## ------------------------------------------------------------------------------------------------------------------------------
import re
import sqlite3
#import mysqldb
import sys

import getpass

try:
    conn = sqlite3.connect('userpass.db')
except:
    print ("Cannot Find Any DB")
    exit()
c = conn.cursor()

##### NOTE
##### First Time Users Uncomment the next two lines, until I make a feature rich installer
#c.execute('''CREATE TABLE if not exists userpass(user text, pass text)''')
#c.execute('''CREATE TABLE if not exixts p_history(user text, pass text)''')

### NOTE (Sankrant):
## This is my first time working with a structured database, I still do not know the left join correctly. So here you have it,
## two redundant tables to solve the triple password problem. A better design would be to make a primary key linked with the user
## on the first table, and use that key instead

## This is also a nightmare in security practice according to some database guys
## I should have created another secure storage for passwords doing some magic fu linking. Will come back to it later

## TODO (Sankrant): maybe hide the password entry by ****

## -------------------------------------------------------------------------------------------------------------------------------


def add_user():
    username = input("Please Enter Username: ")
    if user_exists(username) == 1:
        print ("User Already Exists")
        exit()
    if check_user(username) == -1:
        print("Wrong Format: Please check formating by typing help as argument")
        exit()
    try:
        c.execute("INSERT INTO userpass VALUES (?, '')", (username,))
        print ("User Added")
    except:
        print("The database query did not work")
    comDB()

## Sometimes Regex is too complex to put your head around
## Good old C style coding

def issym(ch):
    if (ch == '!' or ch == '@' or ch == '#' or ch == '$' or ch == '%' or ch == '^' or ch == '&' or ch == '*' or ch == '(' or ch == ')' or ch == '-' or ch == '=' or ch == '+' or ch == '[' or ch == ']' or ch == '{' or ch == '}' or ch == '<' or ch == '>' or ch == '.' or ch == '?' or ch == ':' or ch == '|' or ch == '\\' or ch == '/' or ch == '`' or ch == '~'):
        return True;
    else:
        return False

def check_7(password):
    for ind,ch in enumerate(password[:-2]):
        if ch.isupper() and password[ind+1].isupper() and password[ind+2].isupper():
            return -1
        elif ch.islower() and password[ind+1].islower() and password[ind+2].islower():
            return -2
        elif ch.isdigit() and password[ind+1].isdigit() and password[ind+2].isdigit():
            return -3
        elif issym(ch) and issym(password[ind+1]) and issym(password[ind+2]):
            return -4
    return 0


def add_pass(username):
    password = input("Enter Password: ")
    check_password_count(username, password)
    if pass_check(password) == -1:
        print ("Follow the Password Rules")
        exit()
    elif pass_check(password) == 0:
        password_re = input("Retype Password: ")
        if (password == password_re):
            try:
                c.execute("INSERT INTO p_history VALUES (?, ?)", (username, password))
                c.execute("UPDATE userpass SET pass = ? WHERE user = ?", (password, username))
            except:
                print("Cant Connect to DB")
                exit()
            comDB()
            print ("Password Added for User")
        else:
            print ("The passwords do not match")
            exit()

def CLIHelp():
    print (
        '''
        Usage : python zdbdemo.py [option] [arg]

        Options
        --------
        useradd : Prompts for adding a Username to the database with an empty password

        passwd : Add a Strong Password and take Username as an argument

        help : Display help

        '''
    )

def Help():
    print('''

    User Rules
    ----------

A) To create new user :

          program_name useradd

    If the user does not exist, it will create a user and then prompt for a password for the user ( the password complexity is defined below ). This user will be stored in a DB or Flat file .
    If the username exists, it will fail letting the user know that the user already exists.
    The username can contain only _ or .  As special character and nothing else .
    The username should not be greater than 10 characters.
    The username should not start with a number .

B) To change password :

         program_name passwd  $username

    The password check will be as below :

    The password length should be more than 8 characters.
    The password length should be less than 15 characters.
    The password should have atleast one [a-z]  ( class 1)
    The password should have atleast one [0-9]  ( class 2)
    The password should have at least one special character ( class 3 )
    The password should not have characters  /,  ~, ` .
     One cannot have more than two same class characters adjacent to each other .
    One cannot use the same password which he/she has used in the past three times . Which means you will have to store the username and corresponding password either in some DB or you can even use a flat file and maintain a history of 3 for each user .
    The password should be entered twice and matched .


    ''')

## HELPER FUNCTIONS
## ----------------------------------------------------------------------------------------------------------------------------------

def purgeOldTable():
    #if __counterCheck__() == 1:
    c.execute("DELETE FROM p_history")

def pass_exists(username, password):
    c.execute("Select count(pass) from p_history where user=? and pass=?", (username, password))
    asdf = c.fetchone()[0]
    c.execute("Select count(pass) from p_history where user=?", (username, ))

    ## TODO(Sankrant): Checking this stuff without pdb is frustating, do some debug

    ## TODO(Sankrant): Encapsulate all of this in clean functions

    qwer = c.fetchone()[0]
    if (qwer >= 3):
        purgeOldTable()
    if (asdf >= 1):
        return 1
    else:
        return 0

def username_exists(password):
    try:
        fetch2 = c.execute("select user from p_history where pass=?", (pass_exists(username, password))).fetchone()[0]
        return fetch2
    except:
        return 0

def user_exists(username):
    try:
        c.execute('''SELECT user FROM userpass WHERE user=?''', (username,))
    except:
        print("Can not execute query")
        exit()
    exists = c.fetchall()
    if exists:
        return 1

def pass_check(password):
    ## TODO (Sankrant): Regular Expressions can be executed in a more elegant manner?????
    if (len(password) < 8 and len(password) > 15):
        flag = -1
    elif not re.search("[a-z]", password):
        flag = -1
    elif not re.search("[A-Z]", password):
        flag = -1
    elif not re.search("[0-9]", password):
        flag = -1
    elif not re.search("[_@$!#$%^&*()\\\]", password):
        flag = -1
    elif re.search("[/~`]", password):
        flag = -1
    elif check_7(password) != 0:
        flag = -1
    else:
        flag = 0

    return flag

def check_user(username):
    if (len(username) > 10):
        flag = -1
    elif re.search("[_.]", username):
        flag = -1
    else:
        flag = 1
    return flag

def check_password_count(username, password):

    if pass_exists(username, password) == 1:
        print("Please Enter Another Password as this has been used before")
        exit()

def comDB():
    conn.commit()

def do_nothing():
    return


## -------------------------------------------------------------------------------------------------------------------------------------------
## MAIN
## DONE (Sankrant): It is trivial to do the entry point function now, I might do it before the final show

def main():
    if(sys.argv[0] == "zdbdemo.py"):
        CLIHelp()

    if(sys.argv[1] == "useradd"):
        add_user()

    elif(sys.argv[1] == "useradd" and sys.argv[2] == "-p"):
        add_user()
        add_pass(username)

    elif(sys.argv[1] == "passwd" and sys.argv[2]):
        if (user_exists(sys.argv[2])):
            add_pass(sys.argv[2])
        else:
            print("No user by the name of %s", %(sys.argv[2]))
        elif(sys.argv[1] == "help"):
            Help()

        else:
            do_nothing()


if __name__ == "__main__":
    main()