如何修复“ TypeError:X为空”

时间:2019-03-30 13:16:30

标签: javascript php html ajax xml

我正在制作一个小的站点/服务器,可以在其中将我的姓名/姓氏/年龄等发送到php文件。 我的问题是,我的JS无法从HTML提取我的输入,然后再更改PHP。

确切的错误是:TypeError:Firefox中的firstName为null

在不同的浏览器上尝试过,但得到的结果相同(如预期)。 还尝试将输入的ID切换为名称,但效果不佳。

<!DOCTYPE html>
<html>
  <head>
     <meta charset="utf-8">
     <meta name="Lucas" content="Nothing important">
     <title>Ajax form_1</title>
  </head>
  <body>
     <h2>Form_2</h2>
     <form>
       <input type = "text"  name = "firstName" placeholder = "voornaam">
       <input type = "text"  name = "lastName" placeholder = "achternaam">
       <input type = "text"  name = "age" placeholder = "leeftijd">
       <input type = "text"  name = "email" placeholder = "email">
       <input type = "button" id = "submitButton" value = "submit">
     </form>
    <div id = "responseHere">Response comes here</div>
    <script src="script.js"></script>
  </body>
</html>
let firstName = document.getElementById("firstName");
let lastName = document.getElementById("lastName");
let age = document.getElementById("age");
let email = document.getElementById("email");
let submitButton = document.getElementById("submitButton");
let responseHere = document.getElementById("responseHere");

submitButton.addEventListener('click', ajax);

function ajax(){
  let xmlhttp = new XMLHttpRequest();
  xmlhttp.onreadystatechange = function(){
    if (this.readyState == 4 && this.status == 200){
      responseHere.innerHTML = this.responseText;
  }
};
let httpString = "form_1.php?firstName=" + firstName.value + "&lastName=" + lastName.value + "&age=" + age.value + "&email=" + email.value;

console.log(httpString);

xmlhttp.open("GET", httpString, true);
xmlhttp.send();
}
    <?php
    $firstName = $_GET['firstName'];
    $lastName = $_GET['lastName']
    $age = $_GET['age'];
    $email = $_GET['email']
    echo "<h2>Response Demo Form</h2><h3>";
    echo "You submitted the following information<br><ul>";
    echo "<li>Name: <strong> $firstName $lastName</strong></li>";
    echo "<li>Age: $age</li>";
    echo "<li>Age: $email</li>";
    echo "</li></ul></h3>";
   ?>

1 个答案:

答案 0 :(得分:1)

您需要在html代码中添加唯一的id属性

 <form>
       <input type = "text" id='firstName'  name = "firstName" placeholder = "voornaam">
       <input type = "text" id='lastName'  name = "lastName" placeholder = "achternaam">
       <input type = "text" id='age' name = "age" placeholder = "leeftijd">
       <input type = "text" id='email' name = "email" placeholder = "email">
       <input type = "button" id = "submitButton" value = "submit">
     </form>