0

我对codeigniter相当陌生。

我的 views/check_availability/index.php 文件中有以下代码：

<?php $type = $_POST['type']; ?>

<a href="/reservation/check_availability/create">Create Reservation</a>

<?php foreach ($check_availability as $check_availability_item): ?>

    <h2><?php echo $check_availability_item['room_type'] ?></h2>
    <div id="main">
        <?php echo $check_availability_item['room_description'] ?>
    </div>
    <p><a href="check_availability/<?php echo $check_availability_item['id'] ?>">View article</a></p>

<?php endforeach ?>

这是我在 models/check_availabiliy_model.php 文件下的代码

public function check_availability($type = FALSE)
{
    if ($type === FALSE)
    {
        $query = $this->db->get('lf_rooms');
        return $query->result_array();
    }

    $query = $this->db->get_where('lf_rooms', array('type' => $type));
    return $query->row_array();
}

这是我的控制器/check_availability.php 文件中的代码

public function index()
{
    $data['check_availability'] = $this->check_availability_model->check_availability();
    $data['title'] = 'Available Roooms';

    $this->load->view('templates/header', $data);
    $this->load->view('check_availability/index', $data);
    $this->load->view('templates/footer');
}

在views/check_availability/index.php 文件下，当我根据POST 方法（如$type = $_POST['type']）调用check_availability 函数时，我想过滤room_type 字段。

目前，这是我用来从 lf_rooms 表中查询所有记录的代码：

<?php foreach ($check_availability as $check_availability_item): ?>

但是如何将 $type 变量的值传递给 check_availability 函数？

4

1 回答 1

1

您应该POST在控制器中获取参数，相应地查询数据库并仅在视图中显示结果。该视图不应包含任何逻辑。

public function index()
{
    $type = $this->input->post('type', TRUE);

    $data['check_availability'] = $this->check_availability_model->
                                  check_availability($type);
}
于 2012-08-12T12:37:25.077 回答