Python：標準英國區的多個字符串測試，爲什麼不工作？

Question

以下是與主程序分離的部分，以便在測試代碼片段時進行測試。 「是」輸出用於測試每個需要查看錯誤出現位置的字符串方法。這似乎不起作用，我無法弄清楚爲什麼，因爲我已經把所有的測試標準英國註冊，檢查字符，因爲他們從輸入顯示。

正如你可以看到一個非常低的水平（Python新手）程序員，請考慮，因爲你提供的解決方案。非常感謝

def main(): 
    reg_number=input("What is your reg number?: ") 
    correct_length = False 
    first_two_char_capital = False 
    next_two_char_digits= False 
    next_char_space = False 
    last_three_char_capital= False 

    first_two_char_capital =reg_number[0:2] 
    print(first_two_char_capital) 
    next_two_char_digits=reg_number[2:4] 
    print (next_two_char_digits) 
    next_char_space=reg_number[4:5] 
    print(next_char_space) 
    last_three_char_capital=reg_number[5:8] 
    print (last_three_char_capital) 

    if len (reg_number) >= 8: 
      correct_length = True 
      print("yes1") 

      for ch in reg_number: 
       if first_two_char_capital.isupper(): 
        first_two_char_capital = True 
        print("yes2") 
       if next_two_char_digits.isdigit(): 
        next_two_char_digits = True 
        print("yes3") 
       if next_char_space.isspace(): 
        next_char_space = True 
        print("yes4") 
       if last_three_char_capital.isupper(): 
        last_three_char_capital = True 
        print("yes5") 
      if correct_length and first_two_char_capital and next_two_char_digits and next_char_space and last_three_char_capital: 
       is_valid = False 
       print(reg_number, "is a non-standard_registration") 
      else: 
       is_valid = True 

       print(reg_number, "is a standard registration") 
       return is_valid 

main() 

Answer 1

這是很容易，只需使用正則表達式在這裏，如：

import re 

if re.match('[A-Z]{2}\d{2} [A-Z]{3}$', input('What is your regnumber:')): 
    print('yeah it matches...')