Question

好的，所以我有

 >> list = Request.find_all_by_artist("someBand")
=> [#<Request id: 1, artist: "someBand", song: "someSong", venue: "Knebworth - Stevenage, United Kingdom", showdate: "2011-07-01", amount: nil, user_id: 2, created_at: "2011-01-01 18:14:08", updated_at: "2011-01-01 18:14:09".............

然后

list.group_by(&:created_at).map {|k,v| [k, v.length]}.sort
=> [[Sat, 01 Jan 2011 18:14:08 UTC +00:00, 10], [Sun, 09 Jan 2011 18:34:19 UTC +00:00, 1], [Sun, 09 Jan 2011 18:38:48 UTC +00:00, 1], [Sun, 09 Jan 2011 18:51:10 UTC +00:00, 1], [Sun, 09 Jan 2011 18:52:30 UTC +00:00, 1], [Thu, 10 Feb 2011 02:22:08 UTC +00:00, 1], [Thu, 10 Feb 2011 20:02:20 UTC +00:00, 1]]

问题是我有几个Sun，1月9日和一对夫妇10日，而不是像这样的一个

这就是我需要的

=> [[Sat, 01 Jan 2011 18:14:08 UTC +00:00, 10], [Sun, 09 Jan 2011 18:34:19 UTC +00:00, 4], [Thu, 10 Feb 2011 20:02:20 UTC +00:00, 2]]

Answer 1

Time是一个非常复杂的分组对象。假设您希望按创建日期进行分组，而不是完整Time，请在模型中开始创建自定义方法以返回组标准。

该方法应该返回创建日期，可能是字符串。

def group_by_criteria
  created_at.to_date.to_s(:db)
end

然后，按该方法分组。

list.group_by(&:group_by_criteria).map {|k,v| [k, v.length]}.sort

Answer 2

我认为这是一个更加优雅和简单的解决方案

list.group_by{|x| x.created_at.strftime("%Y-%m-%d")}

Answer 3

有一个宝石：groupdate。

用法（来自文档）：

User.group_by_day(:created_at).count
# {
#   2013-04-16 00:00:00 UTC => 50,
#   2013-04-17 00:00:00 UTC => 100,
#   2013-04-18 00:00:00 UTC => 34
# }

Answer 4

Ipsum的回答实际上很好，可能是最好的：

在Arel：

requests = Arel::Table.new(:requests)
query = requests.project("COUNT(*), CAST(requests.created_at AS DATE) as created_at")
query = query.group("CAST (requests.created_at AS DATE)")
Request.find_by_sql(query.to_sql)

Answer 5

没有额外宝石的小组：

def self.group_by_day items
   data = items.group_by{|x| x.created_at.to_date}
   chart_data = {}

   data.each do |a,b|
     chart_data.merge!({a => b.count})
   end

   return chart_data
end

Answer 6

你可以在MySQL中使用GROUP BY DATE(created_at)

关于ruby代码，你可以像这样使用

list.group('DATE(created_at)').map {|k,v| [k, v.length]}.sort

我如何按天而不是按日期分组？

6 个答案: