Javascript沒有將值傳遞到我的PHP代碼中運行

Question

當我運行表格和使用action="addCustomer.php"的sql時，它工作正常，但是當我使用js時，數據未通過，並且控制台中未出現錯誤，它表示正在接聽來自原始頁面的時間軸中的變量，但是js不是嗎？ 誰能看到原因？

我正在使用以下形式：

  <form id="dataForm" method="GET">
  <div>
    <h2 id="formheader">Add Order Details:</h2>
    <label>First Name:</label>
    <input class="inputForm" id="inputFirstName" type="text" name="firstname">
  </div>
  <div>
    <label>Last Name:</label>
    <input class="inputForm" id="inputLastName"  type="text" name="lastname">
  </div>
  <div>
    <label>Address:  </label>
    <input class="inputForm" id="inputAdress" type="text" name="address">
  </div>
  <div>
    <label>Post Code:</label>
    <input class="inputForm" id="inputPostcode" type="text" name="postcode">
  </div>
  <div>
  <section class="inputForm">
  <label>Delievery Type:</label>
  <select name="deliverytype">
  <option ="3-5 Days">3-5 Days</option>
  <option ="Next Day">Next Day</option>
  </select> 
  </section> 
  <input type="hidden" id="calc" value="" name="calc">
  </div>
  <div id="theSubmit">
      <button id="checkoutButton">Submit</button>
    </div>
  </form>

和下面的js：

function addCustomer(){
  var xmlhttp = new XMLHttpRequest();
  var firstName = document.getElementById("inputFirstName").value;
  var lastName = document.getElementById("inputLastName").value;
  var address = document.getElementById("inputAdress").value;
  var postcode = document.getElementById("inputPostcode").value;
  var delievery = document.getElementById("deliverytype").value;
  if(firstName != "" && lastName != "" && address !="" && postcode !=""){
    var url = "addCustomer.php?FIRSTNAME=" + firstName + "&LASTNAME=" + lastName + "&ADDRESS=" + address + "&POST_CODE=" + postcode + "&DELIVERY_TYPE=" + delievery;

    xmlhttp.open("GET", url, false);
    xmlhttp.send();
     }
   else{
    alert("enter Some Data");
   }
}

submitButton = document.getElementById("checkoutButton");
submitButton.addEventListener("click", addCustomer);

和以下php sql代碼：

$name = $_GET['firstname'];
     $lastname = $_GET['lastname'];
     $userAddress = $_GET['address'];
     $userPostCode = $_GET['postcode'];
     $delivery = $_GET['deliverytype'];
     $totalCost = $_GET['calc'];

     if($name !="" && $lastname !="" && $userAddress !="" && $userPostCode !="" ){
     $sql = "INSERT INTO USERS (FIRSTNAME, SECONDNAME, ADDRESS, POST_CODE, DELIVERY_TYPE) VALUES ('$name', '$lastname', '$userAddress', '$userPostCode', '$delivery') ";
     $conn->exec($sql);
     echo "worked";
     }

Answer 1

嘗試以小寫形式發送變量：

var url = "addcustomer.php?firstname=" + firstName + "&lastname=" + lastName + "&address=" + address + "&post_code=" + postcode + "&delivery_type=" + delievery;

您以任何方式發送變量的任何更改都應該在數據庫中發生，而不是在瀏覽器的控制台日志中發生，請閱讀有關SQL注入的信息 ...