Sql 更快的数据库，用于Rails可用性日历_Sql_Ruby On Rails_Ruby

Sql 更快的数据库，用于Rails可用性日历

sql ruby-on-rails ruby

Sql 更快的数据库，用于Rails可用性日历,sql,ruby-on-rails,ruby,Sql,Ruby On Rails,Ruby,我正在构建一个RubyonRails旅馆床位预订系统，我正在寻找一种更好的方式来加载我的可用日历。下面的代码非常适合显示床的可用性（见附图），但我觉得多个数据库调用对服务器来说太麻烦了，必须有一种更快的方法来处理数据型号： user.rb bed.rb（用户id、名称） guest.rb（用户id、姓名、电话、地址、电子邮件） reservation.rb（用户id、床id、客人id、日期）使用者是旅馆的主人。他们创造尽可能多的床，并给他们起个名字（床1、床2、双阁楼、主楼2楼等）每个用户

我正在构建一个RubyonRails旅馆床位预订系统，我正在寻找一种更好的方式来加载我的可用日历。下面的代码非常适合显示床的可用性（见附图），但我觉得多个数据库调用对服务器来说太麻烦了，必须有一种更快的方法来处理数据

型号：

user.rb

bed.rb（用户id、名称）

guest.rb（用户id、姓名、电话、地址、电子邮件）

reservation.rb（用户id、床id、客人id、日期）

使用者是旅馆的主人。他们创造尽可能多的床，并给他们起个名字（床1、床2、双阁楼、主楼2楼等）

每个用户都有许多客人，这些人在特定日期睡在他们的旅馆里

使用simple_calendar gem，我获得当月的所有预订并处理它们以显示每周视图

排序逻辑：

第1步。在一周的第一天迭代当前用户的床位。（简单日历提供每天的预订）

第2步。如果床上有当天的预订（！=nil），请填写床名和预订客人的姓名

第3步。如果床上没有预约，请填写床名和“空”

第4步。完成一周的第一天后，移动到第二天并重复。（简单的日历会处理这个问题）

正如我所说，下面的代码是有效的，但是当一个用户有50张床位时，它每天会打50个WHERE电话，所以7天的日历会打350个WHERE电话（如果需要找到客人的名字，还会打更多）

如果您需要更多信息或解释，请告诉我。谢谢

= week_calendar events: @monthly_reservations, attribute: :date do |date, appts|
  - current_user.beds.each do |bed|
    - if bed.reservations.where(date: date).first != nil
      - guest_name = bed.reservations.where(date: date).first.guest.abbreviated_name
        %p{style: "border: 1px solid black; padding: 5px; font-size: 10px;"}
          = "#{bed.name}: #{guest_name}"
    - else
      %p{style: "border: 1px solid black; padding: 5px; font-size: 10px;"}
        = "#{bed.name}: Empty"

编辑： 理想的日历布局如下图所示：

然而，问题是使床位名称与正确的床位预订相符。如果5月2日只预订了两张床，即1号床和4号床，则代码不知道在最后插入4号床的预约之前保留两张空床。有道理吗

编辑#2:

@大卫德里奇，类似这样的：

像这样的

reservs={“[2017-05-01，床位1]”=>“guest_23”、“[2017-05-01，床位2]”=>“empty”}

编辑3 下面是加载一个只有5张床位的旅馆时日志的样子

Bed Load (0.3ms)  SELECT "beds".* FROM "beds" WHERE "beds"."user_id" = $1  [["user_id", 1]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 1], ["date", "2017-07-02"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 2], ["date", "2017-07-02"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 3], ["date", "2017-07-02"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 4], ["date", "2017-07-02"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 5], ["date", "2017-07-02"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 1], ["date", "2017-07-03"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 2], ["date", "2017-07-03"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 3], ["date", "2017-07-03"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 4], ["date", "2017-07-03"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 5], ["date", "2017-07-03"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 1], ["date", "2017-07-04"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 2], ["date", "2017-07-04"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 3], ["date", "2017-07-04"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 4], ["date", "2017-07-04"]]
  Reservations Load (0.1ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 5], ["date", "2017-07-04"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 1], ["date", "2017-07-05"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 2], ["date", "2017-07-05"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 3], ["date", "2017-07-05"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 4], ["date", "2017-07-05"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 5], ["date", "2017-07-05"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 1], ["date", "2017-07-06"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 2], ["date", "2017-07-06"]]
  Reservations Load (0.1ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 3], ["date", "2017-07-06"]]
  Reservations Load (0.1ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 4], ["date", "2017-07-06"]]
  Reservations Load (0.3ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 5], ["date", "2017-07-06"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 1], ["date", "2017-07-07"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 2], ["date", "2017-07-07"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 3], ["date", "2017-07-07"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 4], ["date", "2017-07-07"]]
  Reservations Load (0.1ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 5], ["date", "2017-07-07"]]
  Reservations Load (0.2ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 1], ["date", "2017-07-08"]]
  Reservations Load (0.6ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 2], ["date", "2017-07-08"]]
  Reservations Load (0.3ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 3], ["date", "2017-07-08"]]
  Reservations Load (0.3ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 4], ["date", "2017-07-08"]]
  Reservations Load (0.3ms)  SELECT  "reservations".* FROM "reservations" WHERE "reservations"."bed_id" = $1 AND "reservations"."date" = $2  ORDER BY "reservations"."id" ASC LIMIT 1  [["bed_id", 5], ["date", "2017-07-08"]]

编辑4 根据@davidardridge建议，我在控制器中创建了一个散列，并将其传递到我的视图中。结果非常好

控制器代码：

@reservations = {}

@monthly_reservations.each do |x|
  @reservations[[x.date,x.bed_id]] = x.guest.try(:name)  
end

以及查看代码：

= week_calendar events: @monthly_reservations, attribute: :date do |date, appts|
  - current_user.beds.each do |bed|
    - if @reservations[[date, bed.id]] != nil
      %p{style: "border: 1px solid black; padding: 5px; font-size: 10px;"}
        = "#{bed.name}: #{@reservations[[date, kennel.id]]}"
    - else
      %p{style: "border: 1px solid black; padding: 5px; font-size: 10px;"}
        = "#{bed.name}:"

Robbie所描述的是批量捕获数据，以便在第二次传递时显示。您可以这样做：

@reservations = Reservations.where(bed_id: [ ... ], date: (from..to)).group_by do |r|
  [r.bed_id, r.date]
end

其中，它返回一个单一结构，该结构应包含所有保留，并且可以使用

bed_id

和日期轻松索引。如果需要，您可以将其转换为两层结构，但通常不是这样

迭代时：

- current_user.beds.each do |bed|
  - reservation = @reservations[[bed.id,date]]
  - if reservation
    # ...

当您处理无法用加载链中的

includes（…）

元素轻松表示的复杂相互依赖关系时，这种选择性但积极地加载记录的做法通常非常有效

另外，请记住，在Ruby中，逻辑上只有两个错误：literal

false

和

nil

。其他所有内容的计算结果在逻辑上都为true，包括0、空字符串、数组和散列。因此，比较

！=nil几乎总是无关的和令人困惑的，特别是如果你像那样做双重否定，除非（x！=nil）

如果您希望能够根据数据库解析任意多个日期+床位对，您可以制作某种类型的预订密钥，该密钥是日期和床位id的组合，那么扫描这些就容易多了。你可以在（…）

中做一个

WHERE-booking\u-token，并将其全部编入索引，性能和简洁。不过，要做好这件事需要一些事先的计划YYYY-MM-DD-bed_id
可以作为第一个通行证。
对于您的查看代码，我会尝试：
= week_calendar events: @monthly_reservations, attribute: :date do |date, appts|
  - current_user.beds.each do |bed|
    %p{style: "border: 1px solid black; padding: 5px; font-size: 10px;"}
      = [bed.name, @reservations[[date, bed.id]]].compact.join(":")

“我觉得多个WHERE数据库调用对服务器来说太麻烦了”。这是直觉，可能正确，也可能不正确。找到答案的唯一方法是仔细查看log/development.log
，查看查询的执行情况。通常，最好始终在窗口中打开tail-f log/development.log
，以了解加载页面和开发功能时代码的效率和丑陋程度。为什么不只查询一次表，以获取当前用户当月的所有记录，然后迭代结果？这正是@monthly\u reservations变量所包含的内容，即当前用户当月的所有预订。然而，WHEREs进来了，因为我需要展示旅馆所有的床位，然后是谁，如果有的话，住在这张床上。塔德曼，我完全同意直觉部分。也许这不会让服务器负担太重，但当我查看日志并看到每次加载页面时都会出现无休止的结果时，我很担心。你应该能够运行一个查询，获取bed、date和user，并使用键[bed，date]
和值guest
构建一个散列。然后你可以按你认为最合理的顺序遍历你的床和日期，从散列中查找每个日期/床组合的任何客人。@davidardridge，我认为这听起来是个好主意。我只是需要一些额外的智慧来帮助我思考其他可能的方法。我的代码可以用，但很臭。谢谢@davidridge。这仅仅是一个重构，还是我缺少某种类型的性能增强？我承认，我的Ruby基础知识并不高。看起来我可能需要再次打开POODR！：）@帕特里克·琼斯，可怜的FTW博士！这是一个消除冗余的重构。