What is the correct syntax for 'else if'?
PythonPython 3.xPython Problem Overview
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.
Python Solutions
Solution 1 - Python
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:'))
Solution 2 - Python
Do you mean elif?
Solution 3 - Python
def function(a):
if a == '1':
print ('1a')
elif a == '2':
print ('2a')
else:
print ('3a')
Solution 4 - Python
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()
Solution 5 - Python
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".
Solution 6 - Python
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.