IF statement will only consider ELIF?

Question

whats up? I'm playing around with my mastermind project for school, only recently started dabbling into Python - and I've ran into a problem I simply cannot figure out? I've looked at other people's questions, who seem to have the same problem as me, but it seems to be more selective, and my code is kind of different. Can anyone tell me why whenever I reply to the question, it immediately skips to "Try again!", even if I know for a fact rnumber == tnumber? (Using Python 3.4.2).

#Generates the random number module
import random
#Creates the variable in which I store my random number
rnumber = random.randint(0,9999)
#Delete this code when complete
print (rnumber)
#Number of tries
numot = 0
#Asks for user input, on what they think the number is
tnumber = input("Guess the four digit number. ")
type(tnumber)
#Compares their number to the random number
if tnumber == rnumber:
    print ("You win!")
elif rnumber != tnumber:
    print ("Try again!")
    numot = numot+1

`tnumber` is a *string*... `rnumber` is an *integer*... they will never compare equal... you need to convert your input first: `tnumber = int(tnumber)`... — Jon Clements, Sep 15 '18 at 14:09

Megan · Accepted Answer · 2018-09-15T19:09:11.447

you need to make your input an int so it considers it a number, try this

#Generates the random number module
import random
#Creates the variable in which I store my random number
rnumber = random.randint(0,9999)
#Delete this code when complete
print (rnumber)
#Number of tries
numot = 0
#Asks for user input, on what they think the number is
tnumber = int(input("Guess the four digit number. "))
#Compares their number to the random number
if tnumber == rnumber:
    print ("You win!")
else rnumber != tnumber:
    print ("Try again!")
    numot = numot+1

IF statement will only consider ELIF?

1 Answers1