Question

我想使用ajax提交表单，我的脚本代码也没有任何错误，但我的php也没有任何错误，但是当我单击“提交”按钮时，什么也没发生，只是表单将为空，否则什么也没发生，数据没有存储在数据库中。我不知道这是什么错误。我不知道如何找到它，有人可以帮我解决这个问题吗？

  <div class="container-fluid">
                <h3 class="alert alert-success" align="center">CHECKOUT FORM</h3>
                <form id="checkout_form">
                <div class="row">

                        <div class="col-md-2">

                        </div>

                        <div class="col-md-4">
                            <div class="form-group">
                              <label for="usr">Name:</label>
                              <input type="text" class="form-control" id="name" name="name">
                            </div>
                            <div class="form-group">
                              <label for="email">Email:</label>
                              <input type="text" class="form-control" id="email" name="email">
                            </div>
                             <div class="form-group">
                              <label for="address">Address:</label>
                              <input type="text" class="form-control" id="address" name="address">
                            </div>
                             <div class="form-group">
                              <label for="city">City:</label>
                              <input type="text" class="form-control" id="city" name="city">
                            </div>
                            <div class="form-group">
                              <label for="city">State:</label>
                              <input type="text" class="form-control" id="state" name="state">
                            </div>
                        </div>
                        <div class="col-md-4">
                            <div class="form-group">
                              <label for="card">Card Name:</label>
                              <input type="text" class="form-control" id="card_name" name="card_name">
                            </div>
                            <div class="form-group">
                              <label for="cvv">CVV:</label>
                              <input type="text" class="form-control" id="cvv" name="cvv">
                            </div>
                             <div class="form-group">
                              <label for="card_num">Card Number:</label>
                              <input type="text" class="form-control" id="card_number" name="card_number">
                            </div>
                             <div class="form-group">
                              <label for="expired_month">Expire Month:</label>
                              <input type="text" class="form-control" id="expired_month" name="expired_month">
                            </div>
                            <div class="form-group">
                              <label for="zip">Zip Code:</label>
                              <input type="text" class="form-control" id="zip_code" name="zip_code">
                            </div>

                            <input type="submit" name="submit" class="btn btn-success" value="CHECKOUT" action = "checkout" id="submit">
                        </div>



                </div>
                </form> 
            </div>

脚本代码

<script>
    $(document).ready(function(){

        $('#submit').click(function(){


        var name = $('#name').val();
        var email = $('#email').val();
        var address = $('#address').val();
        var city = $('#city').val();
        var state = $('#state').val();
        var card_name = $('#card_name').val();
        var cvv = $('#cvv').val();
        var card_number = $('#card_number').val();
        var expired_month = $('#expired_month').val();
        var zip = $('#zip_code').val();

        var data_string ='name='+ name + '&email=' + email + '&address=' + address + '&city=' + city + '&state=' + state + '&card_name=' + card_name + '&cvv=' + cvv + '&card_number=' + card_number + '&expired_month' + expired_month + '&zip' + zip_code ;


        $.ajax({
            url : "action1.php",
            method : "POST",
            data :data_string,
            success:function(data){
                $('#checkout_form').html("<div id='message'></div>");
                $('#message').val("form submitted");
            }
        });
    });
});
    </script>

action1.php

<?php

include('database_connection.php');

if(isset($_POST['submit']){

        $data = array(
            ':name' => $_POST['name'],
            ':email' => $_POST['email'],
            ':address' => $_POST ['address'],
            ':city' => $_POST['city'],
            ':state' => $_POST['state'],
            ':card_name' => $_POST ['card_name'],
            ':cvv' => $_POST['cvv'],
            ':card_number' => $_POST['card_number'],
            ':expired_month' => $_POST['expired_month'],
            ':zip_code'=> $_POST['zip_code']
        );

        $query = "INSERT INTO checkout (name,email,address,city,state,card_name,cvv,card_number,expired_month,zip_code) VALUES (:name,:email,:address,:city,:state,:card_name,:cvv,:card_number,:expired_month,:zip_code)";

        $statement = $connect ->prepare($query);

        if($statement->execute($data))
        {
            echo "form submitted";
        }
}
?>

Answer 1

单击提交，您的表单页面将刷新。这样做：

$('#submit').click(function(e){
     e.preventDefault;

如评论中所述，

echo $query;
exit; //This exit is causing the query not to run because the script stops running.

您的字段名称前面有分号。删除它们：

$data = array(
            'name' => $_POST['name'],
            'email' => $_POST['email'],
            'address' => $_POST ['address'],
            'city' => $_POST['city'],
            'state' => $_POST['state'],
            'card_name' => $_POST ['card_name'],
            'cvv' => $_POST['cvv'],
            'card_number' => $_POST['card_number'],
            'expired_month' => $_POST['expired_month'],
            'zip_code'=> $_POST['zip_code']
        );

Answer 2

欢迎来到Stackoverflow @ prabhu-msd

在此代码下将e.preventDefault;添加为第一行：

$('#submit').click(function(){

它防止刷新页面并运行Ajax。在成功下，在location.reload();行之后添加$('#message').val("form submitted");，以在提交数据后刷新页面（如果要刷新页面），否则可以执行任何操作。

在您的PHP文件中，您正在使用退出，这会阻止执行查询。请删除该exit。

使用ajax提交结帐表单

2 个答案: