小编gau*_*tam的帖子

当用户尝试在我的网站上使用 google 进行注册时，Google 会向用户显示上述 3 个弹出窗口，但是当我尝试注册时，它不会向我显示相同的弹出窗口。我看到的就是这个——

这 3 个独立的弹出窗口导致我的网站大量流失。如何删除这 3 个不同的权限弹出窗口？

Google 登录按钮详细信息 - 使用React Google 登录

    <div className="manager-signup-container">
      <div className="content">Join Peoplebox as a Manager to have productive 1:1s that drive actions.</div>
      <GoogleLogin
      className={`google-btn-${this.state.loginbtnStatus}`}
      clientId={GOOGLE_CLIENT_ID}
      buttonText="Sign up with Google"
      onSuccess={(response) => this.responseOauth(response, "google")}
      onFailure={this.responseGoogleFail}
      responseType="code"
      scope="profile email https://www.googleapis.com/auth/calendar.events https://www.googleapis.com/auth/calendar.readonly"
      accessType="offline"
      prompt="consent"
      disabled={this.state.loginDisabled}
    />

我用来'active_model_serializers', '~> 0.10.6'渲染我的 API 响应。为了我的index行动，我正在这样做 -

render json: @items, root: 'data', each_serializer: ItemsSerializer

但在我的回复中，我没有得到root key-data

[
  {
    "id": 85,
    "title": "B",
    "source": "manager_added",
    "shared": true,
    "status": "suggested",
    "item_type": "action_item",
    "manager": {
        "id": 2614,
        "full_name": "Calvin H",
        "first_name": "Calvin"
    },
    "reportee": {
        "id": 2614,
        "full_name": "Calvin H",
        "first_name": "Calvin"
    }
  },
  {
    "id": 87,
    "title": "D",
    "source": "manager_added",
    "shared": true,
    "status": "suggested",
    "item_type": "action_item",
    "manager": {
        "id": 2614,
        "full_name": "Calvin H",
        "first_name": "Calvin"
    }, …

在 Rails 6 应用程序中，我有以下代码。

class Reservation < ApplicationRecord
  after_create     :publish_creation

  def publish_creation
    Publishers::Reservations::CreateJob.perform_later(self)
  end
end

class Publishers::Reservations::CreateJob < ApplicationJob
  queue_as :default

  def perform(reservation)
    puts reservation.inspect #dummy code for testing
  end
end

大多数时候我创建一个Reservation（记录总是在数据库中创建），我收到以下错误

2020-02-14T09:54:03.707Z pid=81787 tid=1xmj 警告： ActiveJob::DeserializationError: Error while trying to deserialize arguments: Couldn't find Reservation with 'id'= 35651cf7-35bc-4da0-bb0d-6285ac093d22

Sidekiq 第一次重试处理作业时，它总能找到Reservation并且一切正常。

Reservation id: "35651cf7-35bc-4da0-bb0d-6285ac093d22", analysis_id: "6b3b167b-1279-49c0-991a-b580c375fd0f", reservable_type: "User", reservable_id: "94f60c16-29d4-4372-983b-7544c393a7e6", reserved_between: 2020-02-10 08:00:00 UTC..2020-02-10 08:10:00 UTC, state: "scheduled", created_at: "2020-02-14 10:02:28", updated_at: "2020-02-14 10:02:28"

我在这里错过了什么吗？这是否与我在development …

"aaabbb".chars.to_a.permutation.map(&:join).uniq 

正确工作并提供所需的输出,即使用给定字符串可以形成的所有单词,但如果字符串是"cdcdcdcdeeeef"或"cdefghmnopqrstuvw"而不是"aaabbb",则相同的代码将失败.我刚试过这三个测试用例,其中三个失败.

"cdefghmnopqrstuvw".chars.to_a.permutation.map(&:join).uniq 

上面的代码行不会返回任何结果.它似乎进入了无限循环.

出了什么问题？