How to update a record in an intermediate table?

Question

Yii

There are tables of news, projects and an intermediate table of news connection with the project and news. It is not entirely clear how this link () works, I would like an example specifically in my case. Now, tables:
news
id | title
projects
id | title
project_news
project_id | news_id
I create a post like this:

$model = new News(['scenario' => News::SCENARIO_CREATE]);
$project = new ProjectNews(); // Модель таблица project_news
$projectsList = Project::find()->all(); // Для вывода списка проектов
....
if($project->load(\Yii::$app->request->post())) {
 $project->news_id = $model->id;
 $project->save();
}
//  Всё ок, запись создаются, в проекте новости выводятся и т.д

But what if you need to change the project to which the news is attached?

$model = News::findOne(['id' => $id]);
$model->scenario = News::SCENARIO_UPDATE;

$project = ProjectNews::findOne(['news_id' => $id]); // Поиск записи в БД
$projectsList = Project::find()->all(); // Для вывода списка проектов

// Если
$projectId = \Yii::$app->request->post('ProjectNews');
$project->project_id = $projectId['project_id'];
$project->update(); // Ругается на отсутсвие PK

How to update a record in an intermediate table?

Answer 1

Use ManyToMany Behavior it solves the tasks you need

Answer 2

Good evening.
In the intermediate table, assign PK to two fields.
When creating a record, write data in both fields of the intermediate table.
When updating, use save(false) instead of update()