Javascript 表单未通过Ajax从wordpress站点提交到数据库

以下是弹出窗口中的表格，我无法将数据提交到数据库。请帮我解决这个问题

<div id="fsModal" class="modal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
<div class="modal-dialog">
    <div class="modal-content">
    <button type="button" id="cross-btn" class="close" data-dismiss="modal" aria-hidden="true">×</button>
        <div id="desk-view" class="modal-body">
            <div id="pop-up" class="row">   
                <div class="arrow_box side-heading col-xs-5 col-sm-5">
                    <h3>Join the Movement, the newsletter that tackles the justice issues that matter to you most.</h3>
                </div>
                <div class="right-form col-xs-7 col-sm-7">
                    <form id="form-search" action="" method="POST" class="form-inline subscribe-form">
                        <div class="form-group">
                            <input id="email-form" type="email" class="email-popup" placeholder="Enter your email" name="email">
                        </div>
                        <a href="#" id="submit-form" class="btn-form">Sign up</a>
                    </form> 
                </div>
            </div>
        </div>

    </div>
</div>

我直接进入模板页面，到newsletter_insert.php提交表单。我的模板php代码如下

$email=$_POST['email'];
if ($_SERVER["REQUEST_METHOD"] == "POST") {
    if (isset($_POST['user_email']) && !empty($_POST['user_email'])) {
        $wpdb->insert('newsletter_table', array(
            "emailaddress" => $email
        ));
        echo "1";
    }else{
        echo "2";
    }
}
if (data == 1) {
        alert("Subscribed Successfully!!");
    } else if(data == 2){
        alert("Something Went Error");
    }

用于将数据字符串化为JSON，然后从ajax传递，如下所示，而不是

数据：{'user_email'：$（“#email form”）.val（），

还要确保在ajax请求中添加

contentType:“application/json”

data: JSON.stringify(data),

---

在wordpress Preference function.php文件中，使用wp_ajax方法： 如下图所示：

您的jquery代码来自wp中的任何文件

        jQuery.ajax({
            url: "http:your_host_name/wp-admin/admin-ajax.php",  
            type: 'POST',
            dataType: 'json',
            data:{ 
                action: 'your_action_name',         
            },
            success : function(response){                   

                console.log(response);

            }
        }); 

在functions.php文件中写入：

add_action("wp_ajax_your_action_name", "test_function");

function test_function(){

  echo "Here";

} 

这样ajax在wordpress中工作

$('#submit-form').click(function(e) {
        $.ajax({
            data:{'user_email':$("#email-form").val()},
            type: "POST",
            url: "http://localhost/xxxx/index.php/news-letter/",
            success: function(data){
                if (data == 1) {
                    alert("Subscribed Successfully!!");
                } else if(data == 2){
                    alert("Something Went Error");
                }
            }
        });
    });

php代码是

if ($_SERVER["REQUEST_METHOD"] == "POST") {
    if (isset($_POST['user_email']) && !empty($_POST['user_email'])) {
        $wpdb->insert('newsletter-table', array(
            "emailaddress" => $_POST['user_email'],
            "date" =>  date('Y-m-d')
        ));
        echo "1";
    }else{
        echo "2";
    }
}

---

您收到的HTTP错误是什么？我没有收到任何错误，但是表单没有提交。我遗漏了一些内容，请查找更新的答案

var data={'user_email'：$（“#email form”）.val（）只需更改disno，它不工作，我想错误在php文件中，我包括您的调试命中php代码？dis是否正确我对此有疑问
$email=$\u POST['email']
应该是
$email=$\u POST['user\u email']？
$('#submit-form').click(function(e) {
        $.ajax({
            data:{'user_email':$("#email-form").val()},
            type: "POST",
            url: "http://localhost/xxxx/index.php/news-letter/",
            success: function(data){
                if (data == 1) {
                    alert("Subscribed Successfully!!");
                } else if(data == 2){
                    alert("Something Went Error");
                }
            }
        });
    });

if ($_SERVER["REQUEST_METHOD"] == "POST") {
    if (isset($_POST['user_email']) && !empty($_POST['user_email'])) {
        $wpdb->insert('newsletter-table', array(
            "emailaddress" => $_POST['user_email'],
            "date" =>  date('Y-m-d')
        ));
        echo "1";
    }else{
        echo "2";
    }
}