Php 试求非对象雄辩关系的性质

我已经在App\Post.php文件中编写了这段代码

public function user(){

    return $this->belongsTo('App\User');

}

我在routes.php中编写了它

Route::get('/{id}', function($id){

    return Post::find($id)->user->name;

});

public function posts(){
    return $this->hasMany(Post::class);
}

public function user(){
    return $this->belongsTo(User::class);
}

public function user()
    {
      return $this->belongsTo('App\User', 'the_foreign_key_Post_in_users', 'primary_key_in_Post');
    }

public function post(){
    return $this->hasMany('App\Post','the_foreign_key_POst_in_users');
  }

当我去测试代码时，我得到了以下错误

正在尝试获取非对象的属性

---

嗯。让我们重新定义您的模型

In-App\User.php

Route::get('/{id}', function($id){

    return Post::find($id)->user->name;

});

public function posts(){
    return $this->hasMany(Post::class);
}

public function user(){
    return $this->belongsTo(User::class);
}

public function user()
    {
      return $this->belongsTo('App\User', 'the_foreign_key_Post_in_users', 'primary_key_in_Post');
    }

public function post(){
    return $this->hasMany('App\Post','the_foreign_key_POst_in_users');
  }

In-App\Post.php

Route::get('/{id}', function($id){

    return Post::find($id)->user->name;

});

public function posts(){
    return $this->hasMany(Post::class);
}

public function user(){
    return $this->belongsTo(User::class);
}

public function user()
    {
      return $this->belongsTo('App\User', 'the_foreign_key_Post_in_users', 'primary_key_in_Post');
    }

public function post(){
    return $this->hasMany('App\Post','the_foreign_key_POst_in_users');
  }

这样您就可以像

//Get User from Post
$user = Post::find($id)->user;

//Get Post collection from User
$posts = User::find($id)->posts

---

请检查您的结果和代码，如下所示：

您的Post：：查找（$id）；返回值还是空？如果它是空的，那么您需要检查您的表，可能该id不存在

检查表中的每个id，并按如下方式优化模型：

App\Post.php

Route::get('/{id}', function($id){

    return Post::find($id)->user->name;

});

public function posts(){
    return $this->hasMany(Post::class);
}

public function user(){
    return $this->belongsTo(User::class);
}

public function user()
    {
      return $this->belongsTo('App\User', 'the_foreign_key_Post_in_users', 'primary_key_in_Post');
    }

public function post(){
    return $this->hasMany('App\Post','the_foreign_key_POst_in_users');
  }

App\User.php

Route::get('/{id}', function($id){

    return Post::find($id)->user->name;

});

public function posts(){
    return $this->hasMany(Post::class);
}

public function user(){
    return $this->belongsTo(User::class);
}

public function user()
    {
      return $this->belongsTo('App\User', 'the_foreign_key_Post_in_users', 'primary_key_in_Post');
    }

public function post(){
    return $this->hasMany('App\Post','the_foreign_key_POst_in_users');
  }

---

Post:：find（$id）

或

Post:：find（$id）->用户

不是object@Scuzzy我怎样才能使它成为一个对象呢？这取决于

Post:：find（$id）

返回的是什么，它实际上是在查找什么，还是返回布尔值false等等？