Question

我正在尝试将$request从控制器中的函数传递给模型中的函数。

这是我的控制器功能：

PostController.php

public function store(Request $request, post $post)
{

    $post->title = $request->title;
    $post->description = $request->description;

    $post->save();

    return redirect(route('post.index'));

}

如何在模型Post.php中保存数据？

我希望控制器仅充当发送信息的角色。信息被发送到模型。所有的计算和存储都在模型中进行

谢谢

Answer 1

您可以使其变得更加容易。 Laravel有自己的助手“ request（）”，可以在代码中的任何地方调用它。

因此，通常，您可以这样做：

PostController.php

    public function store()
    {
        $post_model = new Post;
        // for queries it's better to use transactions to handle errors
        \DB::beginTransaction();
        try {
           $post_model->postStore();
           \DB::commit(); // if there was no errors, your query will be executed
        } catch (\Exception $e) {
            \DB::rollback(); // either it won't execute any statements and rollback your database to previous state
            abort(500);
        }
        // you don't need any if statements anymore. If you're here, it means all data has been saved successfully
        return redirect(route('post.index'));
    }

Post.php

    public function postStore()
    {
        $request = request(); //save helper result to variable, so it can be reused
        $this->title = $request->title;
        $this->description = $request->description;
        $this->save();
    }

我将向您展示有关更新和创建的完整最佳实践示例：

web.php

    Route::post('store/post/{post?}', 'PostController@post')->name('post.store');

yourform.blade.php -可用于更新和创建

    <form action='{{ route('post.store', ['post' => $post->id ?? null]))'>
       <!-- some inputs here -->
       <!-- some inputs here -->

    </form>

PostController.php

    public function update(Post $post) {
        // $post - if you sent null, in this variable will be 'new Post' result
        // either laravel will try to find id you provided in your view, like Post::findOrFail(1). Of course, if it can't, it'll abort(404)

        // then you can call your method postStore and it'll update or create for your new post.
        // anyway, I'd recommend you to do next

        \DB::beginTransaction();
        try {
          $post->fill(request()->all())->save();
          \DB::commit();
        } catch (\Exception $e) {
          \DB::rollback();
           abort(500);
        }
      return redirect(route('post.index'));
    }

Answer 2

您只需简单地实例化即可创建新模型：

$post = new Post;  //Post is your model

然后将内容记录下来

$post->title = $request->title;
$post->description = $request->description;

，最后将其保存到数据库中：

$post->save();

要使用create方法将所有数据保存在模型中。使用create并在模型的fillable属性中设置列时需要设置质量分配。

  protected $fillable = [ 'title', 'description' ];

然后使用输入内容调用

    $post = Post::create([ 'parametername'  => 'parametervalue' ]);

and if request has unwanted entries like token then us except on request before passing.

    $post = Post::create([ $request->except(['_token']) ]);

希望这会有所帮助。

Answer 3

根据描述，不确定要确切确定什么，但假设您想要一个干净的控制器和模型。这是一种方法

模型-发布

class Post {
    $fillable = array(
        'title', 'description'
    );
}

PostController

class PostController extend Controller {

    // store function normally don't get Casted Objects as `Post`
    function store(\Request $request) {

        $parameters = $request->all(); // get all your request data as an array

        $post = \Post::create($parameters); // create method expect an array of fields mentioned in $fillable and returns a save dinstance
        // OR
        $post = new \Post();
        $post->fill($parameters);
    }
}

我希望对您有帮助

Answer 4

我发现可以回答我的问题：

将$request传递到模型Post.php中的my_method：

PostController.php：

    public function store(Request $request)
    {
        $post_model = new Post;
        $saved = $post_model->postStore($request);

        //$saved = response of my_method in model

        if($saved){
            return redirect(route('post.index'));
        }
    }

并将数据保存在模型中：

Post.php

我们可以将实例或布尔值返回给控制器。

我将bool（保存方法响应）返回给控制器：

    public function postStore($request)
    {
        $this->title = $request->title;
        $this->description = $request->description;

        $saved = $this->save();

        //save method response bool

        return $saved;

    }

通过这种方式，所有计算和存储都在模型中执行（将数据保存在MVC中的最佳方式）

Answer 5

 public function store(Request $request)
{
    $book = new Song();
    
    $book->title = $request['title'];
    $book->artist = $request['artist'];
    $book->rating = $request['rating'];
    $book->album_id = $request['album_id'];
    
    $result=  $book->save();
 }

laravel 5.7如何传递控制器的请求进行建模和保存

5 个答案:

模型-发布

PostController