What is the correct syntax for 'else if'?

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Chapters
00:00 Question
00:44 Accepted answer (Score 454)
01:18 Answer 2 (Score 21)
01:29 Answer 3 (Score 13)
01:40 Answer 4 (Score 10)
02:22 Thank you

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Full question
https://stackoverflow.com/questions/2395...

Answer 1 links:
[elif]: http://docs.python.org/3.1/tutorial/cont...

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https://meta.stackexchange.com/help/lice...

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Tags
#python #python3x

#avk47

ACCEPTED ANSWER

Score 460

In python "else if" is spelled "elif".
Also, you need a colon after the elif and the else.

Simple answer to a simple question. I had the same problem, when I first started (in the last couple of weeks).

So your code should read:

def function(a):
    if a == '1':
        print('1a')
    elif a == '2':
        print('2a')
    else:
        print('3a')

function(input('input:'))

ANSWER 2

Score 21

Do you mean elif?

ANSWER 3

Score 13

def function(a):
    if a == '1':
        print ('1a')
    elif a == '2':
        print ('2a')
    else:
        print ('3a')

ANSWER 4

Score 10

since olden times, the correct syntax for if/else if in Python is elif. By the way, you can use dictionary if you have alot of if/else.eg

d={"1":"1a","2":"2a"}
if not a in d: print("3a")
else: print (d[a])

For msw, example of executing functions using dictionary.

def print_one(arg=None):
    print "one"

def print_two(num):
    print "two %s" % num

execfunctions = { 1 : (print_one, ['**arg'] ) , 2 : (print_two , ['**arg'] )}
try:
    execfunctions[1][0]()
except KeyError,e:
    print "Invalid option: ",e

try:
    execfunctions[2][0]("test")
except KeyError,e:
    print "Invalid option: ",e
else:
    sys.exit()