应用`raw_input`时如何使用`if`？

Question

I am somewhat new to Python and coding in general, and I require some help working with raw_input and an if statement. My code is as follows;

    age = raw_input ("How old are you? ")
    if int(raw_input) < 14:
    print "oh yuck"
    if int(raw_input) > 14:
    print "Good, you comprehend things, lets proceed"

Answer 1

Issues

There are three issues with your code:

• Python uses indentation for creating blocks.
• You've assigned the input to the variable age , so use age .
• In Python 3, you have to use print(...) instead of print ...

Correct solution

age = raw_input("How old are you? ")

if int(age) < 14:
    print("oh yuck")
else:
    print("Good, you comprehend things, lets proceed")

Note that this is not equivalent to your code. Your code skips the case age == 14 . If you want this behaviour, I suggest:

age = int(raw_input("How old are you? "))

if age < 14:
    print("oh yuck")
elif age > 14:
    print("Good, you comprehend things, lets proceed")

Learning Python

• http://www.codecademy.com/tracks/python - A tutorial that introduces you to programming / Python step by step
• Manual - Looking things up and discovering new stuff, eg raw_input() and int()

Answer 2

if int(raw_input) < 14:

Should be int(age) , and same for the other if . raw_input is the function you called, but you stored the result from it in the variable age . You can't turn a function into an integer.

Rather than converting the age to an integer repeatedly, you could just do it once, when you do the input:

age = int(raw_input("How old are you? "))

Then you can just do if age > 14 and so on, since it's already an integer.

I assume the indentation problems (the line following each if should be indented at least one space, and preferably four) are just a formatting issue.