Hi guys been having problem with the if state code on pydroid app...

Question

age =70

if age <=18 & age<=45
print("you are welcome to the party")

elif age>45:
print("you are very welcom")
else:
print("go home")

print(20).
THIS IS THE ERROR MESSAGE.
Traceback (most recent call last):
File "/data/user/0/ru.iiec.pydroid3/files/accomp_files/iiec_run/iiec_run.py", line 31, in <module>
start(fakepyfile,mainpyfile)
File "/data/user/0/ru.iiec.pydroid3/files/accomp_files/iiec_run/iiec_run.py", line 30, in start
exec(open(mainpyfile).read(), __main__.__dict__)
File "<string>", line 3
if age <=18 & age<=45
^
SyntaxError: invalid syntax

[Program finished]

Answer 1

In python we use keyword "and" exept &

if age <= 18 and age <= 45:
print("something")

You maybe make typo also, you check are age less or equal to 18 and check are age less or equal 45.
You probably mean greater than 18
age >=18 not age<=18

Also some indentation is wrong, i am not sure what print(20) do and this is also not indented well.
And you miss ":" after if statement

Answer 2

Here is difference between "and" and "&"
https://www.geeksforgeeks.org/difference-between-and-and-in-JUMP_LINK__&&__python__&&__JUMP_LINK/amp/