- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
我一直在努力处理 ActiveRecord 查询,但我可以完成我想要的。我有以下模型和关系:
class User < ActiveRecord::Base
has_many :orders
end
class Order < ActiveRecord::Base
belongs_to :user
belongs_to :clinic
end
class Clinic < ActiveRecord::Base
has_many :orders
end
而且我需要让用户按时间段下订单。 上个月、过去3 个月、过去6 个月 和过去12 个月 使用created_at
属性。
12 months
列中的总数是前 3 个时间段的总和时间加上一年中剩下的 6 个月。我一直在尝试一些解决方法,获得它的最佳方法是首先获得具有相应诊所的用户列表。然后我遍历每个用户并执行以下查询以获得每个时间段的总数:
users = User.joins(:clinics).select('user.id, clinic.name')
users.each do |user|
month = user.orders.where('created_at > ? and created_at < ?', Time.current.beginning_of_month, Time.current.end_of_month).count
month_3 = user.orders.where('created_at > ? and created_at < ?', Time.current.beginning_of_month - 2.months, Time.current.end_of_month).count
month_6 = user.orders.where('created_at > ? and created_at < ?', Time.current.beginning_of_month - 5.months, Time.current.end_of_month).count
month_12 = user.orders.where('created_at > ? and created_at < ?', Time.current.beginning_of_month - 11.months, Time.current.end_of_month).count
end
I'm wondering if there's a better way to do it without iterating over each and every user record and perform the queries there!
有人知道吗?提前致谢。
最佳答案
未经测试但应该像下面这样工作:
beginning_of_month = Time.current.beginning_of_month
end_of_month = Time.current.end_of_month
# Let's get first all orders count group by user + clinic
users_per_clinic_orders_count_since_beginning_of_month = Order.where('created_at >= ? AND created_at < ?', , beginning_of_month, end_of_month).group(:user_id, :clinic_id).count
users_per_clinic_orders_count_since_three_months_ago = Order.where('created_at >= ? AND created_at < ?', , beginning_of_month - 2.months, end_of_month).group(:user_id, :clinic_id).count
users_per_clinic_orders_count_since_six_months_ago = Order.where('created_at >= ? AND created_at < ?', , beginning_of_month - 5.months, end_of_month).group(:user_id, :clinic_id).count
users_per_clinic_orders_count_since_last_year = Order.where('created_at >= ? AND created_at < ?', , beginning_of_month - 11.months, end_of_month).group(:user_id, :clinic_id).count
# Now Let's get all users JOINED with clinics into memory
users = User.joins(:clinics).select('id', 'clinics.name')
data = {}
# Set data with (key == [user_id, clinic_id]) with the corresponding table columns
# All of this should be fast as all of these objects including user and clinic records are already in memory
# data[[user_id, clinic_id]][0] means "id" column
# data[[user_id, clinic_id]][1] means "clinic name" column
# data[[user_id, clinic_id]][2] means "last order at" column
# data[[user_id, clinic_id]][3] means "orders 1 Month" column
# data[[user_id, clinic_id]][4] means "orders 3 Months" column
# data[[user_id, clinic_id]][5] means "orders 6 Months" column
# data[[user_id, clinic_id]][6] means "orders 12 Months" column
users_per_clinic_orders_count_since_beginning_of_month.each do |(user_id, clinic_id), orders_count_since_beginning_of_month|
if data[[user_id, clinic_id]].nil?
user_in_memory = users.detect { |user| user.id == user_id }
next unless user_in_memory # ignore if this user_id is not part of `users` (joined with clinics)
clinic_in_memory = user_in_memory.clinic
next unless clinic_in_memory # ignore if this clinic_id is not part of `users` (joined with clinics)
end
data[[user_id, clinic_id]] ||= []
data[[user_id, clinic_id]][0] ||= user_in_memory.id
data[[user_id, clinic_id]][1] ||= clinic_in_memory.name
data[[user_id, clinic_id]][3] = orders_count_since_beginning_of_month
end
users_per_clinic_orders_count_since_three_months_ago.each do |(user_id, clinic_id), orders_count_since_three_months_ago|
if data[[user_id, clinic_id]].nil?
user_in_memory = users.detect { |user| user.id == user_id }
next unless user_in_memory # ignore if this user_id is not part of `users` (joined with clinics)
clinic_in_memory = user_in_memory.clinic
next unless clinic_in_memory # ignore if this clinic_id is not part of `users` (joined with clinics)
end
data[[user_id, clinic_id]][0] ||= user_in_memory.id
data[[user_id, clinic_id]][1] ||= clinic_in_memory.name
data[[user_id, clinic_id]][4] = orders_count_since_three_months_ago
end
users_per_clinic_orders_count_since_six_months_ago.each do |(user_id, clinic_id), orders_count_since_six_months_ago|
if data[[user_id, clinic_id]].nil?
user_in_memory = users.detect { |user| user.id == user_id }
next unless user_in_memory # ignore if this user_id is not part of `users` (joined with clinics)
clinic_in_memory = user_in_memory.clinic
next unless clinic_in_memory # ignore if this clinic_id is not part of `users` (joined with clinics)
end
data[[user_id, clinic_id]][0] ||= user_in_memory.id
data[[user_id, clinic_id]][1] ||= clinic_in_memory.name
data[[user_id, clinic_id]][5] = orders_count_since_six_months_ago
end
users_per_clinic_orders_count_since_last_year.each do |(user_id, clinic_id), orders_count_since_last_year|
if data[[user_id, clinic_id]].nil?
user_in_memory = users.detect { |user| user.id == user_id }
next unless user_in_memory # ignore if this user_id is not part of `users` (joined with clinics)
clinic_in_memory = user_in_memory.clinic
next unless clinic_in_memory # ignore if this clinic_id is not part of `users` (joined with clinics)
end
data[[user_id, clinic_id]][0] ||= user_in_memory.id
data[[user_id, clinic_id]][1] ||= clinic_in_memory.name
data[[user_id, clinic_id]][6] = orders_count_since_last_year
end
# Lastly we need to get the "last order at" to be inserted into each data[[user_id, clinic_id]][2]
# this is the only unnecessarily slower part as this do a N-times more sql per find_by
# (but couldnt think of a quick solution yet to eager load the last order created_at)
# Feel free to change this if you know a faster way
data.each do |(user_id, clinic_id), columns|
last_order = Order.order(created_at: :desc).find_by(user_id: user_id, clinic_id: clinic_id)
data[[user_id, clinic_id]][2] = last_order.created_at
end
# sort the table by [user_id, clinic_id] and then puts and show the table
puts data.sort.map do |(user_id, clinic_id), columns]
columns
end
# => [
# [1, Providence, 2019-08-14T15:51:12.342Z, 2, 6, 12, 24],
# [1, Joseph Center, 2019-08-22T16:26:29.151Z, 1, 3, 6, 12],
# [2, Eubanks, 2019-08-22T16:26:29.151Z, 1, 4, 8, 18],
# ...
# ]
关于ruby-on-rails - 按日期范围对记录进行分组并在 Rails 中对其进行计数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57517892/
我有一个网站。 必须登录才能看到里面的内容。 但是,我使用此代码登录。 doc = Jsoup.connect("http://46.137.207.181/Account/Login.aspx")
我正在尝试为我的域创建一个 SPF 记录并使我的邮件服务器能够对其进行评估。我在邮件服务器上使用 Postfix 并使用 policyd-spf (Python) 来评估记录。目前,我通过我的私有(p
我需要为负载平衡的 AWS 站点 mywebsite.com 添加 CName 记录。记录应该是: @ CNAME mywebsite.us-east-1.elb.amazon
我目前正在开发一个相当大的多层应用程序,该应用程序将部署在海外。虽然我希望它在解聚后不会折叠或爆炸,但我不能 100% 确定这一点。因此,如果我知道我可以请求日志文件,以准确找出问题所在以及原因,那就
我使用以下命令从我的网络摄像头录制音频和视频 gst-launch-0.10 v4l2src ! video/x-raw-yuv,width=640,height=480,framerate=30/1
我刚刚开始使用 ffmpeg 将视频分割成图像。我想知道是否可以将控制台输出信息保存到日志文件中。我试过“-v 10”参数,也试过“-loglevel”参数。我在另一个 SO 帖子上看到使用 ffmp
我想针对两个日期查询我的表并检索其中的记录。 我这样声明我的变量; DECLARE @StartDate datetime; DECLARE @EndDate datetime; 并像这样设置我的变量
在 javascript 中,我可以使用简单的 for 循环访问对象的每个属性,如下所示 var myObj = {x:1, y:2}; var i, sum=0; for(i in myObj) s
最近加入了一个需要处理大量代码的项目,我想开始记录和可视化调用图的一些流程,让我更好地理解一切是如何组合在一起的。这是我希望在我的理想工具中看到的: 每个节点都是一个函数/方法 如果一个函数可以调用另
如何使用反射在F#中创建记录类型?谢谢 最佳答案 您可以使用 FSharpValue.MakeRecord [MSDN]创建一个记录实例,但是我认为F#中没有任何定义记录类型的东西。但是,记录会编译为
关闭。这个问题不满足Stack Overflow guidelines .它目前不接受答案。 想改善这个问题吗?更新问题,使其成为 on-topic对于堆栈溢出。 3年前关闭。 Improve thi
我是 Sequelize 的新手并且遇到了一些语法问题。我制作了以下模型: // User sequelize.define('user', { name: { type: DataTyp
${student.name} Notify 这是我的output.jsp。请注意,我已经放置了一个链接“Notify”以将其转发到 display.jsp 上。但我不确定如何将 Stud
例如,这是我要做的查询: server:"xxx.xxx.com" AND request_url:"/xxx/xxx/xxx" AND http_X_Forwarded_Proto:(https O
我一直在开发大量 Java、PHP 和 Python。所有这些都提供了很棒的日志记录包(分别是 Log4J、Log 或logging)。这在调试应用程序时有很大帮助。特别是当应用程序 headless
在我的Grails应用程序中,我异步运行一些批处理过程,并希望该过程记录各种状态消息,以便管理员以后可以检查它们。 我考虑过将log4j JDBC附加程序用作最简单的解决方案,但是据我所知,它不使用D
我想将进入 MQ 队列的消息记录到数据库/文件或其他日志队列,并且我无法修改现有代码。是否有任何方法可以实现某种类似于 HTTP 嗅探器的消息记录实用程序?或者也许 MQ 有一些内置的功能来记录消息?
如果我有一条包含通用字段的记录,在更改通用字段时是否有任何方法可以模仿方便的 with 语法? 即如果我有 type User = // 'photo can be Bitmap or Url {
假设我有一个名为 Car 的自定义对象。其中的所有字段都是私有(private)的。 public class Car { private String mName; private
当记录具有特定字段时,我需要返回 true 的函数,反之亦然。示例: -record(robot, {name, type=industrial, ho
我是一名优秀的程序员,十分优秀!