javascript - PHP Post and and print to innerHTML -


i trying post employee sql , display employee data after submit has button. can post , display not both @ same time. way should jsbin: https://jsbin.com/nenupupifo/edit?html,js,output

when implement php code post employee database refreshes form , not display output information. there better way this? 

var employees=[];   function process() { 'use strict';  var firstname = document.getelementbyid('firstname').value; var lastname = document.getelementbyid('lastname').value; var department = document.getelementbyid('department').value;  // 2) create employeeid variable , store randomly generated 8 digit employee id number. make sure id number unique. var employeeid = parseint(math.random() * 100000000);  // reference output goes: var output = document.getelementbyid('output');  // 3) add employeeid property employee object , set value employeeid variable. // 4)  make employee object json object , send end json object. var employee = {     firstname: firstname,     lastname: lastname,     department: department,     employeeid: employeeid,     date: new date() };  // 5)  add each newly created employee employees array (check duplicate before adding). employees.push(employee);  // 6)  can debug things easier, use console.log output employees array console after creating new employee. console.log(employees);   // 7)  display emplodyee in output html. var message = '<h2>employee added</h2>name: ' + employee.lastname + ', ' + employee.firstname + '<br>'; message += 'department: ' + employee.department + '<br>' + 'employee id: ' + employee.employeeid + '<br>' + 'hire date: ' + employee.date.todatestring() + '<br>' + // 8)  add 1 last line message display total number of employees have been added array. 'number of employees: ' + employees.length;  // display employee object: output.innerhtml = message;  addtosql2(employee) return false; }   function addtosql2(employee) { <?php    $servername = "localhost"; ?> <?php    $username = "root"; ?> <?php    $password = ""; ?> <?php     $dbname = "employees"; ?>   <?php  $conn = mysqli_connect($servername, $username, $password, $dbname); ?> <?php if (!$conn) {      die("connection failed: " . mysqli_connect_error());  }      $firstname = employee.firstname;     $lastname = employee.lastname;     $department =  employee.department;      $id = employee.id;       $sql = "insert employee (firstname, lastname, department, id ,dateadded)      values ('$firstname', '$lastname', '$department', '$id', now())";    if (mysqli_query($conn, $sql)) {   } else {      echo "error: " . $sql . "<br>" . mysqli_error($conn);  }   mysqli_close($conn); ?>  return false; }  function init() { 'use strict'; document.getelementbyid('theform').onsubmit = process; }  window.onload = init;


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