Python unexpected indent

Question

Error:

Traceback (most recent call last):
File "python", line 2
if Number == "50" and "51" and "52" and "53" and "54" and "55" and "56 "and "57" and "58" and "59" and "60":
 ^
IndentationError: unexpected indent

Code:

Number = input ("Please enter a number between 50 and 60: ")
    if Number == "50" and "51" and "52" and "53" and "54" and "55" and "56 "and "57" and "58" and "59" and "60":
print ("The number ") + (Number) + (" is with in range")

I'm trying to run this python code but i keep getting error message "unexpected indent". I'm not sure whats wrong. The spacing seem to be fine. Any idea's?

Your `if` statement is also wrong (Not the cause of the Error). It should be `if Number == "50" and Number == "51" ....`and so on. Check this for reference http://stackoverflow.com/questions/15112125/how-do-i-test-one-variable-against-multiple-values — Bhargav Rao, Apr 20 '15 at 15:58
In addition to the syntax error pointed out by Bhargav, a number can't be 50 *and* 51. — Bill the Lizard, Apr 20 '15 at 16:00
You are almost certainly mixing tabs and spaces. Don't do that, configure your editor to expand tabs to 4 spaces, and ask it to convert any existing tabs to spaces. Other than that, see [How do I test one variable against multiple values?](https://stackoverflow.com/q/15112125) to avoid the other mistake in your code. — Martijn Pieters, Apr 20 '15 at 16:00

score 3 · Answer 1 · answered Apr 20 '15 at 16:00

3

Your indentation is backwards. The line after the if should be indented.

Number = input ("Please enter a number between 50 and 60: ")
if Number == "50" and "51" and "52" and "53" and "54" and "55" and "56 "and "57" and "58" and "59" and "60":
    print ("The number ") + (Number) + (" is with in range")

answered Apr 20 '15 at 16:00

Kevin

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I can't in good conscience upvote this knowing you didn't address the faults in the condition itself. – Sterling Archer Apr 20 '15 at 16:03
1

For the record, I 100% endorse Bhargav's comment linking to http://stackoverflow.com/questions/15112125/how-do-i-test-one-variable-against-multiple-values :-) – Kevin Apr 20 '15 at 16:04
i'm new to python i don't know how to code but i'm trying my best to learn it thank for the help. – Josh Hunt Apr 20 '15 at 16:12

Alex · Answer 2 · 2015-04-20T16:06:10.890

2

Have you indented the line after if with four spaces ? Please note there is a logic flaw in your code; your if statement will never be true. Did you mean to use or instead? In that case you can implement it more efficient using this code snippet:

Number = input ("Please enter a number between 50 and 60: ")
if 50 <= Number <= 60:
    print ("The number ") + (Number) + (" is with in range")

edited Apr 20 '15 at 16:06

answered Apr 20 '15 at 16:01

Alex

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You need to convert the return value of `input()` to an integer, in Python 3. Not that that `print()` line would work in Python 3... – Martijn Pieters Apr 20 '15 at 16:05

Deacon · Answer 3 · 2015-04-20T16:24:40.970

0

You have your if statement indented and your print not indented. It should be the other way around. Also, as previous folks have noted, your if statement is incorrect, which will generate another error.:

Number = input("Please enter a number between 50 and 60: ")
if Number in ['50', '51', '52', '53', '54', '55', '56', '57', '58', '59', '60']:
    print("The number %s is within range" % Number)

I'm not sure why you're treating the number as a string, though. I'd do it as:

try:
    Number = int(input("Please enter a number between 50 and 60: "))
except ValueError:
    print("You were supposed to enter a number!")
else:
    if 50 <= Number <= 60:
        print("The number %d is within range" % Number)

edited Apr 20 '15 at 16:24

answered Apr 20 '15 at 16:03

Deacon

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`if 50 <= int(Number) <= 60:`... (and handle the potential `ValueError` exception). – Martijn Pieters Apr 20 '15 at 16:04
@Martijn - Thanks. I just thought of that myself. – Deacon Apr 20 '15 at 16:09

Python unexpected indent

3 Answers3