I have the following code;
// check phone exist or not
$query = "SELECT * FROM user WHERE phone_number=".$phone;
$result = mysqli_query($conn,$query);
$count = mysqli_num_rows($result);
if($count!=0){
$error = true;
$phoneError = "Provided Phone Number($phone) is already in use.";
}
How can i send the result of the query as json feed to my colleague who we are working on the same project?