@past sphinx
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#🔒 Checking character length of string
rose harborBOT
Python help channel opened
deft bane
!code
rose harborBOT
deft bane
What's the problem?
past sphinx
I want to check when the user inputs a character that exceeds 40 characters and if it does, it re-prompts them to enter in a character less than 40.
When they enter in the menu_list that equals add
kind zenith
I want to check when the user inputs a character that exceeds 40 characters and ...
>>> some_str = 'abcdef'
>>> len(some_str)
6
len(...) can be used to check length
past sphinx
here is a more detailed explanation...
For option “Add”, the program searches the list to see whether the selected year is already there. If it isn’t, the
user is prompted to enter a title, and category, then adds the [year, title, category] to the list. If already there, the
user is asked if they want to replace it with a new title and category. If yes, the program prompts for a title and
category and replaces it. The movie list is kept in year order.
The values are validated by your program as follows:
year – must be an integer between 1927 and 2022, inclusive
title – must be a string of size less than 40 (What i am currently stuck with)
category – must be one of these values: (‘drama’, ‘western’, ‘historical’, ‘musical’, ‘comedy’, ‘action’,
‘fantasy’, ‘scifi’, ‘suspense’, ‘comedy-drama’)
Option “Scat” lists the valid categories.
past sphinx
```pycon
>>> some_str = 'abcdef'
>>> len(some_str)
6
```
`len(...)` can be used ...
so considering that i am checking the string size of the movie... would i use str()?
kind zenith
so considering that i am checking the string size of the movie... would i use st...
mm why would u need str()
past sphinx
mm why would u need `str()`
just throwing ideas out there because ive been stuck on this lol
i attempted using while and if, rearranged the order of the code but can't seem to figure it out.
kind zenith
well, title is entered by a user, right?
past sphinx
yes
kind zenith
and it is stored in a variable
kind zenith
len(movie_1) will get the length of that string
past sphinx
yeah i did that
kind zenith
which you can inside of if statement to compare it against your limit length
past sphinx
well i had it as an if statement before and changed it to a while
kind zenith
while True:
if len(movie_1) > 40:
print('length must be less than 40, please try again')
continue # Skips all code below and goes back to while True
break # Ends the loop
# ^ note the amount of spaces. This break is outside the if statement (just to make sure u know that sorry)
also instead of continue u can use else: ... block
while True:
if len(movie_1) > 40:
print('length must be less than 40, please try again')
else:
break # Ends the loop
while True is simply required to keep asking the question over and over until the user input satisfies the program
kind zenith
anyway, last thing i gotta say about len(...), is that it can be used for a lot of objects, that u can measure the length of.
>>> some_list = [10, 20, 50, 150, 999]
>>> len(some_list)
5
for example u can use len(...) on a list, tuple, etc
makes sense i guess 🗣️
kind zenith
that way of writing a program isnt the best practice, but its alright for now
but u can easily make a mistake and confuse the if statements, for and while loops
by using wrong amount of spaces or something
or putting else statements in wrong places
past sphinx
i think maybe the indentations is the problem here possibly.
kind zenith
also im gonna say a few things apart from the problem
past sphinx
sure
btw i didnt do this all in one sitting, but i came back occassionally to view my code to make sure i wasnt missing anything.
but im all ears tho
kind zenith
wait...
while True:
year_1 = int(input('Type in a year (between 1927 and 2022), or type "0" to quit: '))
if year_1 == 0:
break
if year_1 > 2022:
print("\nIt must be between 1927 and 2022")
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
if year_1 < 1927:
print("\nIt must be between 1927 and 2022")
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
found = False
for s in movies:
if year_1 == s[0]:
replace = input(str(s[0]) + ' is in the list, do you want to replace ' + s[1] + ' (y/n)?')
found = True
that looks like a logic mistake
when u ask for year_1
past sphinx
hmm?
kind zenith
if its outside of range, u enter one of if statements
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
inside you have this
and for example, u enter 1920 again, which is outside of range
and because there are no more if statements down below
the program will continue with that wrong year_1 value
that needs to be a while loop by itself
past sphinx
no is its outside of the range it tells the user that its outside of the range.
of 1927 and 2022
kind zenith
yeah and then it hits the line year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
and then it goes down to found = False
etc
kind zenith
so, yeah, it has to be while True loop by its own
u remember that example right?
while True:
if len(movie_1) > 40:
print('length must be less than 40, please try again')
else:
break # Ends the loop
yes.
kind zenith
thats the problem with if statements!
the way u did them to be exact
year_1 = int(input('Type in a year (between 1927 and 2022), or type "0" to quit: '))
if year_1 == 0:
break
if year_1 > 2022:
print("\nIt must be between 1927 and 2022")
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
if year_1 < 1927:
print("\nIt must be between 1927 and 2022")
...(other code)
lets go line by line
we enter year_1
past sphinx
ok
kind zenith
we enter 2023
past sphinx
yes
kind zenith
so year = 2023
if year_1 == 0: doesnt work
>>> if year_1 > 2022:
print("\nIt must be between 1927 and 2022")
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
then we're in the first if statement, which worked, because 2023 > 2022
and we enter year_1 again
past sphinx
mhm
kind zenith
# year_1 = 2023
# if year_1 == 0:
# break
>>> if year_1 > 2022:
# print("\nIt must be between 1927 and 2022")
>>> year_1 = 2023
if year_1 < 1927:
print("\nIt must be between 1927 and 2022")
we entered 2023 again
whats the next line?
if year_1 < 1927: if its less than 1927
and, 2023 < 1927 is False
past sphinx
hmmm
kind zenith
so we skip over that if statement and go down
and down below is all the other code
here's the fix tho
u can combine those two if statements together
into a single condition
past sphinx
i guess i should change the print statements to be more concise and logically it needs to make sense
yeah i could use "or"
but
just for step by step purposes
i intentionally left it like that
kind zenith
i intentionally left it like that
yeah but as u can see
its worse like that
cuz it allows for such "hack"
past sphinx
yep
kind zenith
and its also a repitition of code
if year_1 > 2022:
print("\nIt must be between 1927 and 2022")
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
if year_1 < 1927:
print("\nIt must be between 1927 and 2022")
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
like you have to write those two lines two times in a rows
and imagine if there was 10 lines
or 50
past sphinx
yeah was abt to put that lol
or
if year_1 > 2022 or year_1 < 1927:
print("\nIt must be between 1927 and 2022")
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
found = False
something like that
kind zenith
and as i said that would be put inside of while loop
the one that i showed u
so it would be something like
past sphinx
oh so
while true:
etc
kind zenith
while True:
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
if year_1 > 2022 or year_1 < 1927:
print("\nIt must be between 1927 and 2022")
continue
break
if u remember, continue just goes to next iteration, so that break is skipped
and it goes back to year_1 = int(...
but if that if statement didnt work
it goes straight into break which ends this loop
and the program continues
kind zenith
well, while True for each... for every single input.. 💀
kind zenith
one for when menu_list == "Add"
and for the year_1
yep
past sphinx
heh
past sphinx
yeah
kind zenith
if u do a while True like that
it will become... a bit hard to end all while True loops
cuz this is a while True inside of another While True
easiest way to exit the program, even though, you would lose that text
like
end of a program
press Enter...
but there is exit() function, that does exactly that
while True:
year_1 = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
if year_1 == 0:
exit()
if year_1 > 2022 or year_1 < 1927:
print("\nIt must be between 1927 and 2022")
continue
break
yea i am not there yet lmao
i wish
kind zenith
but as u can see
if year_1 will be equal to 0
exit() will just close the program
u can put the text u want inside the if statement, tho
past sphinx
yeah, i understand but i haven't learned that yet.
kind zenith
but you'd need to do it in a lot of places... which is annoying
past sphinx
year_1 will equal 0 but it closes the program
kind zenith
but i guess with your knowledge it has to be like that
kind zenith
year_1 will equal 0 but it closes the program
yep
also, just a tip
better to not name stuff like year_1
cuz its not clear what is year_1 supposed to represent
first year?
past sphinx
ik butttttt
kind zenith
in ur case, i guess, its like input_year
past sphinx
i already have a variable called year
past sphinx
ok
kind zenith
other variations would be fine as well, like user_year etc
but i prefer the first one
past sphinx
yeah ill do input_year
past sphinx
how so
kind zenith
but it will get very bad once you will want to check for something longer
like, yes can be spelled yes, Yes, YEs, YES, yES, ... you get it
kind zenith
here's the fix if u wanna know:
>>> 'aAA'.lower()
'aaa'
>>> 'BFwsiFF'.lower()
'bfwsiff'
str.lower() gives u a new string where all characters are lowercase
if replace.lower() == 'yes':
...
sorry that this is out of the topic tho
kind zenith
🗣️
kind zenith
i guess both
past sphinx
yeah
past sphinx
oh yeah
it still doesn't prompt the user to re-enter after 40 characters have been entered even after implementing the while True 😬
kind zenith
it still doesn't prompt the user to re-enter after 40 characters have been enter...
show the while loop where u enter those 40 characters
past sphinx
alright
while True:
if input_year > 2022 or input_year < 1927:
print("\nIt must be between 1927 and 2022")
input_year = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
found = False
for s in movies:
if input_year == s[0]:
replace = input(str(s[0]) + ' is in the list, do you want to replace ' + s[1] + ' (y/n)?')
found = True
if replace in 'nN':
break
if replace in 'yY':
movie_1 = input('Type in a new movie: ')
while len(movie_1) > 40:
movie_1 = input('Type in a movie less than 40 characters: ')
else:
category_1 = input('Type in a new category: ')
if category_1 not in valid_categories:
print('\nEnter in a valid category')
category_1 = input('Type in a new category: ')
s[1] = movie_1
s[2] = category_1
break
while True:
if input_year > 2022 or input_year < 1927:
print("\nIt must be between 1927 and 2022")
input_year = int(input('\nType in a year (between 1927 and 2022), or type "0" to quit: '))
that must be a fully separated loop
i'll write an example where u have to enter age and name, with similiar limitations to ur program
while True:
age = int(input('Enter your age (from 0 to 100) > '))
if age < 0 or age > 100:
print('Age must be a number from 0 to 100. Try again.')
continue
break
while True:
name = input('Enter your name (no more than 20 chars.) > ')
if len(name) > 20 or len(name) == 0:
print('Name must not be empty and cannot be longer than 20 characters. Try again.')
continue
break
print(f'Hello {name}, looks like you are {age} years old!!!')
as u can see... for each single input u must re-ask if that input doesnt satisfy the requirements
so each input is inside its own while True loop
past sphinx
yes lemme check mines rq
kind zenith
and obviously, if u have different commands
it becomes insanely difficult!
i'll write an example with two commands, too
hol' up a minute
past sphinx
ok
kind zenith
print('Available commands:\n'
'say_hi\n'
'add_numbers\n'
'exit\n')
while True:
command = input('Command > ').lower()
if command == 'say_hi':
while True:
name = input('Enter your name (no more than 20 chars.) > ')
if len(name) > 20 or len(name) == 0:
print('Name must not be empty and cannot be longer than 20 characters!')
continue
break
print(f'Hello there, {name}!\n')
elif command == 'add_numbers':
while True:
num_1 = int(input('Enter any number (from 1 to 100) > '))
if num_1 < 1 or num_1 > 100:
print('Please enter a number from 1 to 100!')
continue
break
while True:
num_2 = int(input('Enter any number (from 1 to 100) > '))
if num_2 < 1 or num_2 > 100:
print('Please enter a number from 1 to 100!')
continue
break
print(f'Sum of {num_1} and {num_2} is {num_1 + num_2}!\n')
elif command == 'exit':
print('Goodbye.')
exit()
else:
print(f'Invalid command {command}.')
i see that you haven't used elif in ur code, so if u dont know what it does, it stands for else if
if condition:
...
else:
if another_condition:
...
can be replaced by elif:
if condition:
...
elif another_condition:
...
u can also try copying the code and running it for yourself, but as u can see, even such a simple thing as having 3 commands gets... quite long and difficult
past sphinx
ok
kind zenith
but basically i hope u understand now
an outer while loop that checks for a command
and while loops inside of each command when needed
past sphinx
yepp!
kind zenith
and also i hope the examples are understandable
rose harborBOT
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