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ruby-on-rails-4 – 如何在rails中使用带有ransack的will_paginate

来源:互联网 收集:自由互联 发布时间:2021-06-23
我正在使用ransack进行搜索,现在我想在我的rails应用程序中实现分页.所以我使用的是will_paginate gem.我面临的问题是,我无法弄清楚如何将paginate放在我当前的控制器代码中,因为它已根据查
我正在使用ransack进行搜索,现在我想在我的rails应用程序中实现分页.所以我使用的是will_paginate gem.我面临的问题是,我无法弄清楚如何将paginate放在我当前的控制器代码中,因为它已根据查询获取结果.

这是我的控制器代码

def search
if params[:search].present? && params[:search].strip != ""
  session[:loc_search] = params[:search]
end
arrResult = Array.new
if session[:loc_search] && session[:loc_search] != ""
  @rooms_address = Room.where(active: true).near(session[:loc_search], 5, order: 'distance')
else
  @rooms_address = Room.where(active: true).all
end

@search = @rooms_address.ransack(params[:q])
@rooms = @search.result

@arrRooms = @rooms.to_a

有人能告诉我如何在这里加分吗?

用日志更新

18:29:40 web.1    |   Room Load (0.8ms)  SELECT  rooms.*, 3958.755864232 * 2 * ASIN(SQRT(POWER(SIN((12.9715987 - rooms.latitude) * PI() / 180 / 2), 2) + COS(12.9715987 * PI() / 180) * COS(rooms.latitude * PI() / 180) * POWER(SIN((77.5945627 - rooms.longitude) * PI() / 180 / 2), 2))) AS distance, MOD(CAST((ATAN2( ((rooms.longitude - 77.5945627) / 57.2957795), ((rooms.latitude - 12.9715987) / 57.2957795)) * 57.2957795) + 360 AS decimal), 360) AS bearing FROM "rooms" WHERE "rooms"."active" = $1 AND (rooms.latitude BETWEEN 12.754501025333727 AND 13.188696374666272 AND rooms.longitude BETWEEN 77.37177993269385 AND 77.81734546730614 AND (3958.755864232 * 2 * ASIN(SQRT(POWER(SIN((12.9715987 - rooms.latitude) * PI() / 180 / 2), 2) + COS(12.9715987 * PI() / 180) * COS(rooms.latitude * PI() / 180) * POWER(SIN((77.5945627 - rooms.longitude) * PI() / 180 / 2), 2)))) BETWEEN 0.0 AND 15)  ORDER BY distance LIMIT 5 OFFSET 5  [["active", "t"]]
18:29:40 web.1    |    (0.3ms)  SELECT COUNT(*) FROM "rooms" WHERE "rooms"."active" = $1 AND (rooms.latitude BETWEEN 12.754501025333727 AND 13.188696374666272 AND rooms.longitude BETWEEN 77.37177993269385 AND 77.81734546730614 AND (3958.755864232 * 2 * ASIN(SQRT(POWER(SIN((12.9715987 - rooms.latitude) * PI() / 180 / 2), 2) + COS(12.9715987 * PI() / 180) * COS(rooms.latitude * PI() / 180) * POWER(SIN((77.5945627 - rooms.longitude) * PI() / 180 / 2), 2)))) BETWEEN 0.0 AND 15)  [["active", "t"]]
您对搜索结果进行了分页,因此在搜索后. 像@rooms = @ search.result.paginate(页面:params [:page],per_page:params [:per_page])之类的东西应该可行.
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