Question

我有这个按钮

 <a class="btn btn-sm btn-success" href="javascript:void(0)" title="info" onclick="delete_province('."'".$info->ID."'".')"><i class="glyphicon glyphicon glyphicon-question-sign"></i> Show File</a>';

我想用这样的东西改变那个按钮功能

<a href = '<?php echo base_url().'Employees/updatepersonalinfo/'.$data; ?>' class = 'btn btn-info pull-right fa fa-edit'>&nbsp Update</a>

请注意这里

echo base_url().'Employees/updatepersonalinfo/'.$data;

我在网址上传递数据，以便我使用函数$this->uri->segment()

如何使按钮像href一样工作？还有其他任何方式，模板不再加载只是一些div？提前谢谢

以及我要传递的.$data来自json？

继续使用附加代码

这是带按钮的控制器

public function index()
{

    $data['content_view'] = 'TaxValues/taxvalues_read';
    $this->templates->admin_template($data);
}

public function ajax_list()
{
    $list = $this->TaxValues_Model->get_datatables();
    $data = array();
    $no = $_POST['start'];
    foreach ($list as $info) {
        $no++;
        $row = array();
        $row[] = $info->RANGE_NO;
        $row[] = $info->TAX;
        $row[] = $info->PERCENTAGE." %";
        $row[] = $info->VALUE;
        $row[] = $info->FREQUENCY;

        //add html for action
        $row[] = '<a class="btn btn-sm btn-primary" href="javascript:void(0)" title="Edit" onclick="edit_provinces('."'".$info->TAX_ID."'".')"><i class="glyphicon glyphicon-pencil"></i> Edit</a>
              <a class="btn btn-sm btn-danger" href="javascript:void(0)" title="Delete" onclick="delete_province('."'".$info->TAX_ID."'".')"><i class="glyphicon glyphicon-trash"></i> Delete</a>';

        $data[] = $row;
    }

    $output = array(
                    "draw" => $_POST['draw'],
                    "recordsTotal" => $this->TaxValues_Model->count_all(),
                    "recordsFiltered" => $this->TaxValues_Model->count_filtered(),
                    "data" => $data,
            );
    //output to json format
    echo json_encode($output);
}

我想传递$info->TAX_ID，例如，我想访问我现有的更新代码，

更新的控制器功能

public function updateinfo(){

    $data['content'] = $this->TaxValues_Model->info();
    $data['content_view'] = 'TaxValues/edit';
    $this->templates->admin_template($data);

}

更新的模型功能

public function info(){

    $query = $this->db->get_where('emp_fi_p', array('ID_NUM' => $this->uri->segment(3)));
    return $query->row();
}

像Iref一样工作的CodeIgniter按钮

0 个答案: