Python: Skip user input and call later on - python

I'm fairly new to Python and am currently just learning by making some scripts to use at work. Real simple, just takes user input and stores it in a string to be called later on. The questions are yes/no answers but I wish for the user to have the option to skip and for the question to be asked again at the end, how would I do this?
Currently this is what I've got:
import sys
yes = ('yes', 'y')
no = ('no', 'n')
skip = ('skip', 's')
power = str(raw_input("Does the site have power? (Yes/No): "))
if power.lower() in yes:
pass
elif power.lower() in no:
pass
elif power.lower() in skip:
pass
else:
print ''
print '%s is an invlaid input! Please answer with Yes or No' % power
print ''
exit()
then at the end of the script after all the questions have been asked I have this:
if power.lower() in skip:
power = str(raw_input("Does the site have power? (Yes/No): "))
if power.lower() in yes:
pass
elif power.lower() in no:
pass
else:
print ''
print '%s is an invlaid input! Please answer with Yes or No' % power
print ''
exit()
else:
pass
if power.lower == 'yes':
print 'Site has power'
else:
print 'Site doesnt have power, NFF.'
I understand this is very messy and I'm just looking for guidance/help.
Regards,
Trap.

Since you're rather new to Python I'll give you some tips:
Store all the questions which receive "skip" as a response into a list.
At the end of all your questions, iterate through (hint: "for" loop) all the questions which the user skipped and asked them again.

# Display the sum
print('The sum of {0} and {1} is {2}'.format(num1, num2, sum))
message=("is this correct")
Print=(message)
#store input
answer = input("yes/no:")
if answer == "yes": print("thank you")
if answer == "no": print("hmm, are you sure?")
#store input
answer = input("yes/no:")
if answer == "yes" : print("please call my suppot hotline: +47 476 58 266")
if answer == "no" : print("ok goodbye:)")
if someone had a solution for line 7 when answerd yes skiped to the end desplaing ok `goodbye:) thank you`

Related

Can't get string user input working (Python)

I've searched the web and this site and been messing around all day, trying 100 ways to get this simple little program working. I'm practicing endless While loops and string user inputs. Can anyone explain what I'm doing wrong? Thank you!
while True:
print("This is the start.")
answer = input("Would you like to continue? (Y/N) ")
answer = answer.islower()
if answer == "n":
print("Ok thank you and goodbye.")
break
elif answer == "y":
print("Ok, let's start again.")
else:
print("You need to input a 'y' or an 'n'.")
your code has one thing wrong answer.islower() will return boolean values True or False but you want to convert it into lower values so correct method will be answer.lower()
while True:
print("This is the start.")
answer = input("Would you like to continue? (Y/N) ")
answer = answer.lower() # change from islower() to lower()
if answer == "n":
print("Ok thank you and goodbye.")
break
elif answer == "y":
print("Ok, let's start again.")
else:
print("You need to input a 'y' or an 'n'.")
You just need one amendment to this line:
Instead of
answer = answer.islower()
Change to
answer = answer.lower()

How do I clear an "if" condition

I'm trying to figure out how to clear an "if" condition and how to fix the result = print(x) part of my code. I'm trying to create a little search code based on the variable data, but I can't figure a few things out:
import time
def start():
data = ["Lucas_Miguel", "João_Batista", "Rafael_Gomes", "Bruna_Santos", "Lucas_Denilson"]
print("1" + " - Check Name")
print("2" + " - Register a New Name")
option = input("Choose an option: ")
if option == "1":
def other():
name = input("Type the first name: ")
for x in data:
if name in x:
result = print(x)
while True:
print("Yes " "or " "No")
confirm = input("Is the name you want in the options?: ")
if confirm == "Yes":
break
if confirm == "No":
print("Yes", " or", " No")
try_again = input("Do you want to write again?: ")
if try_again == "Yes":
return other()
other()
else:
print("Option not available")
time.sleep(1)
return start()
start()
The first problem is in the result = print(x) part. It works, but when the answer is more than one name, only the first one appear and I don't know how to fix it.
The second problem is in the "confirm = input" part. Basically, if the person answered with "No", when they go back, the answer will still be saved and the input will run twice, the first time with the saved answer and the second with the new answer. So I want to be able to clear that before the person answer it again.
I want to apologize already if the code is ugly or weird, but I started a few days ago, so I'm still learning the basics. Also thanks in advance for the help.
There is quite a bit here to unpack and like the comment on the question suggests you should aim to look at how to ask a more concise question.
I have some suggestions to improve your code:
Split the other into its own function
Try to use more accurate variable names
As much as you can - avoid having multiple for loops happening at the same time
Have a look at list comprehension it would help a lot in this case
Think about whether a variable really belongs in a function or not like data
What you're asking for is not immediately clear but this code should do what you want - and implements the improvements as suggested above
import time
data = ["Lucas_Miguel", "João_Batista", "Rafael_Gomes", "Bruna_Santos", "Lucas_Denilson"]
def other():
name_input = input("Type the first name: ")
matches = [name for name in data if name_input in name]
if len(matches) == 0:
print ("No matches")
for name in matches:
print(name)
while True:
print("Yes " "or " "No")
confirm = input("Is the name you want in the options?: ")
if confirm == "Yes":
break
if confirm == "No":
print("Yes", " or", " No")
try_again = input("Do you want to write again?: ")
if try_again == "Yes":
return other()
else:
return
def start():
print("1" + " - Check Name")
print("2" + " - Register a New Name")
option = input("Choose an option: ")
if option == "1":
other()
else:
print("Option not available")
time.sleep(1)
return start()
start()
The first problem will be solved when you remove 8 spaces before while True:.
The second problem will be solved when you add return (without arguments) one line below return other() at the indentation level of if try_again == "Yes":
Everybody can see that you are just learning Python. You don't have to apologize if you think, your code is "ugly or weird". We all started with such small exercises.

How can I create a function with Yes or No?

How can I go about creating a input and output with Y\N (yes or no) function in a question?
My example; if my question is Would you like some food? (Y \ N):, how can I do this and have the answers show Yes, please. or No, thank you. for either choice and then proceed to the next question with the same function?
I thought about using this: valid=("Y": True, "y": True, "N": False, "n": False) but that only shows up as True or False for me, or is there a way to change from True \ False to Yes \ No? or this one:
def user_prompt(yes_no):
while True:
user_input=input(yes_no)
But I'm really not sure how else to proceed with this one, or if there's any other easier solution to this.
I think what you're looking for is a conditional print statement rather than a true/false return statement on the function.
For example:
def user_prompt():
while True:
user_input = input("Would you like some food? (Y \ N)")
print ("Yes, please" if user_input == 'Y' else "No, thank you")
Or, more readable:
def user_prompt():
while True:
user_input = input("Would you like some food? (Y \ N)")
if (user_input == 'Y'):
print("Yes, please")
elif (user_input == 'N'):
print("No, thank you")
I hope I understood your question correctly, you basically check the first letter (incase user enters yes/no) every time the user enters a value and try to verify that if it's Y/N you break the loop if not you keep asking the same question.
def user_prompt(yes_no):
while True:
user_input=input(yes_no)
if user_input[0].lower() == 'y':
print("Yes, please.")
break
elif user_input[0].lower() == 'n':
please("No, thank you.")
break
else:
print("Invalid, try again...")
Not sure if this is the best way, but I have a class based implementation of same
""" Class Questions """
class Questions:
_input = True
# Can add Multiple Questions
_questions = [
'Question 1', 'Question 2'
]
def ask_question(self):
counter = 0
no_of_question = len(self._questions)
while self._input:
if counter >= no_of_question:
return "You have answred all questions"
user_input = input(self._questions[counter])
self._input = True if user_input.lower() == 'y' else False
counter += 1
return "You have opted to leave"
if __name__ == '__main__':
ques = Questions()
print(ques.ask_question())
Firstly there is many ways you can go around this, but I am guessing you found the solution yourself already but here is one that is the best.
def get_yes_no_input(prompt: str) -> bool:
allowed_responses = {'y', 'yes', 'n', 'no'}
user_input = input(prompt).lower()
while user_input not in allowed_responses:
user_input = input(prompt).lower()
return user_input[0] == 'y'
continue = get_yes_no_input('Would you like to proceed? [Y/N]: ')
And there we go.

New Programmer, How do you use If and Elif and Else statements? Its giving me NameError: name 'No' is not defined

answer = input('Hi! Would you like to say something? (No or Yes)')
if answer == No or no:
print('Okay then, have a good day!')
elif answer == Yes or yes:
answertwo = input('What would you like to say?')
print(answertwo, 'Hmmmmmm, Intresting.')
**if answer == No or no:
NameError: name 'No' is not defined**
I would suggest you using .lower() for every input. This ensures that if some types "nO" or "YeS" it takes the lowercase equivalent. Example:
ui = input("Type \"Hi\"").lower()
Next, you should really add an else option to your code. This is for answers you don't want them to type. This should be held in a while loop Example:
while(True):
ui = input("Type \"Hi\"").title()
if(ui == "Hi"):
print("Hello")
break
else:
print("That isn't a choice!")

Why isn't this code repeating? Python 3.3

This is a code that I have used when repeating a sequence I have used but it doesnt seem to be working can anyone see any problems?The code is for a currency converter. Im using Python 3.3
userDoAgain = input("Would you like to use again? (Yes/No)\n")
if userDoAgain == "Yes":
getChoice()
elif userDoAgain == "No":
print("Thankyou for using this program, Scripted by PixelPuppet")
import time
time.sleep(3)
else:
print("Error: You entered invalid information.")
doagain()
Edit,This is the rest of the code:
if userChoice == "1":
userUSD = float(input("Enter the amount of USD you wish to convert.\n"))
UK = userUSD * 0.62
print("USD", userUSD, "= ", UK, "UK")
elif userChoice == "2":
UK = float(input("Enter the amount of UK Currency you wish to convert.\n"))
userUSD = UK * 1.62
print("UK", UK, "= ", userUSD, "USD")
def doagain():
userDoAgain = raw_input("Would you like to use again? (Yes/No)\n")
if userDoAgain == "Yes":
getChoice()
elif userDoAgain == "No":
print("Thankyou for using this program, Scripted by PixelPuppet")
import time
time.sleep(3)
else:
print("Error: You entered invalid information.")
doagain()
Generally speaking, using recursion to handle a repeated control flow in Python is a bad idea. It's much easier, and less problematic to use loops instead. So, rather than defining a function doagain to ensure you get an answer to your question about running again, I suggest using a while loop. For the larger function that you'll be repeating, I suggest using a loop as well.
def repeat_stuff():
while True: # keep looping until told otherwise
# do the actual stuff you want to do here, e.g. converting currencies
do_stuff_once()
while True: # ask about doing it again until we understand the answer
userDoAgain = input("Would you like to use again? (Yes/No)\n")
if userDoAgain.lower() == "yes":
break # go back to the outer loop
elif userDoAgain.lower() == "no":
print("Thank you for using this program")
return # exit the function
else:
print("Error: You entered invalid information.")
Note that I've changed the checks of the yes/no input strings to be case insenstive, which is a rather more user friendly way to go.
You are using recursion (the function calls itself) while it may be much nicer to just wrap the code you want to repeat in a while loop.
Example of this usage:
userContinue = "yes"
while (userContinue == "yes"):
userInput = input("Type something: ")
print("You typed in", userInput)
userContinue = input("Type in something else? (yes/no): ").lower()
Probably you need to use the function "raw_input" instead of only input.

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