Python字符串格式化

Question

给定一个至少包含一个空格字符的字符串。输出位于源字符串的第一个和第二个空格之间的子字符串。如果字符串只包含一个空格，则输出一个空字符串。我的尝试:但是输入不正确，例如:user_input=Hello World my name，输入是:World my，我不知道为什么，你能帮我吗?

user_input = input("Enter your string: ")

space_counter = 0
for char in user_input:
if char == " ":
space_counter += 1
if space_counter > 1:
start_space_index = None
for i in range(len(user_input)):
if user_input[i] == " ":
start_space_index = i
break
second_space_index = None
for i in range(len(user_input)-1, -1, -1):
if user_input[i] == " ":
second_space_index = i
break
print(user_input[start_space_index+1: second_space_index])
else:  
print("Empty string")

Answer 1

示例:1

假设输入如下:

hello my name is abc

输出应为

hello my

示例2:输入

hello my

输出

None

代码:

a = 'hello my name is abc'
obj = a.split(" ") #this splits like ['hello', 'my', 'name', 'is',   'abc']
if len(obj) > 2:
print(obj[0], obj[1])
else:
print None

Answer 2

这里是

user_input = input("Enter your string: ")
Lst = user_input.split(" ")
space_counter = 0
for char in user_input:
if char == " ":
space_counter += 1
if space_counter > 1:
start_space_index = None
for i in range(len(user_input)):
if user_input[i] == " ":
start_space_index = i
break
second_space_index = None
for i in range(len(user_input)-1, -1, -1):
if user_input[i] == " ":
second_space_index = i
break
if user_input[0] == " ":
print(Lst[0])
else:  
print(Lst[1])