Question

我有一个注册表，我正在使用ajax

< form id="form" class="form" action="form.php" method="post" >

在form.php单击按钮注册时，验证以表格形式发生就像用户名存在和密码匹配然后它注册它在form.php中工作正常但是对于ajax我必须创建另一个页面check.php，它还检查用户是否存在ajaxusername = username然后它返回用户存在于前面用户名字段的确定，但是当我按下注册时，即使用户存在，它也会注册用户。

我使用了ajax：

     ---------------- form . php -------------

    function aj(){
    $("document").ready(function (){

    var username = $("#username").val();


    $.ajax({
      url: 'check.php',
      type: 'POST',
      data: {ajaxusername:username},

      success: function(data) {
        $('#existu').html(data);
        }
      });


            });
    };



  <?php
 session_start();
 $con=mysqli_connect("localhost","user","","db") or 
 die(mysql_error());

 if(isset($_POST['register'])){
 $name=$_POST['username'];
 $email=$_POST['email'];
 $password=$_POST['password'];
 $confirmpassword=$_POST['confirmpassword'];

 $q=mysqli_query($con,"SELECT * FROM userreg WHERE username='$name'");
 $query="INSERT INTO regform(username,email,password,confirmpassword) 
 VALUES('$name','$email','$password','$confirmpassword')";

 if(mysqli_num_rows($q) > 0){

 $userexist="User already exists!";

 }
elseif ($password==$confirmpassword){


  mysqli_query($con,$query);
  $a=header("location:welcome.php");

 }
 else{

 $error="Password does not match!";

}

}





 ?>

check.php

session_start();
$con=mysqli_connect("localhost","user","","db") or die(mysql_error());
if(isset($_POST['ajaxusername'])){
$query="SELECT * FROM userreg WHERE username='".$_POST["ajaxusername"]."'";
$result=mysqli_query($con,$query);
if(mysqli_num_rows($result)>0){
echo "exist";
}
else 
{
echo "does not exist";
}
}
?>

表单验证+ ajax问题

0 个答案: