Question

我希望能够将“params”传递给“is_following”方法，如下所示：

       respond_to |format|
          format.json { render json: @user, :methods => [:is_following params] }
       end

但是，当我在方法名称后加上“params”时，会抛出错误：

syntax error, unexpected tIDENTIFIER, expecting ']'

当我渲染JSON（没有is_following方法）时，它看起来像这样：

[{"id":81,"image_url":"https://graph.facebook.com/123213123/picture?type=large","full_name":"John Johnson"},{"id":85,"image_url":"https://graph.facebook.com/123123123/picture?type=large","full_name":"Bill Nye"}]

我想添加is_following方法，以便像这样的JSON loos：

[{"id":81,"image_url":"https://graph.facebook.com/123213123/picture?type=large","full_name":"John Johnson",'is_following'=>4},{"id":85,"image_url":"https://graph.facebook.com/123123123/picture?type=large","full_name":"Bill Nye",'is_following'=>9}]

修改

def friends
  @user = User.first
  respond_to do |format|
    format.json { render json: @user.friends.order("first_name ASC"), :methods => [:is_following], :only => [:id] }
  end
end

模型中的

和is_following方法是：

def is_following params
    return Friendship.where("user_id = ? AND friend_id = ?", params[:user_id], self.id).count
end

任何人都知道如何解决这个问题？我会给你一百万美元。

Answer 1

布赖恩，

只需更改此代码：

:methods => [:is_following params]

到

:is_following => is_following(params)

修改

无论如何，这是一种更好的方法：

def friends
  @users = User.find_by_sql("SELECT users.*, 
      count(distinct friendships.friend_id) as is_following FROM users
      LEFT JOIN friendships ON users.id = friendships.user_id
      WHERE users.id = " + params[:user_id] + "
      GROUP BY users.id
      ORDER BY users.name ASC")

  respond_to do |format|
    format.json { render json: @users }
  end
end

添加参数来渲染方法

1 个答案: