会不会因Rails 5.1而分页？

Question

自从升级到Rails 5.1以来，在尝试对未过滤（但未过滤）的用户表结果进行分页时，出现以下错误：

我使用will-paginate 3.1.6和Rails 5.1.0。我什至不知道是哪条线引起了错误。大概是这条线

ReservationService

由于错误上方，我看到一个数据库调用，要求偏移25限制为25的用户。

我什至尝试为（标准ActiveRecord）@WebMvcTest @RunWith(SpringRunner.class) public class MvcTest { @Autowired private MockMvc mockMvc; @MockBean(name = "reservation") private ReservationService reservationService; //my chnage here @Test public void postReservation() throws Exception { // You need that to specify what should your mock return when getReservation() is called. Without it you will get null when(reservationService.getReservation()).thenReturn(new Reservation()); //my chnage here mockMvc.perform(MockMvcRequestBuilders.post("/reservation")) .andExpect(MockMvcResultMatchers.content().contentType(MediaType.APPLICATION_JSON_UTF8_VALUE)) .andExpect(MockMvcResultMatchers.status().isOk()); } } 模型创建format.html { @members = @members.page(params[:page]); render } 方法，但这将错误更改为抱怨，因为它不是字符串。

我需要将呼叫更改为will_paginate吗？我需要更早版本的will_paginate吗？如何使其与Rails 5.1一起使用？

编辑：这是@members填充的方式：

empty?

Answer 1

错误很可能是在视图中，您将User的单个实例传递给了page_entries_info而不是集合（您没有提供视图和整个控制器动作，因此很难分辨）

可执行错误示例如下：

#!/usr/bin/env ruby
# frozen_string_literal: true

require "bundler/inline"

install = true
gemfile(install) do
  source "https://rubygems.org"
  gem "rails", "5.1.0"
  gem "will_paginate", '3.1.6'
  gem "sqlite3"
end

require "active_record"
require "action_controller/railtie"
require "will_paginate"
require "logger"
require "rack/test"

# we are skipping initializers so include manually:
require 'will_paginate/active_record'
require 'will_paginate/view_helpers/action_view'

ActiveRecord::Base.establish_connection(adapter: "sqlite3", database: ":memory:")
ActiveRecord::Schema.define do
  create_table :users, force: true do |t|
    t.string :country
  end
end

class User < ActiveRecord::Base
  self.per_page = 25
end

class TestApp < Rails::Application
  config.root = __dir__
  config.session_store :cookie_store, key: "cookie_store_key"
  secrets.secret_key_base = "secret_key_base"

  config.logger = Logger.new($stdout)
  ActiveRecord::Base.logger = config.logger
  Rails.logger  = config.logger

  routes.draw{ get "/" => "members#index" }
end

class MembersController < ActionController::Base
  include Rails.application.routes.url_helpers

  def index    
    @members = User.order("id DESC")
    @members = @members.page(params[:page])
    respond_to do |format|
      format.html{
        # passing single instance instead of collection:
        render inline: "<%= page_entries_info(@members.first) %>"
      }
    end
  end
end

require "minitest/autorun"

class BugTest < Minitest::Test
  include Rack::Test::Methods

  def test_returns_success
    User.create!
    get "/?page=1"
    assert last_response.ok?
    puts last_response.body
  end

  private
    def app
      Rails.application
    end
end

Answer 2

我相信你应该改变

@members = @members.page(params[:page])

到

@members = User.page(params[:page])

但当然取决于@members的设置。如果以前根据某些搜索条件将@members设置为User的数组，则不应实例化User对象，而应将关系保留在@members中。

Answer 3

更改您认为导致此问题的行（在您的控制器中）：

format.html { @members = @members.page(params[:page]); render }

对此（如果您也使用JavaScript）：

@members = Member.paginate(page: params[:page]).order("id DESC")
respond_to do |format|
  format.html
  format.js
end

或者这个（如果您愿意的话，可以使用原始版本）：

format.html { @members = Member.paginate(page: params[:page]); render }

在您看来，请尝试以下操作：

<% if current_user.admin_level < 8 %>
  <%= will_paginate @members.where("country = ?", current_user.country) %>
<% end %>

Answer 4

可能不是这种情况，但是I've been reading the gem in question，不是吗？？：

format.html { @members = @members.paginate(page: params[:page]); render }