在CodeIgniter中验证表单后未设置参数

Question

我通过$idURL 传递一个变量,并以表单形式加载数据,但是当我要验证表单时,$id无法识别该值.

function myfunction() {
    $id = $this->uri->segment(4); 
    echo $id;   //yes, print the value  

    $data_user = $this->admin_model->query_data_user($id);
    $direccion = $data_user[0]->address;
    $phone = $data_user[0]->phone;

    $this->data['address'] = array(
            'name' => 'address',
            'id' => 'address',
            'value' => $address,
            'class' => 'input',
    );

    $this->data['phone'] = array(
            'name' => 'phone',
            'id' => 'phone',
            'value' => $phone,
            'class' => 'input',
    );

    echo $id;  //yes, print the value

    $this->form_validation->set_rules('address', 'Address', 'xss_clean|max_length[100]');
    $this->form_validation->set_rules('phone', 'Phone', 'required|xss_clean|max_length[20]|is_natural_no_zero');

    if ($this->form_validation->run() == true) {
        echo $id;  //here NOT PRINT $id
        $data = array(
                'address' => $this->input->post('address'),
                'phone' => $this->input->post('phone'),
        );

        $id = $this->uri->segment(4); 
        echo $id;  //here not print the value $id

        $this->ion_auth->update_user($id, $data); 
        $this->load->view('includes/template_mensajes', $data);
    }
    $this->load->view('users/update_user', $this->data);
}

在$id验证我的表单时,无法识别该值,我的错误是什么？

Answer 1

你在哪里提交表格.您确定要在表单的action属性的URI中传递ID.

例如,表单打开标记看起来应该是这样的

<form action="/admin/myfunction/<?= $id ?>">

或使用codeigniter助手

echo form_open('admin/myfunction/' . $id);