Keypad Passcode

[CookieGamez2018]

Hello,
i need some help with setting up a pass-code with my raspberry pi. i would like it to be a four digit code that you enter on the keypad and it prints "access granted" if right and "access denied" if wrong. any help would be great!

my code so far, it prints the button you hit:

Code: Select all

import RPi.GPIO as GPIO
import time

GPIO.setmode(GPIO.BOARD)
GPIO.setwarnings(False)

MATRIX = [
     [1,2,3,'A'],
     [4,5,6,'B'],
     [7,8,9,'C'],
     ['*',0,'#','D']
]

ROW = [3,5,8,10]
COL = [19,21,23,24]

for j in range(4):
    GPIO.setup(COL[j], GPIO.OUT)
    GPIO.output(COL[j], 1)

for i in range(4):
    GPIO.setup(ROW[i], GPIO.IN, pull_up_down = GPIO.PUD_UP)

try:
    while(True):
        for j in range(4):
            GPIO.output(COL[j],0)
            
            for i in range(4):
                if GPIO.input(ROW[i]) == 0:
                    print (MATRIX[i][j])
                    time.sleep(0.2)
                    while(GPIO.input(ROW[i]) == 0):
                        pass


            GPIO.output(COL[j],1)    
except KeyboardInterupt:
    GPIO.cleanup()
    

[Andyroo]

Welcome to the forum.

What keypad do you have?

Could please edit your post so the code is in the 'code' tag:

[ c o d e ] Python goes here [ / c o d e ]
(without the spaces by using the </> button>)

Then we can see the indents in the program - these are vital for Python

[scotty101]

Code: Select all

import RPi.GPIO as GPIO
import time

GPIO.setmode(GPIO.BOARD)
GPIO.setwarnings(False)

MATRIX = [
     [1,2,3,'A'],
     [4,5,6,'B'],
     [7,8,9,'C'],
     ['*',0,'#','D']
]

ROW = [3,5,8,10]
COL = [19,21,23,24]

for j in range(4):
    GPIO.setup(COL[j], GPIO.OUT)
    GPIO.output(COL[j], 1)

for i in range(4):
    GPIO.setup(ROW[i], GPIO.IN, pull_up_down = GPIO.PUD_UP)

try:
    while(True):
        for j in range(4):
            GPIO.output(COL[j],0)
            
            for i in range(4):
                if GPIO.input(ROW[i]) == 0:
                    print (MATRIX[i][j])
                    time.sleep(0.2)
                    while(GPIO.input(ROW[i]) == 0):
                        pass


            GPIO.output(COL[j],1)    
except KeyboardInterupt:
    GPIO.cleanup()

Does your code currently work? As in does it print out the correct key when you press a button?

If so, all you need to do is add each key pressed to a list and compare the contents of the list to another list that contains the correct password.

Might look something like

Code: Select all

#Start with a blank list
userEntry = []
# Define the correct answer
correctKey = [1,2,3,4]
...........
#Add the pressed key to the list
userEntry.append(MATRIX[i][j])
#compare the length of the list of entered keys against the correct answer
if len(userEntry) == len(correctKey):
    #If they match, check to see if the value is correct
    if userEntry == correctKey:
        print("Correct")
    else:
        print("Incorrect!!")
    #Clear the list for next time.
    userEntry.clear()
        

[Andyroo]

@scotty101 - I hope your indents match his as that could be part of his problem

[scotty101]

@scotty101 - I hope your indents match his as that could be part of his problem

Not sure what you mean.

[Andyroo]

@scotty101 - I hope your indents match his as that could be part of his problem

Not sure what you mean.

Apologies - spent part of the weekend sorting a working program where my indentation did not match the original and that was the issue - only when it was posted in code blocks did I find it

[CookieGamez2018]

Code: Select all

import RPi.GPIO as GPIO
import time

GPIO.setmode(GPIO.BOARD)
GPIO.setwarnings(False)

MATRIX = [
     [1,2,3,'A'],
     [4,5,6,'B'],
     [7,8,9,'C'],
     ['*',0,'#','D']
]

ROW = [3,5,8,10]
COL = [19,21,23,24]

for j in range(4):
    GPIO.setup(COL[j], GPIO.OUT)
    GPIO.output(COL[j], 1)

for i in range(4):
    GPIO.setup(ROW[i], GPIO.IN, pull_up_down = GPIO.PUD_UP)

try:
    while(True):
        for j in range(4):
            GPIO.output(COL[j],0)
            
            for i in range(4):
                if GPIO.input(ROW[i]) == 0:
                    print (MATRIX[i][j])
                    time.sleep(0.2)
                    while(GPIO.input(ROW[i]) == 0):
                        pass


            GPIO.output(COL[j],1)    
except KeyboardInterupt:
    GPIO.cleanup()

Does your code currently work? As in does it print out the correct key when you press a button?

If so, all you need to do is add each key pressed to a list and compare the contents of the list to another list that contains the correct password.

Might look something like

Code: Select all

#Start with a blank list
userEntry = []
# Define the correct answer
correctKey = [1,2,3,4]
...........
#Add the pressed key to the list
userEntry.append(MATRIX[i][j])
#compare the length of the list of entered keys against the correct answer
if len(userEntry) == len(correctKey):
    #If they match, check to see if the value is correct
    if userEntry == correctKey:
        print("Correct")
    else:
        print("Incorrect!!")
    #Clear the list for next time.
    userEntry.clear()
        

Yes it does print the correct key that I press. also where would i put the code that you provided

[CookieGamez2018]

Welcome to the forum.

What keypad do you have?

Could please edit your post so the code is in the 'code' tag:

[ c o d e ] Python goes here [ / c o d e ]
(without the spaces by using the </> button>)

Then we can see the indents in the program - these are vital for Python

i am using a 4x4 keypad

[scotty101]

Did you try using the code example I provided?

[CookieGamez2018]

Did you try using the code example I provided?

Yes it did work thank you so much

[CookieGamez2018]

Thank you Scotty101 so much for helping me and solving the problem