控制在Python

Question

一個循環我目前正在對這個代碼：

import time 
import webbrowser 

while True: 
    user_inputq3 = raw_input("Which model of phone is it? ") 
    import csv 
    words=user_inputq3.split(" ") 
    reader=csv.reader(open("task3worksheet.csv")) 
    problemlist=[row for row in reader] 
    for x in range(0,len(words)): 
     for y in range(0,len(problemlist)-1,2): 
      if(words[x].lower() in problemlist[y]): 
       print(problemlist[y+1][0]) 
       break 
      elif user_inputq3 == "exit": 
       print("The programme will now shut down.") 
       time.sleep(5) 
       exit(0) 
       break 
      else: 
       print("This troubleshooting programme will only support iPhone 3-7.") 
       print("Make sure you type in iphone then the model of iPhone. (e.g iphone4)") 
       print("If you do not have these versions of the iPhone, please type in exit when prompted with the question again.") 
       break 

的「task3worksheet」是一個Excel工作表，這是充滿關鍵字和響應。因此，例如，當輸入「iPhone4」時，應該出現「下一個問題」。但是，當我輸入「iPhone4」時，它會提出下一個問題並再次回到問題。我該如何做到這一點，以便它不會再次回到問題中，我希望它能夠在輸入有效輸入時結束當前循環，因此我可以繼續使用其他代碼。

Answer 1

我添加了一個變量，告訴我如果我已經回答了問題，如果我沒有打破外部循環。 改良版在評論一個建議：

import time 
import webbrowser 

answered = False 
while not answered: 
    user_inputq3 = raw_input("Which model of phone is it? ") 
    import csv 
    words=user_inputq3.split(" ") 
    reader=csv.reader(open("task3worksheet.csv")) 
    problemlist=[row for row in reader] 
    for x in range(0,len(words)): 
     if answered: 
      break 
     for y in range(0,len(problemlist)-1,2): 
      if(words[x].lower() in problemlist[y]): 
       print(problemlist[y+1][0]) 
       answered = True 
       break 
      elif user_inputq3 == "exit": 
       print("The programme will now shut down.") 
       time.sleep(5) 
       exit(0) 
       break 
      else: 
       print("This troubleshooting programme will only support iPhone 3-7.") 
       print("Make sure you type in iphone then the model of iPhone. (e.g iphone4)") 
       print("If you do not have these versions of the iPhone, please type in exit when prompted with the question again.") 
       break