OPS435 Python Assignment 2 A
Contents
Overview
When making back up of data files or log files, it is a very common practice to name the backup directories and/or files based on the date the backup was done. In order to restore or locate the directory/file, we often need to find out the backup date from today's date.
The task for this assignment is to write a python program which will take a date in the "YYYYMMDD" format and the number of day before or after the given date as the command line arguments, and output the actual date which is the number of day before or after the given date in the same format.
Instruction
Program Name and valid command line arguments
Name your python3 program as dbda.py
. The program should accept two command line parameters, the first one is the date in "YYYYMMDD" format, and the second one is the number of day from the given date, a position value indicates the number of days after the given date, and a negative value indicates the number of days before the given date. For examples:
-
python3 dbda.py 20180101 1
, and the output should be
20180102
-
python3 dbda.py 20180101 -1
, and the output should be
20171231
-
python3 dbda.py 20180101 2
, and the output should be
20180103
If there is too few or too many command line parameters given, display the proper usage.
Program structure
Your program code should all be in a single python file with at least two functions, one called yesterday() and the other called tomorrow():
- The yesterday() function will take a date in "YYYYMMDD" format and return the date of the previous day in the same format.
- The tomorrow() function will take a date in "YYYYMMDD" format and return the date of the next day in the same format. Next paragraph is a sample python code for the tomorrow() function.
#!/usr/sbin/env python3 import ... def tomorrow(today): .... return next_day def yesterday(today): .... return previous_day .... if __name__ == __main__: .. processing command line arguments .. .. set loop based on the number of days given .. call tomorrow() or yesterday() as appropriate ... output the expected date
Sample code for the tomorrow() function
# Return the date in YYYYMMDD after the given day # def tomorrow(today): if len(today) != 8: return '00000000' else: year = int(today[0:4]) month = int(today[4:6]) day = int(today[6:]) lyear = year % 4 if lyear == 0: feb_max = 29 # this is a leap year else: feb_max = 28 # this is not a leap year lyear = year % 100 if lyear == 0: feb_max = 28 # this is not a leap year lyear = year % 400 if lyear == 0: feb_max = 29 # this is a leap year tmp_day = day + 1 # tomorrow's day mon_max = { 1:31, 2:feb_max, 3:31, 4:30, 5:31, 6:30, 7:31, 8:31, 9:30, 10:31, 11:30, 12:31} if tmp_day > mon_max[month]: to_day = tmp_day % mon_max[month] # if tmp_day > this month's max, reset to 1 tmp_month = month + 1 else: to_day = tmp_day tmp_month = month + 0 if tmp_month > 12: to_month = 1 year = year + 1 else: to_month = tmp_month + 0 next_date = str(year)+str(to_month).zfill(2)+str(to_day).zfill(2) return next_date