麻烦简单的Python代码

Question

I'm learning Python, and I'm having trouble with this simple piece of code:

a = raw_input('Enter a number: ')

if a > 0:
    print 'Positive'
elif a == 0:
    print 'Null'
elif a < 0:
    print 'Negative'

It works great, apart from the fact that it always prints 'Positive', no matter if i enter a positive or negative number or zero. I'm guessing there's a simple solution, but i can't find it ;-)

Thanks in advance

Answer 1

That's because a is a string as inputted. Use int() to convert it to an integer before doing numeric comparisons.

a = int(raw_input('Enter a number: '))
if a > 0:
    print 'Positive'
elif a == 0:
    print 'Null'
elif a < 0:
    print 'Negative'

Alternatively, input() will do type conversion for you.

a = input('Enter a number: ')

Answer 2

Because you are using raw_input you are getting the value as a String, which is always considered greater than 0 (even if the String is '-10')

Instead, try using input('Enter a number: ') and python will do the type conversion for you.

The final code would look like this: a = input('Enter a number: ') if a > 0: print 'Positive' elif a == 0: print 'Null' elif a < 0: print 'Negative' However, as a number of folks have pointed out, using input() may lead to an error because it actually interprets the python objects passed in. A safer way to handle this can be to cast raw_input with the desired type, as in:

a = int( raw_input('Enter a number: '))

The final code would look like this: a = input('Enter a number: ') if a > 0: print 'Positive' elif a == 0: print 'Null' elif a < 0: print 'Negative' However, as a number of folks have pointed out, using input() may lead to an error because it actually interprets the python objects passed in. A safer way to handle this can be to cast raw_input with the desired type, as in:

The final code would look like this: a = input('Enter a number: ') if a > 0: print 'Positive' elif a == 0: print 'Null' elif a < 0: print 'Negative' However, as a number of folks have pointed out, using input() may lead to an error because it actually interprets the python objects passed in. A safer way to handle this can be to cast raw_input with the desired type, as in:

 a = int( raw_input('Enter a number: ')) 

But beware, you will still need to do some error handling here to avoid trouble!

Answer 3

Expanding on my comment on the accepted answer , here's how I would do it.

value = None
getting_input = True

while getting_input:
    try:
        value = int(raw_input('Gimme a number: '))
        getting_input = False
    except ValueError:
        print "That's not a number... try again."

if value > 0:
    print 'Positive'
elif value < 0:
    print 'Negative'
else:
    print 'Null'

Answer 4

raw_input 

返回一个字符串，所以你需要转换a是一个字符串为整数第一： a = int(a)

Answer 5

raw_input is stored as a string, not an integer.

Try using a = int(a) before performing comparisons.

Answer 6

raw input will return a string, not an integer. To convert it, try adding this line immediately after your raw_input statement:

a = int(a)

This will convert the string to an integer. You can crash it by giving it non-numeric data, though, so be careful.