Python - Basic Coding

Question

I am new to coding in Python. I am trying to make my code so that if I enter the age a series of text will be printed. However, my code only works if i follow it line by line. For example, when i input the age 2000+ immediately nothing will happen. I need to first input an integer less than 12, followed by an integer over 2000.

print('Please input name')
if input() == 'Alice':
 print('Hi, Alice.Please input age')
if int(input()) < 12:
 print('You are not Alice, kiddo.')
elif int(input()) > 2000:
 print('Unlike you, Alice is not an undead, immortal vampire.')
elif int(input()) == 100:
 print('You are not Alice, grannie.')
elif 12 < int(input()) < 99:
 print('You are Alice!.')

• PRMoureu: don't repeat input each time, assign it to a variable, and compare the variable
• cSharma: If you got your answer then close it.

Answer 1

var = input('Please input name ')
if var == 'Alice':
 var = int(input('Hi, Alice.Please input age '))
 if var < 12:
 print('You are not Alice, kiddo.')
 elif var > 2000:
 print('Unlike you, Alice is not an undead, immortal vampire.')
 elif var == 100:
 print('You are not Alice, grannie.')
 elif 12 < var < 99:
 print('You are Alice!.')
else:
 print ("Invalid Name")

This code works because it asks one time and tries to see if some conditions are true, instead of asking each time.

Answer 2

Here I wrote code for your understanding purpose. Take new variable, so that no need to repeat input() method several times. Also, Age validation code keeps inside the first condition and it will be executed when the 1st condition will be true.

print('Please input name')
var = input()
if var == 'Alice':
 print('Hi, Alice.Please input age')
 var = input()
 try:
 if int(var) < 12:
 print('You are not Alice, kiddo.')
 elif int(var) > 2000:
 print('Unlike you, Alice is not an undead, immortal vampire.')
 elif int(var) == 100:
 print('You are not Alice, grannie.')
 elif 12 < int(var) < 99:
 print('You are Alice!.')
 except Exception as ex:
 print('Invalid Data: Error: ' + ex)
else:
 print ("Invalid Name")

• SethMMorton: I think we can assume the OP is using Python 3 based on the print function. Therefore, it is likely that raw_input is not available and input is in fact correct.
• cSharma: oh yes, above query related on P3 but defining of logic is right so just use input() method instead of raw_input() and rest of all are same.
• SethMMorton: If I am a beginner, and I see your answer and decide to copy/paste it into my code and run it to better understand what is going on, I will get a SyntaxError where you call print with no parenthesis, and a NameError on raw_input. I will end up being more confused than before I saw this answer. Rather than leaving your answer as-is and just saying what needs to be fixed in a comment, you should edit the answer so that is correct.
• cSharma: Now, I think all should be right. I tested all scenarios on Python 3 shell. Actually, I have Python 2.7 so I suggested according to that way.

Answer 3

Every time you go to another branch in you if you are asking user to enter another age! Instead do the following:

age = int(input())
if age < 12:
 print('You are not Alice, kiddo.')
elif age > 2000:
 print('Unlike you, Alice is not an undead, immortal vampire.')
elif age == 100:
 print('You are not Alice, grannie.')
elif 12 < age < 99:
 print('You are Alice!.')

Answer 4

print('Please input name')
if input() == 'Alice':
 print('Hi, Alice.Please input age')
age = int(input()) # take input and assign it on a variable
if age < 12:
 print('You are not Alice, kiddo.')
elif age > 2000:
 print('Unlike you, Alice is not an undead, immortal vampire.')
elif age == 100:
 print('You are not Alice, grannie.')
elif 12 < age < 99:
 print('You are Alice!.')

Answer 5

input is invoked every time when followed by (). So the multiple input()'s in the if elif's are not necessary.

store the result of input() like age = int(input()), then use age in the if and elif parts instead.

Answer 6

The input() function returns a string. Quoting the docs (emphasis mine):

The function then reads a line from input, converts it to a string (stripping a trailing newline), and returns that.

So, in each if when you call input(), you have to enter a new string. Thus, you have to first enter an integer below 12.

To fix this problem, you need to store the original input in a variable. Now, as the docs say, input() returns a string. So, either you can cast (using int()) the integer in each case, by doing:

if int(age) < 12:

and storing the variable as a string.

Though, unless you do not have any specific reason to keep the age as a string, I'd recommend you to convert the string while storing the age in the variable in the first place:

age = int (input())

In this case, age will have an int.

Answer 7

Hopefully, this is what you are looking for:

while True:
name = input("Please ENTER your name: ")
if name == "Alice":
 print("Hi Alice!")
 break
print("Sorry, your name isn't correct. Please re-enter.")
age = False
while age != True:
 age = int(input("Please ENTER your age: ")
 age = True
 if age < 12:
 print("You're not Alice, kiddo.")
 age = False
 elif age > 2000:
 print("Unlike you, Alice is not an undead, immortal vampire.")
 age = False
 elif age == 100:
 print("You're not Alice, Granny!")
 age = False
 else:
 print("You are Alice!")
 age = True