如何在 Codeigniter 4 的单个控制器中访问两个不同的模型?
how to access two different model in single controller in Codeigniter 4?
我是 Codeigniter 4 的新手和初学者,所以我问这个问题。
其实我有两个不同的模型。
首先是tbl_user.
里面有很多字段 tbl_user(id, name,duty_station,enlistment_date,dob) //这里id为主键.
其次是usermos.
usermos有四个字段[id, user_id, mos_id, additional_mos_id] //这里id是主键,user_id是外键到tbl_user
我生成了get_profile的控制器...
我想将 usermos
模型访问到 get_profile
控制器中。
以下是我的代码
Headers
<?php namespace App\Controllers;
use CodeIgniter\RESTful\ResourceController;
use App\Models\MosModel;
use App\Models\Additional_mosModel;
use App\Models\Api_auth_model;
use App\Models\Api_Usermos_Model;
控制器
public function get_profile()
{
if (($this->request->getMethod() == 'post') && ($_SERVER['PHP_AUTH_USER'] == AUTHUSER_NAME) && ($_SERVER['PHP_AUTH_PW'] == AUTH_PASSWORD)) {
if(!empty($this->request->getPost('id'))){
$user_api_mos_model = new Api_Usermos_Model();
$usermos = $this->user_api_mos_model->findAll();
print_r($usermos);
exit;
if($profile){
if($usermos){
}
$selected_mos = $this->getMosFromID($usermos[0]['mos_id']);
$selected_add_mos = $this->getAdd_MosFromID($usermos[0]['additional_mos_id']);
$profile['badge'] = $this->getBadgeFromID($profile['badge_id']);
// exit;
return $this->respond([
"status" => "Success",
"message" => "Profile found.",
"Common" => ["Title" => "Load Profile API", 'version' => '1.0', 'Description' => 'Load Profile API', 'Method' => 'POST'],
"Response" => ["Userdata" => $profile,"mos" => $selected_mos,"additonal_mos" => $selected_add_mos]
]);
}else{
return $this->respond([
"status" => "Fail",
"message" => "Profile Not found.",
"Common" => ["Title" => "Load Profile API", 'version' => '1.0', 'Description' => 'Load Profile API', 'Method' => 'POST'],
"Response" => ["Value" => 'Profile Not found.']
]);
}
}
}
}
感谢您的宝贵时间。
只需在控制器页面 header 部分的顶部定义模型的命名空间,然后在该控制器页面中使用。
例如:
<?php namespace App\Controllers;
use app\models\User;
use app\models\Usermos;
Class UserController extend controller {
public function actionProfile (){
$user = new User();
$user_mos = new Usermos();
}
}
我是 Codeigniter 4 的新手和初学者,所以我问这个问题。
其实我有两个不同的模型。
首先是tbl_user.
里面有很多字段 tbl_user(id, name,duty_station,enlistment_date,dob) //这里id为主键.
其次是usermos.
usermos有四个字段[id, user_id, mos_id, additional_mos_id] //这里id是主键,user_id是外键到tbl_user
我生成了get_profile的控制器...
我想将 usermos
模型访问到 get_profile
控制器中。
以下是我的代码
Headers
<?php namespace App\Controllers;
use CodeIgniter\RESTful\ResourceController;
use App\Models\MosModel;
use App\Models\Additional_mosModel;
use App\Models\Api_auth_model;
use App\Models\Api_Usermos_Model;
控制器
public function get_profile()
{
if (($this->request->getMethod() == 'post') && ($_SERVER['PHP_AUTH_USER'] == AUTHUSER_NAME) && ($_SERVER['PHP_AUTH_PW'] == AUTH_PASSWORD)) {
if(!empty($this->request->getPost('id'))){
$user_api_mos_model = new Api_Usermos_Model();
$usermos = $this->user_api_mos_model->findAll();
print_r($usermos);
exit;
if($profile){
if($usermos){
}
$selected_mos = $this->getMosFromID($usermos[0]['mos_id']);
$selected_add_mos = $this->getAdd_MosFromID($usermos[0]['additional_mos_id']);
$profile['badge'] = $this->getBadgeFromID($profile['badge_id']);
// exit;
return $this->respond([
"status" => "Success",
"message" => "Profile found.",
"Common" => ["Title" => "Load Profile API", 'version' => '1.0', 'Description' => 'Load Profile API', 'Method' => 'POST'],
"Response" => ["Userdata" => $profile,"mos" => $selected_mos,"additonal_mos" => $selected_add_mos]
]);
}else{
return $this->respond([
"status" => "Fail",
"message" => "Profile Not found.",
"Common" => ["Title" => "Load Profile API", 'version' => '1.0', 'Description' => 'Load Profile API', 'Method' => 'POST'],
"Response" => ["Value" => 'Profile Not found.']
]);
}
}
}
}
感谢您的宝贵时间。
只需在控制器页面 header 部分的顶部定义模型的命名空间,然后在该控制器页面中使用。 例如:
<?php namespace App\Controllers;
use app\models\User;
use app\models\Usermos;
Class UserController extend controller {
public function actionProfile (){
$user = new User();
$user_mos = new Usermos();
}
}