Python, unable to convert input() to int()

问题

I am trying to convert input() data to int() with the following code:

prompt_text = "Enter a number: "
try:
  user_num = int(input(prompt_text))
except ValueError:
  print("Error")

for i in range(1,10):
  print(i, " times ", user_num, " is ", i*user_num)

even = ((user_num % 2) == 0)

if even:
  print(user_num, " is even")
else:
  print(user_num, " is odd")

I get the following odd error when I enter asd2 for example:

Enter a number: asd2 Error 
Traceback (most recent call last):   File "chapter3_ex1.py", line 8, in <module>
    print(i, " times ", user_num, " is ", i*user_num) 
NameError: name 'user_num' is not defined

What am I doing wrong?

回答1:

The problem that you are facing is that the interpreter raises the error in the try and executes the except block. After that it will start to execute everyline. This will throw the NameError

You can overcome that by putting the rest of the program into the else block.

prompt_text = "Enter a number: "

try:
    user_num = int(input(prompt_text))  

except ValueError:
    print("Error")

else:
    for i in range(1,10):
      print(i, " times ", user_num, " is ", i*user_num)

    even = ((user_num % 2) == 0)

    if even:
      print(user_num, " is even")
    else:
      print(user_num, " is odd")

Quoting from the Python tutorial

The try ... except statement has an optional else clause, which, when present, must follow all except clauses. It is useful for code that must be executed if the try clause does not raise an exception.

Another way is to use a sentinel value

prompt_text = "Enter a number: "
user_num = 0 # default value
try:
    user_num = int(input(prompt_text))
except ValueError:
    print("Error")

This will also work. However the results may not be as expected.

Protip - Use 4 spaces to indent

回答2:

The problem isn't with the conversion to an int. user_num doesn't get a value if an exception is thrown, but it's used later.

prompt_text = "Enter a number: "
try:
  user_num = int(input(prompt_text)) # this fails with `asd2`
except ValueError:
  print("Error") # Prints your error

for i in range(1,10):
  print(i, " times ", user_num, " is ", i*user_num) # user_num wasn't assigned because of the error

even = ((user_num % 2) == 0)

if even:
  print(user_num, " is even")
else:
  print(user_num, " is odd")

You can fix this by putting the code that uses user_num in the try-block. I'll also add create a function to clean things up.

def is_even(num):
  return num%2 == 0

prompt_text = "Enter a number: "
try:
  user_num = int(input(prompt_text))
  for i in range(1,10):
    print(i, " times ", user_num, " is ", i*user_num)
  if is_even(user_num):
    print(user_num, " is even")
  else:
    print(user_num, " is odd")
except ValueError:
  print("Error")

See the ideone here.

回答3:

This may not be the cleanest solution but it addresses the problem, in your code user_num is not initialized unless it is a number.

prompt_text = "Enter a number: "
user_num = "no Input"
try:
  user_num = int(input(prompt_text))
except ValueError:
  print("Error")

if str(user_num).isnumeric():
  for i in range(1,10):
    print(i, " times ", user_num, " is ", i*user_num)

  even = ((user_num % 2) == 0)

  if even:
    print(user_num, " is even")
  else:
    print(user_num, " is odd")
else:
  print("You did not enter a number")

来源：https://stackoverflow.com/questions/33586244/python-unable-to-convert-input-to-int

标签

python

integer

explicit