How do you get a string of characters in one line?

Question

I'm trying to make a random generator of numbers and letters go into one line after being in a string, for example from:

a
b
c

Like so...:

To "abc"

I was wondering if there was any way of doing that. The code I use is:

import random
import time
amount = int(input("How many digits do you want the password to be?\n"))
time.sleep(0.5)
print("============================================================\n")
for count in range(amount):
    time.sleep(0.1)
    num = random.randint(1,62)
    if num == 1:
        a = print("1")
    if num == 2:
        print("2")
    if num == 3:
        print("3")
    if num == 4:
        print("4")
    if num == 5:
        print("5")
    if num == 6:
        print("6")
    if num == 7:
        print("7")
    if num == 8:
        print("8")
    if num == 9:
        print("9")
    if num == 10:
        print("a")
    if num == 11:
        print("b")
    if num == 12:
        print("c")
    if num == 13:
        print("d")
    if num == 14:
        print("e")
    if num == 15:
        print("f")
    if num == 16:
        print("g")
    if num == 17:
        print("h")
    if num == 18:
        print("i")
    if num == 19:
        print("j")
    if num == 20:
        print("k")
    if num == 21:
        print("l")
    if num == 22:
        print("m")
    if num == 23:
        print("n")
    if num == 24:
        print("o")
    if num == 25:
        print("p")
    if num == 26:
        print("q")
    if num == 27:
        print("r")
    if num == 28:
        print("s")
    if num == 29:
        print("t")
    if num == 30:
        print("u")
    if num == 31:
        print("v")
    if num == 32:
        print("w")
    if num == 33:
        print("x")
    if num == 34:
        print("y")
    if num == 35:
        print("z")
    if num == 36:
        print("A")
    if num == 37:
        print("B")
    if num == 38:
        print("C")
    if num == 39:
        print("D")
    if num == 40:
        print("E")
    if num == 41:
        print("F")
    if num == 42:
        print("G")
    if num == 43:
        print("H")
    if num == 44:
        print("I")
    if num == 45:
        print("J")
    if num == 46:
        print("K")
    if num == 47:
        print("L")
    if num == 48:
        print("M")
    if num == 49:
        print("N")
    if num == 50:
        print("O")
    if num == 51:
        print("P")
    if num == 52:
        print("Q")
    if num == 53:
        print("R")
    if num == 54:
        print("S")
    if num == 55:
        print("T")
    if num == 56:
        print("U")
    if num == 57:
        print("V")
    if num == 58:
        print("W")
    if num == 59:
        print("X")
    if num == 60:
        print("Y")
    if num == 61:
        print("Z")

This basically creates the string of how many characters you want but I can't get them all on one line.

Answer 1

You can use the string module.

Example Code:

import string
print(string.ascii_letters+string.digits)

You can then get a certain number of random letters and numbers using this code:

import random
import string
number_of_letters_and_digits = 5
print(''.join([random.choice(string.ascii_letters+string.digits) for i in range(number_of_letters_and_digits)]))

Completed Code:

import random
import string
import time
amount = int(input("How many digits do you want the password to be?\n"))
time.sleep(0.5)
print("============================================================\n")
print(''.join([random.choice(string.ascii_letters+string.digits) for i in range(amount)]))

If you need to use the code that you showed, you could append the letters to the string (replace print("letter") with listName.append("letter")). Then, after the for loop, you could run print(''.join(listName)).