Group rows by day/week/month using inserted_at column.
Example : If first row is created before 1 week then group by days and if any data is not created in middle date then add o as value.
[
{"21 Jul" => 12},{"22 Jul" => 0},{"23 Jul" => 3}, {"24 Jul" => 1}...
{"5 Aug" => 0}
]
Thanks in advance.
Let's suppose you have Event schema with DateTime field called starts_at.
To get, for example, events count by date you can make such function:
def get_events_count_grouped_by_date(query) do
query
|> group_by([e], fragment("date(?)", e.starts_at))
|> select([e], %{date: fragment("date(?)", e.starts_at), count: count(e.id)})
|> order_by([e], asc: fragment("date(?)", e.starts_at))
end
Then use it like this:
Event |> get_events_count_grouped_by_date() |> Repo.all()
It's not quite clear from the question exactly what is required, but hopefully the following query helps: It will give you the count of records grouped by the date part of inserted_at.
defmodule CountByDateQuery do
import Ecto.Query
#doc "to_char function for formatting datetime as dd MON YYYY"
defmacro to_char(field, format) do
quote do
fragment("to_char(?, ?)", unquote(field), unquote(format))
end
end
#doc "Builds a query with row counts per inserted_at date"
def row_counts_by_date do
from record in SomeTable,
group_by: to_char(record.inserted_at, "dd Mon YYYY"),
select: {to_char(record.inserted_at, "dd Mon YYYY"), count(record.id)}
end
end
Usage:
row_counts_by_date() |> Repo.all() |> Map.new()
%{"05 Aug 2017" => 2}
Related
I am trying to pull data (some transaction related data) from DB. To pull the data, I am passing start date and end date as an argument to the query.
Here I need to pull the data of last 2 months. i.e., Start time would be Jun 01, 2022 and End time would be Aug 01, 2022.
Below is the script:
#!/usr/bin/perl
use strict;
use warnings;
use DateTime;
use Date::Format;
use Date::Parse;
my $nowdate = DateTime->now(time_zone => 'local');
my ($month, $year) = ($nowdate->month, $nowdate->year);
my $date = DateTime->new(
year => $year,
month => $month,
day => 1,
);
my $end_time = str2time($date);
print "END:$end_time\n";
print time2str("%d-%m-%Y %T", $end_time)."\n"; #printing just to see in human readable format
my $start_time = $date->clone;
$start_time->add( months => 1 )->subtract( days => 92 );
$start_time = str2time($start_time);
print "START:$start_time\n";
print time2str("%d-%m-%Y %T", $start_time)."\n"; #printing just to see in human readable format
I have two issues over here:
I am using DateTime object two times to get end time. Can it be done in one shot?
$start_time->add( months => 1 )->subtract( days => 92 ); In this line of code, I have to explicitly mention subtract 92 days, which wouldn't be right always. Since some months have 30 days or 31 days or even 29 days. How can I get 2 month's beginning day date?
Another example: Lets assume if we are in September 2022, then Start time and End time would be Jul 01, 2022 and Sep 01, 2022 respectively.
Note: Perl version is 5.16.3
It would be good If I can do it with Core modules which comes with 5.16.3
You could simplify it by using truncate(to => 'month') to get to the first in the current month:
my $end = DateTime->now(time_zone => 'local')
->truncate(to => 'month');
This may however fail on a day without a midnight so this may be an option:
my $end = DateTime->now(time_zone => 'local')
->set_time_zone('floating')
->truncate(to => 'month');
Then subtract the number of months to get the start date:
my $start = $end->clone->subtract( months => 2 );
Then:
my $start_time = str2time($start);
my $end_time = str2time($end);
print "START:$start_time\n";
print time2str("%d-%m-%Y %T", $start_time)."\n";
print "END:$end_time\n";
print time2str("%d-%m-%Y %T", $end_time)."\n";
Possible output:
START:1654034400
01-06-2022 00:00:00
END:1659304800
01-08-2022 00:00:00
**DATE FROM:**
def format=new java.text.SimpleDateFormat("yyyyMMdd")
def cal=Calendar.getInstance()
cal.get(Calendar.YEAR);
cal.set(Calendar.MONTH, 0);
cal.set(Calendar.DAY_OF_MONTH, 31);
[format.format(cal.getTime())]
**DATE TO:**
def format=new java.text.SimpleDateFormat("yyyyMMdd")
def cal=Calendar.getInstance()
cal.add(Calendar.DAY_OF_MONTH,-cal.get(Calendar.DAY_OF_MONTH))
[format.format(cal.getTime())]
when year changes (2020 - 2021) - it confuses January of previous year with January of this year
I have to correct so that in January (December reporting) it extracts data for period 31.01 - 31.12. of previous year.
The job was wrong because it extracted data from 31.01.2021 to 31.12.2020
// retrieve details of the current date
def cal = Calendar.instance;
def currentYear = cal.get(Calendar.YEAR);
def currentMonth = cal.get(Calendar.MONTH);
// set the instance to the start of the previous month
if ( currentMonth == 0 ) {
cal.set(currentYear-1, 11, 1);
} else {
cal.set(currentYear, (currentMonth-1), 1);
}
// extract the date, and format to a string
Date previousMonthStart = cal.time;
String previousMonthStartFormatted = previousMonthStart.format('yyyy-MM-dd');
If all you are looking for is the start of the previous year as in your title then the following code:
import java.time.*
def startOfPreviousYear = LocalDate.now()
.withDayOfMonth(1)
.withMonth(1)
.minusYears(1)
println startOfPreviousYear
def againStartingFromJanuary = LocalDate.of(2021, 1, 15)
.withDayOfMonth(1)
.withMonth(1)
.minusYears(1)
println againStartingFromJanuary
demonstrates one way to accomplish this. When run, this prints (with now being today's date of 2021.Mar.10):
─➤ groovy solution.groovy
2020-01-01
2020-01-01
updated after comments
You can get the end of previous and current months with something like this:
import java.time.*
def endOfPreviousMonth = LocalDate.now()
.withDayOfMonth(1)
.minusDays(1)
def endOfCurrentMonth = LocalDate.now()
.withDayOfMonth(1)
.plusMonths(1)
.minusDays(1)
println "end of last month: ${endOfPreviousMonth}"
println "end of current month: ${endOfCurrentMonth}"
which with current date prints:
end of last month: 2021-02-28
end of current month: 2021-03-31
or if we are in january:
def endOfPreviousMonth = LocalDate.of(2021, 1, 15)
.withDayOfMonth(1)
.minusDays(1)
def endOfCurrentMonth = LocalDate.of(2021, 1, 15)
.withDayOfMonth(1)
.plusMonths(1)
.minusDays(1)
println "end of last month: ${endOfPreviousMonth}"
println "end of current month: ${endOfCurrentMonth}"
which prints:
─➤ groovy solution.groovy
end of last month: 2020-12-31
end of current month: 2021-01-31
In general you should try to, when possible, stay away from using manual date arithmetic when dealing with dates if your target is based on the current date (as in, previous month, next month, three months ago, etc). Use the api:s handed to you by java. The date classes take care of rolling years, rolling months, rolling days, leap years, etc, all that stuff that you really do not want to spend time solving yourself.
How can I get the previous month without using a package or library in elixir?
For example, if the current date is 2018-01-25, I will get 2017-12-25.
Or If the current date is 2018-03-31, I will get 2018-02-28 (2018 is not a leap year)
The answer by #Sheharyar is almost there, the only difference you need to subtract the maximum of days in both months:
defmodule Dating do
def previous_month(%Date{day: day} = date) do
days = max(day, (Date.add(date, -day)).day)
Date.add(date, -days)
end
end
Works for all cases:
iex|1 ▶ Dating.previous_month(~D[2018-03-31])
#⇒ ~D[2018-02-28]
iex|2 ▶ Dating.previous_month(~D[2018-03-01])
#⇒ ~D[2018-02-01]
iex|3 ▶ Dating.previous_month(~D[2018-01-02])
#⇒ ~D[2017-12-02]
Use Timex library
iex(1)> ~D[2018-01-25] |> Timex.shift(months: -1)
~D[2017-12-25]
iex(2)> ~D[2018-03-31] |> Timex.shift(months: -1)
~D[2018-02-28]
I'm trying to format a DateTime object to a String by using the Dates.format() function.
I'd like to get the week number for a specific time, however I can't find right formatting option in the docs.
I know I could get this by running Dates.week but I would really like to know if it's possible via format to have cleaner code (and crack a code golfing challenge...).
Here's some actual code:
julia> my_time = now()
2018-03-22T08:16:15.601
julia> Dates.week(my_time)
12
julia> Dates.format(my_time, "Y m d H:M:S")
"2018 3 22 8:16:15"
In R, I could do "%V" to get the weeks formatted. Is there a similar way in Julia?
No, there isn't a format specifier like %V:
julia> Base.Dates.CONVERSION_SPECIFIERS
Dict{Char,Type} with 12 entries:
'm' => Base.Dates.Month
'd' => Base.Dates.Day
'M' => Base.Dates.Minute
'Y' => Base.Dates.Year
'e' => Base.Dates.DayOfWeekToken
's' => Base.Dates.Millisecond
'U' => Base.Dates.Month
'E' => Base.Dates.DayOfWeekToken
'H' => Base.Dates.Hour
'y' => Base.Dates.Year
'u' => Base.Dates.Month
'S' => Base.Dates.Second
The in the code for these specifiers there's a note that there should be a way for packages to add more.
I'm trying to format the Timex module to look a certain way. I'm trying to get today's date. but I want it formatted like this:
2017/12/12.
year/mn/day
In ruby I would go to the strftime class but I'm not sure how to do this with Elixir:
Current attempt:
Timex.local => #DateTime<2017-12-12 19:57:17.232916-05:00 EST America/Detroit>
How can I take that and format it how I specified?
Elixir 1.11 has Calendar.strftime/3 built-in for your strftime needs.
Calendar.strftime(~U[2019-08-26 13:52:06.0Z], "%y-%m-%d %I:%M:%S %p")
"19-08-26 01:52:06 PM"
Timex is a third-party library that was created in the era when Elixir had no good support for dates/times. Nowadays, there is DateTime native class in the core, so I am unsure why do you want to use Timex at all.
In any case, DateTime is a struct:
iex|1 ▶ today = DateTime.utc_now
#⇒ #DateTime<2017-12-13 07:22:58.290075Z>
iex|2 ▶ [today.year, today.month, today.day]
#⇒ [2017, 12, 13]
iex|3 ▶ Enum.join [today.year, today.month, today.day], "/"
#⇒ "2017/12/13"
To pad with leading zeroes for "2018/1/1":
iex|4 ▶ with {:ok, today} <- Date.new(2018, 1, 1) do
...|4 ▶ [today.year, today.month, today.day]
...|4 ▶ |> Enum.map(&to_string/1)
...|4 ▶ |> Enum.map(&String.pad_leading(&1, 2, "0"))
...|4 ▶ |> Enum.join("/")
...|4 ▶ end
#⇒ "2018/01/01"
If you want to do this without an external library, you can use io_lib:format/2 to pad the integers with zeroes where necessary like this:
iex(1)> date = Date.utc_today
~D[2017-12-13]
iex(2)> :io_lib.format("~4..0B/~2..0B/~2..0B", [date.year, date.month, date.day]) |> IO.iodata_to_binary
"2017/12/13"
iex(3)> {:ok, date} = Date.new(2018, 1, 1)
{:ok, ~D[2018-01-01]}
iex(4)> :io_lib.format("~4..0B/~2..0B/~2..0B", [date.year, date.month, date.day]) |> IO.iodata_to_binary
"2018/01/01"
iex(5)> {:ok, date} = Date.new(1, 1, 1)
{:ok, ~D[0001-01-01]}
iex(6)> :io_lib.format("~4..0B/~2..0B/~2..0B", [date.year, date.month, date.day]) |> IO.iodata_to_binary
"0001/01/01"
You can do this to add zeros
Timex.local |> Timex.format!("{YYYY}/0{M}/0{D}") => "2017/01/01"
So it appears the Timex Module has a format!/2 function which will return a string of what ever date you pass to it.
Here is what I came up with:
Timex.local |> Timex.format!("{YYYY}/{M}/{D}") => "2017/12/12"
The answer to add 0 padding is incorrect and will always pad 0s even if it does not require it. The correct way to pad 0s is as follows:
Timex.local |> Timex.format!("{YYYY}/{0M}/{0D}") => "2017/01/01"
As of Elixir 1.11 you can do this sanely with a date formatter in the standard library's Calendar module:
https://hexdocs.pm/elixir/1.13.0/Calendar.html#strftime/3
DateTime.utc_now()
|> Calendar.strftime("%Y/%m/%d")
"2022/01/12"