whats wrong with my code

Question

please ell me whats wrong with my code

when i run this code , i get an error :

print("welcome to my first app")
code=input("enter a number betwin 7 and 9")
if code==7 :
print("your number is: 7")
else code<=8 :
print("your nmber is greater or eqal 8")
else code==9 :
print ("your number is 9")

please tell me

thanks

Answer 1

you should use elif instead of else

if code 7

elif code 8

elif code 9

Answer 2

well you should share what error did you get

Answer 3

i got it
you need to make an indent block like this

if a:
A
elif b:
B

if you use the plyground here, you can see tab button the left on top of keyboard

Answer 4

you need to make a space after if-else statement.
if number == 1
print "number one"
elif number == 2
print "number two"

I guess you just dont know what indentation means

Answer 5

just checkout this
http://www.diveintopython.net/getting_to_know_python/indenting_code.html

Answer 6

It's not
if
else
else

It is:
if
ELIF
else

Answer 7

your first mistake is INDENTATION,
and also since u want a number between 7 to 9, u use while loop to literate from 7 down to 9. easier than using if and else.
I THINK THIS IS WHAT U WANT
CHECK MY VERSION OF IT: copy and paste it into playground.

print("welcome to myfirst app");
CODE=int(input('Enter a number between 7 to 9'));
if CODE<=9:
print("your number is",(CODE))
if CODE==8:
print('your number is',(CODE))
elif CODE==9:
print("your number is",(CODE))
elif CODE<=6:
print("It is a number below 7:",(CODE))
else:
if CODE>9:
print('it is a number above 9:',(CODE))

print("END")

in case someone type in a number below 7 or above 9 your code will crash, so I have handled that with my version check it out.

Answer 8

this is the error which it's gives me:

ERROR: Traceback (most recent call last):
File "/home/mataeoh/buildozer/.buildozer/android/app/PyInterpreter/PyInterpreter.py", line 65, in run_script
File "<string>", line 4
print("your number is: 7")
^
IndentationError: expected an indented block

thanks for your help

Answer 9

thanks a lot for helping me but i still not get it

what exactly should i done with this code:

print("welcome to my first app")
code=input("enter a number betwin 7 and 9")
if code==7 :
print("your number is: 7")
elif code<=8 :
print("your nmber is greater or eqal 8")
elif code==9 :
print ("your number is 9")

thank you

Answer 10

thanks a lot for helping me

my problem solved

thanks

Answer 11

THE FULLY WRIGHT CODE:

print("welcome to my first app")

code=input("enter a number between 7 and 9")

if code==7 :
#Use indentation of 4 spaces or 1 TAB
print("your number is: 7")

#Use elif instead of else
elif code<=8 :
#again, 4 spacer or 1 TAB
print("your nmber is greater or eqal 8")

#Again, elif
elif code==9 :
print ("your number is 9")
-----------------------------------------------------------------------
p.s. i would use
x = input()
int (x)
if x == some_number:
pass

Answer 12

whitespaces ...

Answer 13

INDENTATION != INDENTATION

Answer 14

after the if , before writing print command always put a white space

Answer 15

THIS ONE WORK properly, make attention to identiations and the use of if/else if/ else:

print("welcome to my first app")
code=input("enter a number betwin 7 and 9")
if code==7 :
print("your number is: 7")
elif code<=8 :
print("your nmber is greater or eqal 8")
else code==9 :
print ("your number is 9")

Answer 16

i change my code again but it give an error again:

print("welcome to my first app")
code=input("enter a number betwin 7 and 9")
if code==7 :
print("your number is: 7")
elif code<=8 :
print("your nmber is greater or eqal 8")
elif code==9 :
print ("your number is 9")

what should i done?

thanks