Question

我的控制器中有两个方法可以找到用户（注意enabled_only范围）：

before_filter :find_user, :only => :show
before_filter :find_any_user, :only => [:edit, :update, :destroy]

def find_user
  @user = User.enabled_only.find(params[:id])
rescue ActiveRecord::RecordNotFound
  flash[:alert] = "The user you were looking for could not be found"
  redirect_to root_path
end

def find_any_user
  @user = User.find(params[:id])
rescue ActiveRecord::RecordNotFound
  flash[:alert] = "The user you were looking for could not be found"
  redirect_to root_path
end

当然，这些可以合并为一种方法，检查是否:action == 'show'，但我无法解决问题。我尝试了类似下面的内容，但它不起作用：

before_filter :find_user, :only => [:show, :edit, :update, :destroy]

def find_user
  @user = if :action == 'show'
    User.enabled_only.find(params[:id])
  else
    User.find(params[:id])
  end
rescue ActiveRecord::RecordNotFound
  flash[:alert] = "The user you were looking for could not be found"
  redirect_to root_path
end

请告知如何做到这一点。

由于

Answer 1

您需要在begin和rescue

之间包装您想要“保护”的代码

before_filter :find_user, :only => [:show, :edit, :update, :destroy]

def find_user
  begin
    @user = if :action == 'show'
      User.enabled_only.find(params[:id])
    else
      User.find(params[:id])
    end
  rescue ActiveRecord::RecordNotFound
    flash[:alert] = "The user you were looking for could not be found"
    redirect_to root_path
  end
end

顺便说一句，你的测试:action == 'show'永远不会成真。 :action是一个符号，其值为:action，其值永远不会改变，'show'相同，其值永远不会改变。我不确定实现这一目标的最佳方式是什么，但你可以这样做，但你可以做if params[:action] == "show"

铁路如果否则检查失败然后救援

1 个答案: