错误插入数据ajax

Question

你好家伙即时尝试创建一个帐户使用Ajax。一切工作正常购买问题是年龄和联系人数不工作注册everydata可以插入除了年龄和联系号码，所以我有一个问题， 我的问题是年龄和联系号码不能插入数据库 上 的index.php

<div class="col-md-4"> 
      <label for="createage">Age</label> 
      <input id="createage" name="createage" class="form-control" type="text" > 
     </div> 


     <div class="col-md-6"> 
      <label for="createcontactnumber">Contact Number</label> 
      <input id="createcontactnumber" 
       name="createcontactnumber" 
       class="form-control" 
       oninput="javascript: if (this.value.length > this.maxLength) this.value = this.value.slice(0, this.maxLength);" 
       type = "number" maxlength = "11"/> 
     </div> 

为register.js

$(document).ready(function(e){ 
$('#register').click(function(){ 

    var Bdate = document.getElementById('createbday').value; 
    var Bday = +new Date(Bdate); 
    Q4A = ~~ ((Date.now() - Bday)/(31557600000)); 
    var theBday = document.getElementById('createage'); 
    theBday.innerHTML = Q4A.val; 


    var createusername = $('#createusername').val(); 
    var createpassword = $('#createpassword').val(); 
    var creategivenname = $('#creategivenname').val(); 
    var createmiddlename = $('#createmiddlename').val(); 
    var createlastname = $('#createlastname').val(); 
    var createbday = $('#createbday').val(); 
    var age = Q4A; 
    var creategender = $('#creategender').val(); 
    var contactnumber = $('#createcontactnumber').val(); 
    var createaddress = $('#createaddress').val(); 
    var createcity = $('#createcity').val(); 
    $.ajax({ 
    type : 'POST', 
    data :{createusername:createusername, 
       createpassword:createpassword, 
       creategivenname:creategivenname, 
       createmiddlename:createmiddlename, 
       createlastname:createlastname, 
       createbday:createbday, 
       age:age, 
       creategender:creategender, 
       contactnumber:contactnumber, 
       createaddress:createaddress, 
       createcity:createcity}, 
    url  :"insert.php", 
    success : function(result){ 
     if(result) 
     { 
     $('#error').html("<span>Success Man</span>"); 
     $('#createusername').val(''); 
     $('#createpassword').val(''); 
     $('#creategivenname').val(''); 
     $('#createmiddlename').val(''); 
     $('#createlastname').val(''); 
     $('#createbday').val(''); 
     $('#createage').val(''); 
     $('#creategender').val(''); 
     $('#createcontactnumber').val(''); 
     $('#createaddress').val(''); 
     $('#createcity').val(''); 
     } 
    } 
    }) 
}); 

}）;

register.php

<?php 

$connect = mysqli_connect("localhost", "root", "", "thaidatabase"); 

    $username = $_POST['createusername']; 
    $password = $_POST['createpassword']; 
    $givenname = $_POST['creategivenname']; 
    $middlename = $_POST['createmiddlename']; 
    $lastname = $_POST['createlastname']; 
    $bday = $_POST['createbday']; 
    $age = $_POST['createage']; 
    $gender = $_POST['creategender']; 
    $contactnumber = $_POST['createcontactnumber']; 
    $address = $_POST['createaddress']; 
    $city = $_POST['createcity']; 

mysqli_query($connect,"insert into account (username,password,givenname,middlename,lastname,bday,age,gender,contactnumber,address,city) 
      values 
      ('$username','$password','$givenname','$middlename','$lastname','$bday','$age','$gender','$contactnumber','$address','$city City')"); 


mysqli_close($connect); 


?> 

Answer 1

你逝去的年龄为年龄在POST请求（JavaScript）的属性，并试图读取createage从PHP _POST阵列。 联系号码相同。

发送的属性必须以相同的名称接收。修复它，它必须解决。

<?php 

$connect = mysqli_connect("localhost", "root", "", "thaidatabase"); 

    $username = $_POST['createusername']; 
    $password = $_POST['createpassword']; 
    $givenname = $_POST['creategivenname']; 
    $middlename = $_POST['createmiddlename']; 
    $lastname = $_POST['createlastname']; 
    $bday = $_POST['createbday']; 

    //Edit This 
    $age = $_POST['age']; 
    $gender = $_POST['creategender']; 

    //Edit This 
    $contactnumber = $_POST['contactnumber']; 

    $address = $_POST['createaddress']; 
    $city = $_POST['createcity']; 

mysqli_query($connect,"insert into account (username,password,givenname,middlename,lastname,bday,age,gender,contactnumber,address,city) 
      values 
      ('$username','$password','$givenname','$middlename','$lastname','$bday','$age','$gender','$contactnumber','$address','$city City')"); 


mysqli_close($connect); 


?>