OPS435 Python Assignment 1 2018 Fall

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 with appropriate functions 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.

Coding Standard

• PEP-8 -- Style Guide for writing Python Code

Due Date

This Assignment is due on Sunday October 14, 2018. Submit on blackboard under the assignments section.

All your Python code for this program must be placed in a single source file. Please include the following declaration by you as a comment in your Python source code file (replace "Student Name" with your own name):

OPS435 Assignment 1 - Fall 2018
Program: [student_id].py (replace student_id with your Seneca User name)
Author: "Student Name"
The python code in this file ([Student_id].py) is original work written by
"Student Name". No code in this file is copied from any other source 
except those provided by the course instructor, including any person, textbook, 
or on-line resource. I have not shared this python program with anyone or 
anything except for submission for grading.  
I understand that the Academic Honesty Policy will be enforced and violators 
will be reported and appropriate action will be taken.

Instruction

Program Name and valid command line arguments

Name your python3 program as [Student_id].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. There is an option called --step that makes the program print out all dates until the final date. If the "YYYYMMDD" format is broken give an appropriate error message. Invalid months (>12) or invalid days of month(different for each month), should be detected and give appropriate error messages. 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

• python3 dbda.py --step 20180101 3, and the output should be
  

    20180102
    20180103
    20180104

• python3 dbda.py 20180701 500, and the output should be
  

    20191113

• python3 dbda.py 20189901 2, and the output should be
  

    Error: wrong month entered

• python3 dbda.py 20180199 2, and the output should be
  

    Error: wrong day entered

• python3 dbda.py 2018 2, and the output should be
  

    Error: wrong date entered

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