在Laravel 5.1中将数据从视图传递到控制器

Question

我想将参数从刀片视图传递到控制器中的函数

index.blade.php

<a href="{{ route('like', [$post->id, 1])  ) }}" class="like">
    Like
</a>

<a href="{{ route('like', [$post->id, -1])  ) }}" class="like">
    Dislike
</a>

PostController.php

  public function getLikePost($post_id, $like_value)
    {
       $post = Post::find($post_id);
       ...
    }

routes.php

Route::get('like', [
        'uses' => 'PostController@getLikePost',
        'as'   => 'like'
      ]);

但我收到一条错误消息

ErrorException in PostController.php line 149:
Missing argument 2 for App\Http\Controllers\PostController::getLikePost()

谁能帮我解决这个问题？

Answer 1

您的路线应为：

Route::get('like/{psot_id}/{like_value}', [
        'uses' => 'PostController@getLikePost',
        'as'   => 'like'

在您看来：

<a href="{{ route('like', ['post_id' => $post->id, 'like_value' => 1]) }}" class="like">
    Like
</a>

<a href="{{ route('like', ['post_id' => $post->id, 'like_value' => -1]) }}" class="like">
    Dislike
</a>

Answer 2

尝试这个：

Route.php

Route::get('like/{post_id}/{like_value}', [
        'uses' => 'PostController@getLikePost',
        'as'   => 'like'
      ]);

index.blade.php

<a href="{{ route('like', ['post_id' => $post->id, 'like_value' => 1]) }}" class="like">
    Like
</a>

<a href="{{ route('like', ['post_id' => $post->id, 'like_value' => -1]) }}" class="like">
    Dislike
</a>

文件

Answer 3

通过空值初始化$ post_id和$ like_value，如下所示。

PostController.php

public function getLikePost($post_id = '', $like_value='')
    {
       $post = Post::find($post_id);
       ...
    }