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社区首页 >问答首页 >Javascript没有将值传递到要运行的php代码中

Javascript没有将值传递到要运行的php代码中
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Stack Overflow用户
提问于 2015-04-17 22:24:28
回答 1查看 66关注 0票数 0

当我使用action="addCustomer.php"运行表单和sql时,它工作得很好,但是当我使用js时,数据没有通过,并且控制台中没有出现错误,它说它正在从原始页面拾取时间线中的变量,但js不是这样吗?有谁知道为什么吗?

我使用的是以下表格:

代码语言:javascript
复制
  <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>

以及下面的js代码:

代码语言:javascript
复制
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);

以及以下php sql代码:

代码语言:javascript
复制
$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";
     }
EN

回答 1

Stack Overflow用户

发布于 2015-04-17 22:35:29

尝试发送小写的变量:

代码语言:javascript
复制
var url = "addcustomer.php?firstname=" + firstName + "&lastname=" + lastName + "&address=" + address + "&post_code=" + postcode + "&delivery_type=" + delievery;

无论如何,您只是发送变量任何更改应该发生在数据库中,而不是在您的浏览器的控制台日志中,请阅读有关SQL Injection

票数 1
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/29701935

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