Python 字符串格式化

Question

Given a string containing at least one space character. Output the substring located between the first and second spaces of the source string. If the string contains only one space, then output an empty string. My attempt: But input is incorrect, for example: user_input=Hello World my name , input is: World my , i don't know why , can you help me?

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

Example: 1

Assuming the input like:

hello my name is abc

Output should be

hello my

Example 2:

input

hello my

output

None

Code:

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

Here, it is

    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])