问题：’else if’的正确语法是什么？
我是一名新的Python程序员，他正在从2.6.4跃升至3.1.1。在尝试使用“ else if”语句之前，一切都进行得很好。解释器在“ else if”中的“ if”之后给了我一个语法错误，原因是我似乎无法弄清。
def function(a):
if a == '1':
print ('1a')
else if a == '2'
print ('2a')
else print ('3a')
function(input('input:'))
我可能缺少一些非常简单的东西；但是，我无法自行找到答案。
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.
回答 0
在python中，“ else if”被拼写为“ elif”。
另外，您还需要在elif和之后加上一个冒号else。
简单回答一个简单的问题。刚开始时（过去几周），我遇到了同样的问题。
因此，您的代码应为：
def function(a):
if a == '1':
print('1a')
elif a == '2':
print('2a')
else:
print('3a')
function(input('input:'))
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:'))
回答 1
回答 2
def function(a):
if a == '1':
print ('1a')
elif a == '2':
print ('2a')
else:
print ('3a')
def function(a):
if a == '1':
print ('1a')
elif a == '2':
print ('2a')
else:
print ('3a')
回答 3
自古以来，if/else ifPython中正确的语法是elif。顺便说一句，如果您有很多if/else.eg ，则可以使用字典。
d={"1":"1a","2":"2a"}
if not a in d: print("3a")
else: print (d[a])
对于msw，使用字典执行函数的示例。
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()
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()
回答 4
这是函数的一点重构（它不使用“ else”或“ elif”）：
def function(a):
if a not in (1, 2):
a = 3
print(str(a) + "a")
@ ghostdog74：Python 3要求在“打印”中加上括号。
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”.
回答 5
def function(a):
if a == '1':
print ('1a')
else if a == '2'
print ('2a')
else print ('3a')
应该更正为：
def function(a):
if a == '1':
print('1a')
elif a == '2':
print('2a')
else:
print('3a')
正如您所看到的，else如果应该更改为elif，则’2’之后应该有冒号，否则，else语句之后应该有一个新行，并关闭print和括号之间的空间。
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.
