php - jQuery/PHP POST 不工作

Question

一切似乎都是正确的，但空白值继续被插入到我的数据库中。它似乎没有使用 $_POST 获取值。我也尝试过 $_GET 和 $_REQUEST ，但这些值似乎仍然没有被获取。我知道由于插入了静态“视频”文本，空白行被添加到数据库中。

HTML:

<form>
    <div class="form-group">
        <input type="text" class="form-control" name="firstName" id="firstName"
               placeholder="First Name">
    </div>
    <div class="form-group">
        <input type="text" class="form-control" name="lastName" id="lastName"
               placeholder="Last Name">
    </div>
    <div class="form-group">
        <input type="text" class="form-control" name="email" id="email"
               placeholder="Email">
    </div>
    <button type="submit" id="view" class="btn bds-button btn-block">View
    </button>
</form>

JavaScript:

$(document).ready(function() {

$('form').validate({
    rules: {
        firstName: {
            required: true
        },
        lastName: {
            required: true
        },
        email: {
            required: true,
            email: true
        }
    },
    messages: {
        firstName: "Please enter your first name",
        lastName: "Please enter your last name",
        email: {
            required: "Please enter your email",
            email: "Please enter a valid email",
        }
    },
    highlight: function(element) {
        $(element).closest('.form-group').addClass('has-error');
    },
    unhighlight: function(element) {
        $(element).closest('.form-group').removeClass('has-error');
    },
    errorElement: "span",
    errorClass: 'help-block',
    errorPlacement: function(error, element) {
        if(element.parent('.input-group').length) {
            error.insertAfter(element.parent());
        } else {
            error.insertAfter(element);
        }
    },
    submitHandler: function(form) {

        var firstName = $("#firstName").val();
        var lastName = $("#lastName").val();
        var email = $("#email").val();

        $.ajax({
            type: "POST",
            url: "php/post-viewer.php",
            data: {'firstName' : firstName, 'lastName' : lastName, 'email' : email},
            dataType: 'json',
            success : function(text){

            }
        });

    }
});

});

PHP:

<?php

$servername = "XXXX";
$username = "XXXX";
$password = "XXXX";
$dbname = "XXXX";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$first_name = $_POST['firstName'];
$last_name = $_POST['lastName'];
$email = $_POST['email'];

$sql = "INSERT INTO viewers (demo, first_name, last_name, email)
VALUES ('Video', '$first_name', '$last_name', '$email')";

if ($conn->query($sql) === TRUE) {
    echo $first_name;
    //echo "New record created successfully";
} else {
    echo $first_name;
    //echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();

?>

Answer 1

在ajax方法上，试试这个...

var vfirstName = $("#firstName").val();
var vlastName = $("#lastName").val();
var vemail = $("#email").val();

data: JSON.stringify({firstName : vfirstName, lastName : vlastName, email : vemail})