What is the correct syntax for ‘else if’?

I’m a new Python programmer who is making the leap from 2.6.4 to 3.1.1. Everything has gone fine until I tried to use the ‘else if’ statement. The interpreter gives me a syntax error after the ‘if’ in ‘else if’ for a reason I can’t seem to figure out.

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

function(input('input:'))

I’m probably missing something very simple; however, I haven’t been able to find the answer on my own.

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

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:'))

Method 2

Do you mean elif?

Method 3

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

Method 4

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()

Method 5

Here is a little refactoring of your function (it does not use “else” or “elif”):

def function(a):
    if a not in (1, 2):
        a = 3
    print(str(a) + "a")

@ghostdog74: Python 3 requires parentheses for “print”.

Method 6

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

Should be corrected to:

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

As you can see, else if should be changed to elif, there should be colons after ‘2’ and else, there should be a new line after the else statement, and close the space between print and the parentheses.

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0