
$wid = $_REQUEST["wid"];
$sql = mysqli_query(
    $conn,
    "INSERT INTO `far_input` (`length`, `breadth`, `width`)
    VALUES
    ('".$len."', '".$bre."', '".$wid."')
    ");
$data = array(
    'len'     => $len,
    'bre'  => $bre,
    'wid'   => $wid
);
echo json_encode($data);
?>
always Ajax response is printing error.But if I comment the insert statement in far-area_m.php I am getting response. Please tell me what is wrong with the code. on sucess of jquery I am not getting any response its throwing error.