Question

我有一个包含表单的模态，当我尝试使用ajax发送数据时，我失败了。这是我的控制器：

public function server()
{
    $name = $_REQUEST['nom'];
    $ip = $_REQUEST['ip'];
    $ftp = $_REQUEST['ftp'];
    $user = $_REQUEST['user'];
    $pass = $_REQUEST['pass'];
    $sql = "INSERT INTO servers (id, name, ip, ftp, user, pass, active) VALUES (NULL, '$name', '$ip', '$ftp', '$user', '$pass', '1');";
    echo $sql;
    $this->db->query($sql);
}

从我的浏览器的consol中我也没有错误，当我试图展示发生了什么时，就像告诉我一切都好。

这是我的观点：

× Ajoute nouveau serveur：喃 IP FTP Nom d'utilisateur Mot de passe ANNULE

    <!-- BEGIN JAVASCRIPT -->
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
    <script src="assets/js/libs/jquery/jquery-migrate-1.2.1.min.js"></script>
    <script src="assets/js/libs/bootstrap/bootstrap.min.js"></script>
    <script src="assets/js/libs/spin.js/spin.min.js"></script>
    <script src="assets/js/libs/autosize/jquery.autosize.min.js"></script>
    <script src="assets/js/libs/nanoscroller/jquery.nanoscroller.min.js"></script>
    <script src="assets/js/core/source/App.js"></script>
    <script src="assets/js/core/source/AppNavigation.js"></script>
    <script src="assets/js/core/source/AppOffcanvas.js"></script>
    <script src="assets/js/core/source/AppCard.js"></script>
    <script src="assets/js/core/source/AppForm.js"></script>
    <script src="assets/js/core/source/AppNavSearch.js"></script>
    <script src="assets/js/core/source/AppVendor.js"></script>
    <script src="assets/js/core/demo/Demo.js"></script>
    <!-- END JAVASCRIPT -->
    <script type="text/javascript">
        $(document).ready(function(){
            $('form.newserverfrm').on('submit', function(form){
                form.preventDefault();
                $.post('/add/server', $('form.newserverfrm').serialize(), function(data){
                    alert(data);
                });
            });
        });
    </script>

Answer 1

您可以更改：

$.post('/add/server', $('form.newserverfrm').serialize(), function(data){...

到

$.post('<?php site_url("add/server") ?>', $('form.newserverfrm').serialize(), function(data){...

Answer 2

你应该采取行动

即

更改

 $.post('/add/server', $('form.newserverfrm').serialize(), function(data){...

要

$.post('server', $('form.newserverfrm').serialize(), function(data){...

Answer 3

请使用这些代码重新放置代码

查看页面

！ - BEGIN FORM MODAL MARKUP - ＆gt; × Ajoute nouveau serveur：喃 IP FTP Nom d'utilisateur Mot de passe ANNULE

<!-- BEGIN JAVASCRIPT -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script src="assets/js/libs/jquery/jquery-migrate-1.2.1.min.js"></script>
<script src="assets/js/libs/bootstrap/bootstrap.min.js"></script>
<script src="assets/js/libs/spin.js/spin.min.js"></script>
<script src="assets/js/libs/autosize/jquery.autosize.min.js"></script>
<script src="assets/js/libs/nanoscroller/jquery.nanoscroller.min.js"></script>
<script src="assets/js/core/source/App.js"></script>
<script src="assets/js/core/source/AppNavigation.js"></script>
<script src="assets/js/core/source/AppOffcanvas.js"></script>
<script src="assets/js/core/source/AppCard.js"></script>
<script src="assets/js/core/source/AppForm.js"></script>
<script src="assets/js/core/source/AppNavSearch.js"></script>
<script src="assets/js/core/source/AppVendor.js"></script>
<script src="assets/js/core/demo/Demo.js"></script>
<!-- END JAVASCRIPT -->
<script type="text/javascript">

功能测试（） {

        $.ajax({
        type : 'POST',
        url : 'server.php',
        data : { },  
        async : false,
        success : function(data)
        {
            alert(data);
        }

        });

    }
</script>

我无法将AJAX请求发送到我的控制器

3 个答案: