php - 如何使用 codeigniter 显示成功消息？

Question

我有一个布局，当我按下提交按钮时，该页面的数据将保存到数据库中。我可以在同一页面中打印我的表单验证错误，但我无法将我的成功消息打印到此页面中，我可以将成功消息打印到另一个页面中，如何将消息显示到同一页面中请帮助。

这是 Controller :

    function addRoom(){
        $data['title'] = 'Add Room';
        $data['main_content'] = 'config/addRoom';
        $data['roomlists'] = $this->config_mdl->get_room_info();
        $this->load->view('_base/layout', $data);
    }

    function roomAdd(){
        $this->form_validation->set_rules('roomType', 'Room Type', 'trim|required|xss_clean');
        $this->form_validation->set_rules('roomName', 'Room Name', 'trim|required|xss_clean');
        $this->form_validation->set_rules('roomDetails', 'Room Details', 'trim|required|xss_clean');
        $this->form_validation->set_error_delimiters('<div class="alert alert-danger" role="alert">', '</div>');

        if ($this->form_validation->run() == FALSE) {

            $this->addRoom();

        } else {
            $roomdata = array('room_type' => $this->input->post('roomType'),
                             'room_name' => $this->input->post('roomName'),
                             'room_details' => $this->input->post('roomDetails'));
            $this->config_mdl->roomAdd($roomdata);

        }           
    }

这是我的模型:

    function roomAdd($roomdata)
    {
        return $this->db->insert('tbl_room_info', $roomdata);
    }

这是我的看法:

    <div class="panel-body">
    <?php echo validation_errors(); ?>
    <?php $attributes = array('class' => 'form-horizontal', 'role' => 'form');
                        echo form_open_multipart('config/roomAdd', $attributes);
                    ?>
    <div class="form-group">
        <label for="roomType" class="col-sm-2 control-label">Room type</label>
        <div class="col-sm-2">
        <select name="roomType" class="form-control">
        <option>Select</option>
        <option value="1">1</option>
        <option value="2">2</option>
        </select>
        </div>
    </div>
    <div class="form-group">
        <label for="roomName" class="col-sm-2 control-label">Room Name</label>
        <div class="col-sm-6">
        <input type="text" class="form-control" name="roomName" placeholder="Type Room Name" />
        </div>
    </div>
    <div class="form-group">
        <label for="roomDetails" class="col-sm-2 control-label">Room Details</label>
        <div class="col-sm-6">
        <textarea class="form-control" name="roomDetails" rows="3"></textarea>
        </div>
    </div>
    <div class="form-group">
        <div class="col-sm-offset-2 col-sm-6">
        <button type="submit" class="btn btn-primary">Submit</button>
        </div>
    </div>
    <?php echo form_close(); ?>
    </div>

Answer 1

只需添加一个flash data插入后(一次 session 使用):

} else {
    $roomdata = array(
        'room_type' => $this->input->post('roomType'),
        'room_name' => $this->input->post('roomName'),
       'room_details' => $this->input->post('roomDetails')
    );

    $this->config_mdl->roomAdd($roomdata);

    // set flash data
    $this->session->set_flashdata('msg', 'Room added');
    redirect('controller_name/addRoom');
}  

然后在 View 中(addRoom):

<?php if($this->session->flashdata('msg')): ?>
    <p><?php echo $this->session->flashdata('msg'); ?></p>
<?php endif; ?>