Htykuut

Why would you do a while True and then break out of this loop while you can also just put your requirements in the while statement since all you want is to stop once you have the age?

age = None

while age is None:

    input_value = input("Please enter your age: ")

    try:

        # try and convert the string input to a number

        age = int(input_value)

    except ValueError:

        # tell the user off

        print("{input} is not a number, please enter a number only".format(input=input_value))

if age >= 18:

    print("You are able to vote in the United States!")

else:

    print("You are not able to vote in the United States.")

This would result in the following:

Please enter your age: *potato*

potato is not a number, please enter a number only

Please enter your age: *5*

You are not able to vote in the United States.

this will work since age will never have a value that will not make sense and the code follows the logic of your "business process"

Though the accepted answer is amazing. I would also like to share a quick hack for this problem. (This takes care of the negative age problem as well.)

f=lambda age: (age.isdigit() and ((int(age)>=18  and "Can vote" ) or "Cannot vote")) or 

f(input("invalid input. Try againnPlease enter your age: "))

print(f(input("Please enter your age: ")))

P.S. This code is for python 3.x.

So, I was messing around with something similar to this recently, and I came up with the following solution, which uses a way of getting input that rejects junk, before it's even checked in any logical way.

read_single_keypress() courtesy https://stackoverflow.com/a/6599441/4532996

def read_single_keypress() -> str:

    """Waits for a single keypress on stdin.

    -- from :: https://stackoverflow.com/a/6599441/4532996

    """



    import termios, fcntl, sys, os

    fd = sys.stdin.fileno()

    # save old state

    flags_save = fcntl.fcntl(fd, fcntl.F_GETFL)

    attrs_save = termios.tcgetattr(fd)

    # make raw - the way to do this comes from the termios(3) man page.

    attrs = list(attrs_save) # copy the stored version to update

    # iflag

    attrs[0] &= ~(termios.IGNBRK | termios.BRKINT | termios.PARMRK

                  | termios.ISTRIP | termios.INLCR | termios. IGNCR

                  | termios.ICRNL | termios.IXON )

    # oflag

    attrs[1] &= ~termios.OPOST

    # cflag

    attrs[2] &= ~(termios.CSIZE | termios. PARENB)

    attrs[2] |= termios.CS8

    # lflag

    attrs[3] &= ~(termios.ECHONL | termios.ECHO | termios.ICANON

                  | termios.ISIG | termios.IEXTEN)

    termios.tcsetattr(fd, termios.TCSANOW, attrs)

    # turn off non-blocking

    fcntl.fcntl(fd, fcntl.F_SETFL, flags_save & ~os.O_NONBLOCK)

    # read a single keystroke

    try:

        ret = sys.stdin.read(1) # returns a single character

    except KeyboardInterrupt:

        ret = 0

    finally:

        # restore old state

        termios.tcsetattr(fd, termios.TCSAFLUSH, attrs_save)

        fcntl.fcntl(fd, fcntl.F_SETFL, flags_save)

    return ret



def until_not_multi(chars) -> str:

    """read stdin until !(chars)"""

    import sys

    chars = list(chars)

    y = ""

    sys.stdout.flush()

    while True:

        i = read_single_keypress()

        _ = sys.stdout.write(i)

        sys.stdout.flush()

        if i not in chars:

            break

        y += i

    return y



def _can_you_vote() -> str:

    """a practical example:

    test if a user can vote based purely on keypresses"""

    print("can you vote? age : ", end="")

    x = int("0" + until_not_multi("0123456789"))

    if not x:

        print("nsorry, age can only consist of digits.")

        return

    print("your age is", x, "nYou can vote!" if x >= 18 else "Sorry! you can't vote")



_can_you_vote()

You can find the complete module here.

Example:

$ ./input_constrain.py

can you vote? age : a

sorry, age can only consist of digits.

$ ./input_constrain.py 

can you vote? age : 23<RETURN>

your age is 23

You can vote!

$ _

Note that the nature of this implementation is it closes stdin as soon as something that isn't a digit is read. I didn't hit enter after a, but I needed to after the numbers.

You could merge this with the thismany() function in the same module to only allow, say, three digits.

def validate_age(age):

    if age >=0 :

        return True

    return False



while True:

    try:

        age = int(raw_input("Please enter your age:"))

        if validate_age(age): break

    except ValueError:

        print "Error: Invalid age."

Try this one:-

def takeInput(required):

  print 'ooo or OOO to exit'

  ans = raw_input('Enter: ')



  if not ans:

      print "You entered nothing...!"

      return takeInput(required) 



      ##  FOR Exit  ## 

  elif ans in ['ooo', 'OOO']:

    print "Closing instance."

    exit()



  else:

    if ans.isdigit():

      current = 'int'

    elif set('[~!@#$%^&*()_+{}":/']+$').intersection(ans):

      current = 'other'

    elif isinstance(ans,basestring):

      current = 'str'        

    else:

      current = 'none'



  if required == current :

    return ans

  else:

    return takeInput(required)



## pass the value in which type you want [str/int/special character(as other )]

print "input: ", takeInput('str')

To edit your code and fix the error:

while True:

    try:

       age = int(input("Please enter your age: "))

       if age >= 18: 

           print("You are able to vote in the United States!")

           break

       else:

           print("You are not able to vote in the United States.")

           break

    except ValueError:

       print("Please enter a valid response")

You can write more general logic to allow user to enter only specific number of times, as the same use-case arises in many real-world applications.

def getValidInt(iMaxAttemps = None):

  iCount = 0

  while True:

    # exit when maximum attempt limit has expired

    if iCount != None and iCount > iMaxAttemps:

       return 0     # return as default value



    i = raw_input("Enter no")

    try:

       i = int(i)

    except ValueError as e:

       print "Enter valid int value"

    else:

       break



    return i



age = getValidInt()

# do whatever you want to do.

Use "while" statement till user enter a true value and if the input value is not a number or it's a null value skip it and try to ask again and so on.
In example I tried to answer truly your question. If we suppose that our age is between 1 and 150 then input value accepted, else it's a wrong value.
For terminating program, the user can use 0 key and enter it as a value.

Note: Read comments top of code.

# If your input value is only a number then use "Value.isdigit() == False".

# If you need an input that is a text, you should remove "Value.isdigit() == False".

def Input(Message):

    Value = None

    while Value == None or Value.isdigit() == False:

        try:        

            Value = str(input(Message)).strip()

        except InputError:

            Value = None

    return Value



# Example:

age = 0

# If we suppose that our age is between 1 and 150 then input value accepted,

# else it's a wrong value.

while age <=0 or age >150:

    age = int(Input("Please enter your age: "))

    # For terminating program, the user can use 0 key and enter it as an a value.

    if age == 0:

        print("Terminating ...")

        exit(0)



if age >= 18 and age <=150: 

    print("You are able to vote in the United States!")

else:

    print("You are not able to vote in the United States.")

While a try/except block will work, a much faster and cleaner way to accomplish this task would be to use str.isdigit().

while True:

    age = input("Please enter your age: ")

    if age.isdigit():

        age = int(age)

        break

    else:

        print("Invalid number '{age}'. Try again.".format(age=age))



if age >= 18: 

    print("You are able to vote in the United States!")

else:

    print("You are not able to vote in the United States.")

You can make the input statement a while True loop so it repeatedly asks for the users input and then break that loop if the user enters the response you would like. And you can use try and except blocks to handle invalid responses.

while True:



    var = True



    try:

        age = int(input("Please enter your age: "))



    except ValueError:

        print("Invalid input.")

        var = False



    if var == True:

        if age >= 18:

                print("You are able to vote in the United States.")

                break

        else:

            print("You are not able to vote in the United States.")

The var variable is just so that if the user enters a string instead of a integer the program wont return "You are not able to vote in the United States."

Use try catch with never ending while loop. To check for blank string use IF statement to check if string is empty.

while True:

    name = input("Enter Your Namen")

    if not name:

        print("I did not understood that")

        continue

    else:

        break



while True:

    try:

        salary = float(input("whats ur salaryn"))

    except ValueError:

        print("I did not understood that")

        continue

    else:

        break



while True:

    try:

        print("whats ur age?")

        age = int(float(input()))

    except ValueError:

        print("I did not understood that")

        continue

    else:

        break



print("Hello "+ name +  "nYour salary is " + str(salary) + 'nand you will be ' + str(age+1) +' in a Year')

This will continue ask user to input the number until they input a valid number:

#note: Python 2.7 users should use raw_input, the equivalent of 3.X's input

while(1):

    try:

        age = int(input("Please enter your age: "))

        if age >= 18: 

            print("You are able to vote in the United States!")

            break()

        else:

            print("You are not able to vote in the United States.")

            break()

    except:

        print("Please only enter numbers ")

One more solution for using input validation using a customized ValidationError and a (optional) range validation for integer inputs:

class ValidationError(ValueError): 

    """Special validation error - its message is supposed to be printed"""

    pass



def RangeValidator(text,num,r):

    """Generic validator - raises 'text' as ValidationError if 'num' not in range 'r'."""

    if num in r:

        return num

    raise ValidationError(text)



def ValidCol(c): 

    """Specialized column validator providing text and range."""

    return RangeValidator("Columns must be in the range of 0 to 3 (inclusive)", 

                          c, range(4))



def ValidRow(r): 

    """Specialized row validator providing text and range."""

    return RangeValidator("Rows must be in the range of 5 to 15(exclusive)",

                          r, range(5,15))

Usage:

def GetInt(text, validator=None):

    """Aks user for integer input until a valid integer is given. If provided, 

    a 'validator' function takes the integer and either raises a 

    ValidationError to be printed or returns the valid number. 

    Non integers display a simple error message."""

    print()

    while True:

        n = input(text)

        try:

            n = int(n)



            return n if validator is None else validator(n)



        except ValueError as ve:

            # prints ValidationErrors directly - else generic message:

            if isinstance(ve, ValidationError):

                print(ve)

            else:

                print("Invalid input: ", n)





column = GetInt("Pleased enter column: ", ValidCol)

row = GetInt("Pleased enter row: ", ValidRow)

print( row, column)

Output:

Pleased enter column: 22

Columns must be in the range of 0 to 3 (inclusive)

Pleased enter column: -2

Columns must be in the range of 0 to 3 (inclusive)

Pleased enter column: 2

Pleased enter row: a

Invalid input:  a

Pleased enter row: 72

Rows must be in the range of 5 to 15(exclusive)

Pleased enter row: 9  



9, 2

Here's a cleaner, more generalized solution that avoids repetitive if/else blocks: write a function that takes (Error, error prompt) pairs in a dictionary and do all your value-checking with assertions.

def validate_input(prompt, error_map):

    while True:

        try:

            data = int(input(prompt))

            # Insert your non-exception-throwing conditionals here

            assert data > 0

            return data

        # Print whatever text you want the user to see

        # depending on how they messed up

        except tuple(error_map.keys()) as e:

            print(error_map[type(e)])

Usage:

d = {ValueError: 'Integers only', AssertionError: 'Positive numbers only', 

     KeyboardInterrupt: 'You can never leave'}

user_input = validate_input("Positive number: ", d)

Building upon Daniel Q's and Patrick Artner's excellent suggestions,
here is an even more generalized solution.

# Assuming Python3

import sys



class ValidationError(ValueError):  # thanks Patrick Artner

    pass



def validate_input(prompt, cast=str, cond=(lambda x: True), onerror=None):

    if onerror==None: onerror = {}

    while True:

        try:

            data = cast(input(prompt))

            if not cond(data): raise ValidationError

            return data

        except tuple(onerror.keys()) as e:  # thanks Daniel Q

            print(onerror[type(e)], file=sys.stderr)

I opted for explicit if and raise statements instead of an assert,
because assertion checking may be turned off,
whereas validation should always be on to provide robustness.

This may be used to get different kinds of input,
with different validation conditions.
For example:

# No validation, equivalent to simple input:

anystr = validate_input("Enter any string: ")



# Get a string containing only letters:

letters = validate_input("Enter letters: ",

    cond=str.isalpha,

    onerror={ValidationError: "Only letters, please!"})



# Get a float in [0, 100]:

percentage = validate_input("Percentage? ",

    cast=float, cond=lambda x: 0.0<=x<=100.0,

    onerror={ValidationError: "Must be between 0 and 100!",

             ValueError: "Not a number!"})

Or, to answer the original question:

age = validate_input("Please enter your age: ",

        cast=int, cond=lambda a:0<=a<150,

        onerror={ValidationError: "Enter a plausible age, please!",

                 ValueError: "Enter an integer, please!"})

if age >= 18: 

    print("You are able to vote in the United States!")

else:

    print("You are not able to vote in the United States.")

Good question! You can try the following code for this. =)

This code uses ast.literal_eval() to find the data type of the input (age). Then it follows the following algorithm:

1. 
   

   Ask user to input her/his age.

   
   
   

   1.1. If age is float or int data type:

   
   
   
   
   

   • Check if age>=18. If age>=18, print appropriate output and exit.

   
   

   • Check if 0<age<18. If 0<age<18, print appropriate output and exit.

   
   

   • If age<=0, ask the user to input a valid number for age again, (i.e. go back to step 1.)

   
   
   
   

   1.2. If age is not float or int data type, then ask user to input her/his age again (i.e. go back to step 1.)

   
   

Here is the code.

from ast import literal_eval



''' This function is used to identify the data type of input data.'''

def input_type(input_data):

    try:

        return type(literal_eval(input_data))

    except (ValueError, SyntaxError):

        return str



flag = True



while(flag):

    age = raw_input("Please enter your age: ")



    if input_type(age)==float or input_type(age)==int:

        if eval(age)>=18: 

            print("You are able to vote in the United States!") 

            flag = False 

        elif eval(age)>0 and eval(age)<18: 

            print("You are not able to vote in the United States.") 

            flag = False

        else: print("Please enter a valid number as your age.")



    else: print("Sorry, I didn't understand that.")