This is my code :
#Choose Report
def chooseReport():
print "Choose a report."
while True:
choice = raw_input("Enter A or B: ")
if choice == 'a' or choice == 'A':
reportA()
break
elif choice == 'b' or choice == 'B':
reportB()
break
else:
continue
When I input either a or b, it just asks me to "Enter A or B" again. It doesn't go to the functions its supposed to.
Any idea why is this?
The code is perfect, except a redundant else, as mentioned in the comment. Are you entering a (a + space) rather than simply a (a without space) ? The problem is in the input that you are providing and not in the code!
def chooseReport():
print "Choose a report."
t=True # t starts at true
while t:
choice = raw_input("Enter A or B: ")
if choice == 'a' or choice == 'A':
reportA()
t=False # t turns false to break out of loop
elif choice == 'b' or choice == 'B':
reportB()
t=False
Try this. It keeps looping when t is true and stops when t is false. The problem might also be in reportA or reportB or how you are calling chooseReport.
The problem is in the raw_input(). It returns a string but maybe this string is "a " or "a\n" though you have entered "a" or "b".
I would do this:
def chooseReport():
print "Choose a report."
while True:
choice = raw_input("Enter A or B: ")
if "a" in choice or "A" in choice:
reportA()
break
elif "b" in choice or "B" in choice:
reportB()
break
else:
continue
Tried your code in the following script, it works fine both on Linux and on Windows.
def reportA():
print "AAAAA"
def reportB():
print "BBBBB"
#Choose Report
def chooseReport():
print "Choose a report."
while True:
choice = raw_input("Enter A or B: ")
if choice == 'a' or choice == 'A':
reportA()
break
elif choice == 'b' or choice == 'B':
reportB()
break
else:
continue
chooseReport();
First, your code works fine, the most probably error is that you are writing a wrong input (e.g: with more characters). To solve that you could use "a" in choice or "A" in choice. But if it isn't working... keep reading.
It's seems that break isn't affecting the while loop, I don't have python 2 so I am not very sure why (in python 3 [after change raw_input to input and print to print()] your code works perfect). So you should use the condition of the while to break it.
while True work theorically for ever because each time the code is executed it checks the condition -True- and because it's true it keeps looping.
You could manipulate that condition in order to break the loop (don't allow execute again its code).
For example you could use this:
#Choose Report
def chooseReport():
print "Choose a report."
allow = True # allow start as True so the while will work
while allow:
choice = raw_input("Enter A or B: ")
if choice.lower().strip() == "a": # This is better. .lower() make "A" > "a", and .strip() delete " a " to "a", and "a/n" to "a".
reportA()
allow = False # allow now is False, the while won't execute again
elif choice.lower().strip() == "b":
reportB()
allow = False # allow now is False, the while won't execute again
# The else is complete redundant, you don't need it
Code is fine. I think you call your chooseReport() function in a loop or your input has extra characters and if conditions didn't satisfied.
Related
I have had an issue where I can't get the While loop to terminate.
userinput = ("")
while userinput != ("Search" or "Add"):
userinput = input("Search or Add?")
if userinput == "Search":
Search()
elif userinput == "Add":
print("run add request")
else: print("please choose from the following two options.")
Edit: I am sorry the changes have worked. I think after I implemented the changes I had an issue with the Shell running the previous version. Sometimes I have no idea what is happening. Thank you all again.
Edit Edit: Placed the original code back in as I did not take into account that it would confuse anyone looking for their own solution. I am quite new in terms of usage of the site. Thanks again for the help
The issue is with your while test. A couple of things:
You can't use or like this. or needs two full conditions that resolve to true or false. Here you have one condition userinput != "Search" and a string "Add". So it's always going to return True since a non-zero value is always True.
As an example:
if "Add": print("true")
>>true
Instead:
userinput != "Search" or userinput != "Add"
or is not correct when testing two negations like !=. One of the two conditions will always return true. For instance if you input "Add" then the condition userinput != "Search" will be True and your while loop will continue since True or False = True. So on and so forth. Instead you want an and.
while userinput != "Search" and userinput != "Add":
As I suggested in my comment though, it's probably just easier to use the not in operator on a list:
while userinput not in ['Search','Add']:
This way as your list grows your test stays nice and small/condense.
Also, while this is just my opinion, I applaud your original pre-edit code where you supplied the condition for breaking your while loop in the while statement instead of doing while True:. Having had many years of bug hunting and feature adding and hotfixing, I know every time I see while True: I'm going to be hunting through hundreds of lines of codes looking for every break. while True: and break has its time and place (I imagine), but I feel like it should be an exception use-case, not the rule.
My solution looks like this:
userinput = ""
while userinput != "Exit":
userinput = input("Search, Add or Exit: ")
if userinput == "Search":
print("run search request")
elif userinput == "Add":
print("run add request")
elif userinput != "Exit":
print("please choose from Search, Add or Exit.")
Notes:
Variable userinput is initialised to "". It doesn't need to be a tuple, ("")
I introduced an 'Exit' option. We loop until we encounter Exit. No need for break or continue
I changed the prompt to a colon and a space, and also show the three options.
We compare the input to the three options. If nothing valid, print an error message
As an alternative, you can use the following, using continue in the else clause, and break at the end of the while block. This way, it is easier to add more elif clauses, without potentially forgetting a break statement.
(It also doesn't require setting a default value for userinput, since that is done as the first thing in the while loop.)
while True:
userinput = input("Search or Add?")
if userinput == "Search":
Search()
elif userinput == "Add":
print("run add request")
else:
print("please choose from the following two options.")
continue
break
Using the word break will break you out of a while loop, like this:
while True:
break
Place break where you want the while loop to end.
The problem is that once the input is received, and when the input meets the end condition of the while loop, and it isn't able to reach the code checking for the condition! Try this:
while True:
userinput = input("Search or Add?")
if userinput == "Search":
Search()
break
elif userinput == "Add":
print("run add request")
break
else: print("please choose from the following two options.")
The format was not correct in the while condition statement part, You can try out this...
userinput = ("")
while userinput not in ["Search", "Add"]:
userinput = input("Search or Add?")
if userinput == "Search":
Search()
elif userinput == "Add":
print("run add request")
else: print("please choose from the following two options.")
Background:
I have made a program that takes a text input, applies encryption (a simple cipher) and saves the output to a list - if desired. Messages can also be decrypted.
The program is navigated through an options menu:
1. Encrypt a message
2. View encrypted messages
3. Decrypt a message
To allow all sections of the program to access the same list (variable) of saved messages, I have written it within a class. Within the class exist def's that call on this list.
Only the 'encrypt a message' bit has been written so far.
Problem:
The user decision flow is made with two Y/N choices.
However, the choices do not work - even if the user types in 'N' - my program thinks they have typed 'Y' in both cases.
def encrypt():
def save_cip():#This function allows the user to save the ciphered message to the ciphered_messages if they choose
choosing = True
while choosing:
save_choice = input("Would you like to save your Ciphered message? (Y/N)\n")
if save_choice == "Y" or "y":
print("You chose yes")
cct.ciphered_messages.append(' '.join(["Message", str(len(cct.ciphered_messages)), ":", cipher]))
choosing = False
elif save_choice == "N" or "n":
print("You chose no")
choosing = False
continue
else:
print("That was not a valid entry, please enter Y or N only")
continue
I think the problem lay within the scope, and that somehow the same variable is not being referenced when setting and reading Y or N. I have been going up and down the same code for about 3 hours and still nothing, declaring all variables in many different places with no luck, so any advice greatly appreciated.
Full executable code:
class cct:
print("Welcome to the CaeserCipher tool v1.0")
menu_state = "main" #This is used to record what state the program is in
unciphered_messages = [] #Decrypted messages are saved here
ciphered_messages = [] #Encrypted messages are saved here
def menu(): #This is the main menu interface
while cct.menu_state == "main": #This while
cct.menu_state = input("What would you like to do? \n 1: Encrypt a Message \n 2: View Encrypted Messages \n 3: Decrypt a message\n")
if cct.menu_state == "1":
cct.encrypt()
elif cct.menu_state == "2":
cct.view()
elif cct.menu_state == "3":
cct.decrypt()
elif cct.menu_state == "main":
print("\n\nWelcome back to the menu!\n\n")
else:
print("You did not enter a valid choice. Please enter a number between 1 and 3.\n")
cct.menu_state = "make_choice"
continue
def encrypt():
def save_cip():#This function allows the user to save the ciphered message to the ciphered_messages if they choose
choosing = True
while choosing:
save_choice = input("Would you like to save your Ciphered message? (Y/N)\n")
if save_choice == "Y" or "y":
print("You chose yes")
cct.ciphered_messages.append(' '.join(["Message", str(len(cct.ciphered_messages)), ":", cipher]))
choosing = False
elif save_choice == "N" or "n":
print("You chose no")
choosing = False
continue
else:
print("That was not a valid entry, please enter Y or N only")
continue
#This while loop continually takes messages, gives the option of saving, and asks if you want to cipher another
while cct.menu_state == "1":
text = input("Enter your message: ") #Enter the message you wish to cipher
cipher = '' #Create a string for the cipher
for char in text: #This for sub-loop will increment each character by 1. e.g. A -> B, T -> U, Z -> A
if not char.isalpha():
continue
char = char.upper()
code = ord(char) + 1
if code > ord('Z'):
code = ord('A')
cipher += chr(code)
print(' '.join(["Your ciphered message is:", cipher]))
save_cip()
print(cct.ciphered_messages)
#This sub-while loop is reponsible for checking if you want to cipher another message
#and making sure the user has made a valid choice
choosing_another = True
while choosing_another == True:
choice_variable = input(' '.join(["You have", str(len(cct.ciphered_messages)), "saved messages \n", "Would you like to Cipher another? (Y/N)\n"]))
if choice_variable == "Y" or "y":
print("You chose yes")
choosing_another = False
elif choice_variable == "N" or "n":
print("You chose no")
cct.menu_state = "main"
choosing_another = False
else:
choice_variable = input("That was not a valid entry, please enter Y or N only: \n")
continue
return
def view():
#TO BE CODED
return
def decrypt():
#TO BE CODED
return
cct.menu()
This is always true:
if save_choice == "Y" or "y":
bacause its interpreted as :
if (condition) or (condition):
the 1st condition is (save_choice == "Y")
the 2nd condition is just ( "y" ) which python interprets as 'True'
So the whole condition is always True.
U probably mean:
if save_choice == "Y" or save_choice == "y":
Or better :
if save_choice.lower() == "y":
Not a python programmer, but i don't think python allows a "implied" subject for the or operator
--> IF save_choice == "Y" OR save_choice == "y" :
I have been working on this code for hours, and I still can solve it.
inside the function build a forever loop (infinite while loop) and inside the loop complete the following
I want it to use a variable to gather input which is supposed to be either an integer of 'q' to quit.
It should check if the input string is a digit (integer) and if it is...
add input integer to report variable.
If the variable is "A" add the numeric character(s) to the item string seperated by a new line.
If the report type is q,
If the report type is "A" print out all the integer items entered and the sum total.
If report type is "T" then print out the sum total only
break out of while loop to end the function after printing the report ("A" or "T").
If not a digit and if not a "Q" then print a message that the "input is invalid".
def adding_report(report=[]):
report = []
at = input("Choose a report type: 'A' or 'T' : ")
while at.lower() != 'a' and at.lower() != 't':
print('what?')
at = input("Choose a report type: 'A' or 'T' : ")
while True:
re = input("print an integer or 'Q' : ")
if re.isdigit() is True:
report.append(re)
report.append('\n')
elif re.startswith('q') is True:
if at.lower() == 'a' is True:
break
print(report)
print(sum(report))
elif at.lower() == 't' is True:
print(sum(report))
break
else:
pass
elif re.isallnum() is False and re.lower().startswith('q') is False:
print('invalid response.')
else:
pass
adding_report(report=[])
If anyone found any way to fix the bugs, could they please tell me. Thanks in advance.
you must put tabulation to code at least here
def adding_report(report=[]): # <-- You have errors here, defined function has nothing in it
report = [] # <-- I suggest tabulating this
this is incorrect usage, and it works
if re.isdigit() is True: # Not recommended
whenever you check if something is True, just Do not do it,
this is better way for simple if statement:
if re.isdigit(): # Recommended, will excecute when it returns any value, except (False and None)
move break after code, if you want it to excecute:
break # <-- this stops loop! you won't print anything that is after it
print(report)
print(sum(report))
I guess you are trying to do something like this maybe? I've changed a bit your code and made some comments where the errors where present.
def adding_report(report=[]):
report = []
at = input("Choose a report type: 'A' or 'T' : ")
while at.lower() != 'a' and at.lower() != 't':
print('what?')
at = input("Choose a report type: 'A' or 'T' : ")
while True:
re = input("print an integer or 'Q' : ")
print(re)
if re.isdigit(): # Remove is true
report.append(int(re)) # Add integer to list. Use int() to convert string to int
elif re.startswith('q'): # Remove is true
if at.lower() == 'a':
print(report) # This two prints MUST be above de break statement in ordered to be executed
print(sum(report))
break
elif at.lower() == 't': # Remove is true
print(sum(report))
break
else:
pass
elif not re.isdigit() and not re.lower().startswith('q'): # Replaced is False with not
print('invalid response.')
else:
pass
adding_report(report=[])
I am stuck on a simple issue. I am attempting to ask the user to choose a desired function from a list. This inputted user string will invoke the chosen function until it finishes running. (It is a lighting sequence). After this sequence ends, I would like to ask the user if he or she wishes to choose another function. If so, continue. If not, exit the code.
I cannot decide if a while true or if statement is the best to achieve this.
Here is my code:
# random functions
def rainbow():
print 'rainbow'
def clover():
print 'clover'
def foo():
print 'eggs'
if __name__ == '__main__':
# here are some random initializations
print 'ctr-c to quit'
user_input = input("choose from the following: ")
if user_input == 'rainbow':
print 'generating'
rainbow()
rainbow()
rainbow()
user_input = input('choose another')
if user_input == 'foo':
clover()
clover()
I would suggest using a while loop here until you get a successful user_input, upon which you'll want to break the loop. Inside the while look you can have your if statements as needed. For example, in your above code, what happens if the user types in "rainboww", it basically just exits the program. It'd be better to have it like this:
while True:
user_input = input('...')
if "good result"
break
else:
continue
while True:
user_input = input("choose from the following: ")
if user_input == "condition a":
do something
elif user_input == "condition b":
do something..
elif any(user_input == keyword for keyword in ["q", "quit"]):
# when meet any quit keyword, use break to terminate the loop
break
else:
# when doesn't find any match, use continue to skip rest statement and goto the beginning of the loop again
continue
while True can meet your requirement. you can use if-elif-else clauses to do different works.
I have a problem with a condition in my python code.
It's a mathematics application, and here's the part of the code that is not working well:
def askNumber():
"""Asks the number to test"""
a=raw_input("Select the number to test (type 'exit' for leaving):")
if len(a)!=0 and a.lower!="exit":
try:
b= int(a)
processing(b)
except ValueError:
print "Your input is not valid. Please enter a 'number'!"
time.sleep(1)
askNumber()
elif len(a)!=0 and a.lower=="exit":
answer()
else:
print "Your input can't be 'empty'"
time.sleep(1)
askNumber()
So, when in the raw_input for "a" I type "exit", the supposed condition to apply is the elif one but ends up applying the if one, ending up printing "Your input is not valid. Please enter a 'number'!" Sorry, if it's something obvious, I'm a begginer, although I tried to find the mistake several times.
You need to call the .lower() function.
if len(a) != 0 and a.lower() != "exit":
# ...
elif len(a) != 0 and a.lower() == "exit":
There is no real need to test for len(a)!=0, simply test for a itself:
if a and a.lower() != "exit":
# ...
elif a and a.lower() == "exit":
Empty strings evaluate to False in a boolean context.
Your program flow is a bit inside out, may I suggest some improvements?
def askNumber():
"""Asks the number to test"""
while True:
a = raw_input("Select the number to test (type 'exit' for leaving):")
if not a:
print "Your input can't be 'empty'"
continue
if a.lower() == "exit":
answer()
break
try:
b = int(a)
except ValueError:
print "Your input is not valid. Please enter a 'number'!"
continue
processing(b)
Actually, the not a branch can be eliminated as well (empty inputs will be handled in except).
You could change the condition for the following one:
if a and a.lower() !="exit":
# .....
elif a and a.lower() == "exit":
answer()
elif a and not a.isdigit(): print "invalid input"
else:
#.............
Please note that yo don't need len(a) != 0 , just by using a will evaluate if it's empty or not.