我有以下三个表:
ADDRESSES
------------
// Usual Address Data Fields
People
-------------
// Usual Personal Info
AddressId - FK
RoleId - FK
Role
--------------
RoleId
Name
一个Role可以有多个People，每个Person只能属于一个Role。至于地址，每个人将被链接到一个地址，每个地址将只分配给一个人。
我想做的是Register a Person。Registration表单必须收集所有三个表的数据，并以一种不会使MySQL抛出错误的方式存储它们。
我的表单看起来不错，我有角色填充下拉菜单。但是我的问题是关于提交表单时到达的POST请求的处理。下面是我的POST处理代码:
public function register()
{
$this->load->library('form_validation');
$this->form_validation->set_rules('FirstName', 'First Name', 'trim|required');
$this->form_validation->set_rules('LastName', 'Last Name', 'trim|required');
$this->form_validation->set_rules('Telephone', 'Telephone', 'trim|required');
$this->form_validation->set_rules('Email', 'Email', 'trim|required|valid_email');
$this->form_validation->set_rules('Password', 'Password', 'trim|required');
$this->form_validation->set_rules('MaxNo', 'MaxNo', 'trim|required');
$this->form_validation->set_rules('FirstLine', 'First Line', 'trim|required');
$this->form_validation->set_rules('SecondLine', 'Second Line', 'trim');
$this->form_validation->set_rules('City', 'City', 'trim|required');
$this->form_validation->set_rules('PostCode', 'Post Code', 'trim|required');
if($this->form_validation->run())
{
echo "Validation Failed";
// something went wrong.
$this->load->model('rolesmodel');
$data['roles'] = $this->rolesmodel->getRolesForUserRegForm();
$this->load->view('shared/header');
$this->load->view('charity/register', $data);
$this->load->view('shared/footer');
} else {
// All is Good
$address = array(
'FirstLine' => $this->input->post('FirstLine'),
'SecondLine' => $this->input->post('SecondLine'),
'City' => $this->input->post('City'),
'PostCode' => $this->input->post('PostCode'),
);
$this->load->model('addresses_model');
$addId = $this->addresses_model->getCurrentCountOfRows();
$this->addresses_model->addNewsAddress($address);
$person = array(
'FirstName' => $this->input->post('FirstName'),
'LastName' => $this->input->post('LastName'),
'Telephone' => $this->input->post('Telephone'),
'Email' => $this->input->post('Email'),
'Password' => $this->input->post('Password'),
'MaxNo' => $this->input->post('MaxNo'),
'AddressId' => (int)$addId + 1,
'RoleId' => $this->input->post('Name') // this is the Roles DropDown Name Property
);
$this->load->model('people_model');
$this->people_model->registerPerson($person);
redirect('/animals/index');
}
}
我知道我必须首先将地址保存到数据库中，但是如果我这样做，我如何才能快速获取其ID?因为它不会有一个ID，除非它首先保存到数据库中。
其次，我如何处理角色下拉选项?
要获得插入后的地址ID，使模型中的函数返回:
return $this->db->insert_id();
为了得到你的下拉值，在你的视图中你会有一个选择。可能是这样的:
<select name="roleId">....</select>
使用这个来获取控制器中的roleId:
$person->RoleId = $this->input->post('roleId')
注意，post()方法中的字符串必须与select
的'name'属性一致将您的代码更改为:
$this->load->model('PeopleModel');
$this->load->model('AddressModel');
$person->AddressId = $this->AddressModel->insert($address);
$this->PeopleModel->register($person);
希望这对你有帮助，不要犹豫，问更多的问题=)
我将简单地这样做:(这必须是你的寄存器函数，只是伪代码)
protected function getAddressId($param = array()){
//save address data here and return it's id;
}
public function register($param = array()){
//1. grab address id
$AdsressId = $this->getAddressId()
//2. save person data + $AddressId + RoleId
//3. All done
}