ok so i have
>> 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".............
and then
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]]
the problem is I have a few Sun, 09 Jan and a couple for the 10th, instead of one like this
this is what i need
=> [[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]]
How Do You Group Data by Week in SQL Server? SQL Server provides a function called DATEPART() , which returns a specified part (year, quarter, month, week, hour, minute, etc.) of a specified date. ORDER BY DATEPART(week, RegistrationDate);
1 Answer. You can use DATE_FORMAT operator. If you are using this you can easily group the date, timestamp or datetime column using whatever format you want.
You can group month and year with the help of function DATE_FORMAT() in MySQL. The GROUP BY clause is also used.
I think this is a much more elegant and simple solution
list.group_by{|x| x.created_at.strftime("%Y-%m-%d")}
Time
is a quite complex object to group by. Assuming you want to group by the creation date, instead of the full Time
, start creating a custom method in your model to return the group criteria.
The method should return the creation date, possibly as string.
def group_by_criteria
created_at.to_date.to_s(:db)
end
Then, group by that method.
list.group_by(&:group_by_criteria).map {|k,v| [k, v.length]}.sort
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