Even "IF" part of this condition is true, Only else part is being executed. Whatever the input is, Only Else executes

Question

Here is my code

inputUSer=input('''
    Please Select The Mode you want to operate The script AS:
    1:  You selected-1 \n
    2:  You selected-2 \n
    3:  You selected-3\n
    4:  You selected-4 \n
''')

if inputUSer==1:
    print("You selected-1")


elif inputUSer==2:
    print("You selected-2")

elif inputUSer== 3:
    print("You selected-3")

elif inputUSer== 4:
    print("You selected-4")

else :
    print("OOPS! Seems like you entered the wrong Input")

I checked all the indentation and it's correct. I am new to python and a Noob. Please Help me... Thank you...

The value returned by `input` in Python 3.x is always a string. So, it is never `1` though it might be `'1'`… — Błotosmętek, Jun 06 '20 at 08:22

score 2 · Accepted Answer · answered Jun 06 '20 at 08:22

The result of input() is a str, it'll never be equal to 1,2,3,4 as their as ints, you need to convert input

inputUSer = int(input('''
    Please Select The Mode you want to operate The script AS:
    1:  You selected-1 \n
    2:  You selected-2 \n
    3:  You selected-3\n
    4:  You selected-4 \n
'''))

Or change all the if to compare to a str

if inputUSer == '1':

score 1 · Answer 2 · answered Jun 06 '20 at 08:24

1

Pass your input into an int()

inputUSer=int(input('''
    Please Select The Mode you want to operate The script AS:
    1:  You selected-1 \n
    2:  You selected-2 \n
    3:  You selected-3\n
    4:  You selected-4 \n
'''))

answered Jun 06 '20 at 08:24

JayC

21
1

score 1 · Answer 3 · answered Jun 06 '20 at 08:24

1

Return value of input is a string. You might want to cast it to an integer using int function. Or you can change your if conditions to compare inputUSer with strings instead.

answered Jun 06 '20 at 08:24

Abhinav Goyal

1,312
7
17

Even "IF" part of this condition is true, Only else part is being executed. Whatever the input is, Only Else executes

3 Answers3