CODEWITHSUNDEEP
pythonpython-3.xif-statementuser-inputcomparison-operators
Ryan
Ryan

Reputation: 23

Numeric comparison with user input always produces "not equal" result

I want to get a number input by the user via input() and compare it with a specific value, i.e., 3.

However, I have the impression my if statement doesn't work. The comparison is always False.

Start = input()
if Start == 3:
     print ("successful")

Upvotes: 1

Views: 104

Answers (4)

Adem Öztaş
Adem Öztaş

Reputation: 21446

Python 3 input function returns string.

Try like this

start = input("-->: ")
if start == "3":
    print("successful")

Upvotes: 3

ellaRT
ellaRT

Reputation: 1366

You can also try this as an alternative:

Adding int will convert the variable to int data type. 

input = int(raw_input("Enter a num: "))
if input == 3:
   print "Successful"

Upvotes: 1

moooeeeep
moooeeeep

Reputation: 32502

Some things you could do on your own to get to the root of the problem:

Ways to get to know the type of the object:

print(type(start)) # prints <class 'str'>
print(repr(start)) # prints '3'

Unlike Python 2.x, the function input() returns a string object (and does not blindly evaluate the expression provided by the user):

input([prompt]):

If the prompt argument is present, it is written to standard output without a trailing newline. The function then reads a line from input, converts it to a string (stripping a trailing newline), and returns that. [...]

This should give an idea how to fix it (compare numbers to numbers).

For further reading:

Upvotes: 1

MikeVelazco
MikeVelazco

Reputation: 905

you should check your tabs (I recomend you to use four spaces, not regular tabs and not to mix them)

start = 0
start = input("-->:")
if start == "3":
    print("Success")

Upvotes: 0

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