I am able to get it to prompt one password, but not all 3. Here is an example of my code:
import getpass
thislist = ["Sun", "Moon", "Star"]
passwds = 'Sun, Moon, Star'
password = input("Please enter a password: ")
if password == "Sun":
print("Welcome, Continue on to the Quiz!")
else:
print("Login failed, Sorry. Please Try Again!")
password = input("Please enter a password: ")^^^ This prompts the password, but when I add the moon/star near "if password ==" it does not work. We are working on adding lists to prompt passwords.
(Sep-21-2019, 06:38 PM)NotPythonQueen Wrote: [ -> ]^^^ This prompts the password, but when I add the moon/star near "if password ==" it does not work.
show the exact code that fails...
(Sep-21-2019, 06:41 PM)buran Wrote: [ -> ] (Sep-21-2019, 06:38 PM)NotPythonQueen Wrote: [ -> ]^^^ This prompts the password, but when I add the moon/star near "if password ==" it does not work.
show the exact code that fails...
import getpass
thislist = ["Sun", "Moon", "Star"]
passwds = 'Sun, Moon, Star'
password = input("Please enter a password: ")
if password == "Sun" "Moon" "Star" :
print("Welcome, Continue on to the Quiz!")
else:
print("Login failed, Sorry. Please Try Again!")
password = input("Please enter a password: ")
if password == "Sun" "Moon" "Star" : is same as
if password == "Sun": if "Sun" == "SunMoonStar":
you can do
if password in ("Sun", "Moon", "Star"): # you can use list instead of tuple...this will also work but the former is better
if password == "Sun" or password == "Moon" or password == "Star":
Anyway to have it use my list above? That works but also just wondering.
One minor nitpick: if "Sun" == "Sun" "Moon" "Star": is actually equivalent to if "Sun" == "SunMoonStar":. Consecutive strings concatenate.
Why doesn't it work when I try to use a dictionary instead of list?
thisdict = {
"1": "sun",
"2": "moon",
"3": "star"
} with the same code?
Because x in thisdict checks against the keys of the dictionary (1, 2, and 3). You would have to check x in thisdict.values(). Note that checking against the keys is very fast in Python, but checking against the values is probably a bit slower than using a list.
(Sep-22-2019, 01:19 AM)ichabod801 Wrote: [ -> ]Because x in thisdict checks against the keys of the dictionary (1, 2, and 3). You would have to check x in thisdict.values(). Note that checking against the keys is very fast in Python, but checking against the values is probably a bit slower than using a list.
okay i will use it like this, but it still didn't work
thisdict = {
"sport": "football",
"drink": "sprite",
"food": "pizza"
}
password = input("Please enter a password: ")
if password x in thisdict.values()
print("Welcome!")