"i is not accessed Pylance". the "i" in for loop doesn't work

Question

The variables in all of the following for loops says that "variable is not accessed Pylance" The font color of the variable is "dead".

if special == "n" and numbers == "n":
            for i in range(l):
                password.append(random.choice(chars))
        elif special == "y" and numbers == "y":
            for c in range(l - nos - non):
                password.append(random.choice(chars))
            for s in range(nos):
                password.append(random.choice(special))
            for n in range(non):
                password.append(random.choice(numerics))

Full code

import random

chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUV"
special = ["@", "#", "$", "%", "&", "*", "."]
numerics = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
password = []

def length(): #ask length
    while True:
        length = int(input("Enter length: "))
        if length > 0:
            return length
        else:
            print("Too short")

def spec(): #ask if want speci char
    while True:
        needsSpecial = input("Do you need special chracter? (y/n): ")
        if needsSpecial == 'y' or needsSpecial == 'n':
            return needsSpecial

def numofspecial(): #ask amount of speci char
    while True:
        specialLen = int(input("How many special charcters?: "))
        if specialLen < 1 or specialLen > l:
            print("wrong input")
        else:
            return specialLen


def num(): #ask if want num
    while True:
        needsNum = input("Do you need numbers? (y/n): ")
        if needsNum == 'y' or needsNum == 'n':
            return needsNum
def numofnum(): #ask for # of num
    while True:
        numofnum = int(input("How many numbers?: "))
        if numofnum < 1 or numofnum > l or (specials == "y" and numofnum > (l - nos)): 
            print("wrong input")
        elif l - nos == 0: 
            return numofnum
        else:
            return numofnum

def main():
    password = []
    global l 
    l = length()
    if l == 1:
        lst = [chars, special, numerics]
        password.append(random.choice(random.choice(lst)))
        password = ''.join(str(e) for e in password)
        print(password)
    else:
        global specials
        specials = spec()
        if specials == "y":
            global nos
            nos = numofspecial()
        numbers = num()
        if numbers == "y":
            global non
            non = numofnum()

        if special == "n" and numbers == "n":
            for i in range(l):
                password.append(random.choice(chars))
        elif special == "y" and numbers == "y":
            for c in range(l - nos - non):
                password.append(random.choice(chars))
            for s in range(nos):
                password.append(random.choice(special))
            for n in range(non):
                password.append(random.choice(numerics))
        print(password)
        random.shuffle(password)
        password = ''.join(str(i) for i in password)
        print(password)
main()

Answer 1

As was mentioned in the comments, the reason for the warning is that you don't use the variable; the convention in python is to use _ for such variables. You can get rid of this problem by using random.choices, passing it a k value of the number of random values you want. I've modified your code to simplify it; note this depends on both non and nos being initialised to 0 and modified (not defined) if the user wants numeric or special characters:

nos = 0
specials = spec()
if specials == "y":
    nos = numofspecial()
non = 0
numbers = num()
if numbers == "y":
    non = numofnum()
# ...
noc = l - nos - non
password = []
password += random.choices(chars, k=noc)
password += random.choices(special, k=nos)
password += random.choices(numerics, k=non)
# make a string
password = ''.join(random.sample(password, k=l))

Notes

• you can simplify the string password generation using random.sample
• you had a typo in your if, you should be use specials, not special. However with this code change it is no longer an issue.