代码点火器中的警报消息成功

Question

我对这个程序有一些问题。我想给模态使用语义ui，但它不起作用。当我点击按钮顺序会显示模式，但我现在的问题是我点击按钮的顺序显示新页面。无法插入数据库。

视图代码：

    <div class = "ui form grid_6 omega">
        <form target="paypal" method="post">
            <div class="field1">
            <div class="field">
                <label>Nama</label>
                <input placeholder="Nama" name="nama" type="text">
            </div>
            <div class="field">
                <label>No. HP</label>
                <input placeholder="No. HP" name="handphone" type="text">
            </div>
            <div class="field">
                <label>Alamat</label>
                <input placeholder="alamat" name="alamat" type="text">
            </div>
            <div class="field">
                <label>Jumlah</label>
                 <div class="selectbox">
                    <select name="jumlah" id="">
                    <?php for ($i=1; $i <= 20; $i++): ?>
                    <option value="<?php echo $i; ?>"><?php echo $i; ?></option>
                    <?php endfor; ?>
                </select>   
                 </div>
            </div>
            <button type="submit" class="ui teal button order-button">Order now</button>
            </div>
        </form>
        </div>
        <div class="ui small modal">
                <i class="close icon"></i>
                <div class="header">
                  Smalls Like Bakin
                </div>
                <div class="content">
                  <p>Thank you for your order</p>
                </div>
                <div class="actions">
                  <div class="ui positive right labeled icon button">
                    <a class="text-white" href="<?php echo site_url('home/order');?>">Back to Home</a>
                    <i class="checkmark icon"></i>
                  </div>
                </div>
              </div>

并编写脚本

    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script type="text/javascript" src="<?php echo base_url(); ?>packaged/javascript/semantic.js"></script>
<script type="text/javascript">
    $('.order-button').click(function(){
        $('.small.modal').modal('show');
    });
</script>

控制器代码；

    function simpanOrder()
    {
        $this->load->model("M_order");
        $data['nama'] = $_POST['nama'];
        $data['handphone'] = $_POST['handphone'];
        $data['alamat'] = $_POST['alamat'];
        $data['jumlah'] = $_POST['jumlah'];
        $result = $this->M_order->insert($data);
    }

你能告诉我我的代码问题在哪里以及如何解决吗？非常感谢。

Answer 1

尝试将模态div的id设置为：

<div id="modaldiv" class="ui small modal">  

并将脚本更改为：

<script type="text/javascript">
$('.order-button').click(function(){
    $('#modaldiv').modal('show');
});  
</script>  

此外，请确保包含了运行模态所需的所有javasrpt文件。