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Python odoo search query to check whether two datetime falls in between other two datetime

initial=2022-04-14 14:00:00
final=2022-04-14 15:30:00
case1:
t1=2022-04-14 13:30:00
t2=2022-04-14 14:30:00
case 2:
t1=2022-04-14 15:00:00
t2=2022-04-14 16:30:00
Want the python odoo search query to check whether initial & final time falls in between t1 &t2

Simple Python example:
from datetime import datetime
dt_1 = datetime.strptime('15 Mar 2022, 19:55 PM', '%d %b %Y, %I:%M %p')
dt_2 = datetime.strptime('30 Apr 2023, 13:45 PM', '%d %b %Y, %I:%M %p')
dt_now = datetime.now()
if dt_1 > dt_2:
print('dt_1 > dt_2')
if dt_1 > dt_now:
print('dt_1 > dt_now')
Example in Odoo: in the "event" module:
from odoo import _, api, fields, models
class EventEvent(models.Model):
_name = 'event.event'
date_end = fields.Datetime(string='End Date')
def _is_event_registrable(self):
return self.date_end > fields.Datetime.now()
def get_date_range_str(self):
today = fields.Datetime.now()
event_date = self.event_begin_date
diff = (event_date.date() - today.date())
if diff.days <= 0:
return _('today')
elif diff.days == 1:
return _('tomorrow')
elif (diff.days < 7):
return _('in %d days') % (diff.days, )

How to convert A "hours and minutes and seconds" string to HH:MM:SS format?

In python, I am trying to make an alarm system where it converts "x Hours y Minutes and z Seconds" to the x:y:z format
For example:
5 hours 20 minutes 6 seconds
05:20:06
1 hour 25 seconds
01:00:25
This is my code, but it seems scuffed:
time_string = '1 hour and 25 seconds'
correction = time_string.replace('and ', '')
duration = correction.replace(' hour', ':').replace(' minute', ':').replace(' second', ':').replace(' ','').replace('s', '')
if 'minute' not in correction and 'second' not in correction:
duration = duration + '00:00'
elif 'minute' not in correction and 'second' in correction:
duration = duration.replace(':',':00:')
elif 'second' not in correction:
duration = duration + '00'
secs = sum(int(x) * 60 ** i for i, x in enumerate(reversed(duration.split(':'))))
How do I improve it?

This returns total number of seconds, which seems to be what you wanted:
def parsex(s):
hh = mm = ss = 0
for word in s.split():
word = word.lower()
if word.isdigit():
save = word
elif word.startswith('hour'):
hh = int(save)
elif word.startswith('minute'):
mm = int(save)
elif word.startswith('second'):
ss = int(save)
return (hh*60+mm)*60+ss
print(parsex('1 hour and 30 seconds'))
print(parsex('2 hours 15 minutes 45 seconds'))

You can use datetime library to convert this type of string to proper formatted string.
from datetime import datetime
def format_time(string):
format_string = ''
if 'hour' in string:
if 'hours' in string:
format_string += '%H hours '
else:
format_string += '%H hour '
if 'minute' in string:
if 'minutes' in string:
format_string += '%M minutes '
else:
format_string += '%M minute '
if 'second' in string:
if 'seconds' in string:
format_string += '%S seconds'
else:
format_string += '%S second'
value = datetime.strptime(string, format_string)
return value.strftime('%H:%M:%S')
string = '5 hours 20 minutes 6 seconds'
print(format_time(string))
string = '1 hour 25 seconds'
print(format_time(string))
string = '1 minute 25 seconds'
print(format_time(string))
Output
05:20:06
01:00:25
00:01:25

from collections import defaultdict
dataexp = [
("5 hours 20 minutes 6 seconds","05:20:06"),
("1 hour 25 seconds","01:00:25")
]
def convert(input_):
words = input_.replace('and','').split()
di = defaultdict(lambda:0)
while words:
num = int(words.pop(0))
unit = words.pop(0)[0]
di[unit] = num
return f"{di['h']:02}:{di['m']:02}:{di['s']:02}"
for inp, exp in dataexp:
got = convert(inp)
msg = "\nfor %-100.100s \nexp :%s:\ngot :%s:\n" % (inp, exp, got)
if exp == got:
print("✅! %s" % msg)
else:
print("❌! %s" % msg)
which outputs:
✅!
for 5 hours 20 minutes 6 seconds
exp :05:20:06:
got :05:20:06:
✅!
for 1 hour 25 seconds
exp :01:00:25:
got :01:00:25:

Calculating time difference from unix $date

I have pairs of timestamps from the unix date:
alvas#ubi:~$ date
Wed May 20 01:04:43 CEST 2015
How can I calculate the time difference between the pairs of timestamp?
Note: I don't have the full datetime stamp from $date but only the time in hr:min:sec left in my logfile.
For example, timediff(start, end):
timediff('11:12:10', '19:58:50')
timediff('15:17:09', '03:11:10')
[out]:
31600
42841
[out2]:
8 hrs 46 mins 40 secs (31600 secs)
11 hrs 54 mins 1 secs (42841 secs)
In the case that end < start, treat it as the next day.
I've tried the script below but is there an easier way to do that?
def timediff(start, end):
start_hr, start_min, start_sec = map(float, start.split(':'))
end_hr, end_min, end_sec = map(float, end.split(':'))
if end_hr < start_hr:
end_hr = end_hr + 24
if end_min < start_min:
end_min = end_min + 60
end_hr -= 1
if end_sec < start_sec:
end_sec = end_sec + 60
end_min -= 1
num_hrs = end_hr - start_hr
num_mins = end_min - start_min
num_secs = end_sec - start_sec
total_seconds = 60*60*num_hrs + 60*num_mins+num_secs
total_time = " ".join([str(num_hrs), 'hrs', str(num_mins), 'mins', str(num_secs), 'secs'])
return total_seconds, total_time
print timediff('11:12:10', '19:58:50')
print timediff('15:17:09', '03:11:10')
[out]:
(31600.0, '8.0 hrs 46.0 mins 40.0 secs')
(42841.0, '11.0 hrs 54.0 mins 1.0 secs')
When I tried dateutil.parser.parse:
>>> from dateutil.parser import parse as dtparse
>>> dtparse('19:58:50') - dtparse('11:12:10')
datetime.timedelta(0, 31600)
>>> dtparse('03:11:10') - dtparse('15:17:09')
datetime.timedelta(-1, 42841)

dateutil.parser(pip install python-dateutil) can parse just about anything ... and its timezone aware
from dateutil.parser import parse as dtparse
print dtparse("Wed May 20 01:04:43 CEST 2015") - dtparse("Wed May 20 00:02:43 CEST 2015")
>>> print dtparse("Fri May 22 01:04:43 CEST 2015") - dtparse("Wed May 20 00:02:43
CEST 2015")
2 days, 1:02:00
>>> print dtparse("15:22:36") - dtparse("12:00:45")
3:21:51 # 3 hours, 21 minutes, 51 seconds
here ... this will work if you dont know which timestamp is later ...
def tdiff(t1,t2):
if t1 > t2:t1,t2 = t2,t1
return t2-t1
print tdiff(dtparse('11:12:10'), dtparse('19:58:50'))
if you want to subtract a day in the event the order is wrong try this
def tdiff(t1,t2):
if t1 > t2:t1 - datetime.timedelta(days=1)
return t2-t1

Probably easier to use datetime:
import datetime
def timediff(t1, t2):
d1=datetime.datetime.strptime(t1, "%H:%M:%S")
d2=datetime.datetime.strptime(t2, "%H:%M:%S")
if d2<d1:
d1-=datetime.timedelta(days=1)
return ((d2-d1).total_seconds(), str(d2-d1))
>>> timediff('11:12:10', '19:58:50')
(31600.0, '8:46:40')
>>> timediff('15:17:09', '03:11:10')
(42841.0, '11:54:01')

string to date in python with the greatest possible year when dow is given

Given:
from datetime import datetime
date = datetime.strptime('Wed 12 Nov', '%a %d %b')
It returns datetime.datetime(1900, 11, 12, 0, 0). What might be the best way to get the greatest past year?, in this case should 2014 instead of 1900.
Any help will be appreciate.

You can get the year from datetime.now() and subtract 1:
from datetime import datetime
date = datetime.strptime('Wed 12 Nov {}'.format(datetime.now().year-1), '%a %d %b %Y')
This will not work for feb 29.
Actually a bug in the first implementation, it was starting on a monday because 12 November 1900 was a Monday:
dte = 'Wed 12 Nov'
start = datetime.strptime('Wed 12 Nov', "%a %d %b")
greatest = None
while start <= datetime.now():
start += timedelta(days=1)
if start.strftime("%a %d %b") == dte:
greatest = start
print(greatest)
2014-11-12 00:00:00
There is also a monthdelta package that you can use to increment by month:
from datetime import datetime
from monthdelta import monthdelta
dte = 'Wed 12 Nov'
start = datetime.strptime('Wed 12 Nov', "%a %d %b")
greatest = None
while start <= datetime.now():
start += monthdelta(1)
if start.strftime("%a %d %b") == dte:
greatest = start
print(greatest)
You can see incrementing by months is much more efficient:
In [1]: from datetime import datetime, timedelta
In [2]: %%timeit
...: dte = 'Wed 12 Nov'
...: start = datetime.strptime('Wed 12 Nov', "%a %d %b")
...: greatest = None
...: while start <= datetime.now():
...: start += timedelta(days=1)
...: if start.strftime("%a %d %b") == dte:
...: greatest = start
...:
1 loops, best of 3: 382 ms per loop
In [3]: from datetime import datetime
In [4]: from monthdelta import monthdelta
In [5]: %%timeit
...: dte = 'Wed 12 Nov'
...: start = datetime.strptime('Wed 12 Nov', "%a %d %b")
...: greatest = None
...: while start <= datetime.now():
...: start += monthdelta(1)
...: if start.strftime("%a %d %b") == dte:
...: greatest = start
...:
100 loops, best of 3: 18.7 ms per loop
Both return pretty quick but if you had many calls to the method then the monthly increase is a better option.
We could also add 30 days and then set the day to 12, there may be bugs as I have not overly tested it:
def match_date(abb_wk_dy, day_date, abb_mon):
dte = "{} {} {}".format(abb_wk_dy.capitalize(), day_date, abb_mon.capitalize())
start = datetime.strptime(dte, "%a %d %b")
greatest = None
while start <= datetime.now():
start += timedelta(days=30)
start = start.strptime("{} {} {}".format(start.year, start.month, day_date), "%Y %m %d")
if start.strftime("%a %d %b") == dte:
greatest = start
return greatest
The last code runs pretty efficiently:
In [12]: timeit match_date("wed","12","nov")
10 loops, best of 3: 34.7 ms per loop
If you only want the year then return greatest.year.
On testing the above code fails for leap years so we need to catch that, we can also just increase the year by 1 each time:
def match_date(abb_wk_dy, day_date, abb_mon):
wkd, dd, ab = abb_wk_dy.capitalize(), day_date, abb_mon.capitalize()
match = "{} {} {}".format(wkd, dd, ab)
try:
dte = "{} {} {} {}".format(1900, wkd, dd, ab)
start = datetime.strptime(dte, "%Y %a %d %b")
except ValueError:
# first leap year since 1900
dte = "{} {} {} {}".format(1904, wkd, dd, ab)
start = datetime.strptime(dte, "%Y %a %d %b")
day, mon = start.day, start.month
greatest = None
while start <= datetime.now():
try:
start = start.strptime("{} {} {}".format(start.year + 1, mon, day), "%Y %m %d")
except ValueError:
start = start.strptime("{} {} {}".format(start.year + 1, 01, 01), "%Y %m %d")
continue
if start.strftime("%a %d %b") == match:
greatest = start
return greatest.year if greatest else "No match"
Which runs in:
In [27]: timeit match_date("Wed","12","Nov")
100 loops, best of 3: 2.63 ms per loop
You would also need to validate that no day > 31 is ever entered and other months and days match up which could be achieved using a dict or calender.monthrange or a dict mapping max day in month to month name.

Python: give start and end of week data from a given date

day = "13/Oct/2013"
print("Parsing :",day)
day, mon, yr= day.split("/")
sday = yr+" "+day+" "+mon
myday = time.strptime(sday, '%Y %d %b')
Sstart = yr+" "+time.strftime("%U",myday )+" 0"
Send = yr+" "+time.strftime("%U",myday )+" 6"
startweek = time.strptime(Sstart, '%Y %U %w')
endweek = time.strptime(Send, '%Y %U %w')
print("Start of week:",time.strftime("%a, %d %b %Y",startweek))
print("End of week:",time.strftime("%a, %d %b %Y",endweek))
print("Data entered:",time.strftime("%a, %d %b %Y",myday))
out:
Parsing : 13/Oct/2013
Start of week: Sun, 13 Oct 2013
End of week: Sat, 19 Oct 2013
Sun, 13 Oct 2013
Learned python in the past 2 days and was wondering if there is a cleaner way to do this.This method works...it just looks ugly and It seems silly to have to create a new time variable for each date, and that there should be a way to offset the given date to the start and end of the week through a simple call but i have been unable to find anything on the internet or documentation that looks like it would work.

Use the datetime module.
This will yield start and end of week (from Monday to Sunday):
from datetime import datetime, timedelta
day = '12/Oct/2013'
dt = datetime.strptime(day, '%d/%b/%Y')
start = dt - timedelta(days=dt.weekday())
end = start + timedelta(days=6)
print(start)
print(end)
EDIT:
print(start.strftime('%d/%b/%Y'))
print(end.strftime('%d/%b/%Y'))

Slight variation if you want to keep the standard time formatting and refer to the current day:
from datetime import date, timedelta
today = date.today()
start = today - timedelta(days=today.weekday())
end = start + timedelta(days=6)
print("Today: " + str(today))
print("Start: " + str(start))
print("End: " + str(end))

Use the pendulum module:
today = pendulum.now()
start = today.start_of('week')
end = today.end_of('week')

you can also use Arrow:
import arrow
now = arrow.now()
start_of_week = now.floor('week')
end_of_week = now.ceil('week')

pip install pendulum
import pendulum
today = pendulum.now()
start = today.start_of('week')
print(start.to_datetime_string())
end = today.end_of('week')
print(end.to_datetime_string())
found from here