Python3：爲負輸入創建錯誤

Question

參加一個介紹課程並需要創建一個過度/不足猜測遊戲。如果有人輸入負數或非整數，我想通過創建錯誤來微調用戶輸入。我正確地報告了非整數錯誤，並且負循環正確返回，但負數不會打印我的錯誤消息。

#Number of plays 
def get_plays(msg): 
    while True: 
     try: 
      x = (int(input(msg))) 

     except ValueError: 
      print ("Integer numbers only please.") 
     except: 
      if x <=0: 
       print ("Positive numbers only please.") 

i = get_plays("\nHow many times would you like to play?") 
print ("The game will play " +str(i)+" times.") 

另外，如果我想使用類似的設置來產生1到20之間的任何負非整數的錯誤，怎麼會變成這樣看？

Answer 1

嘗試：

def get_plays(msg): 
    while True: 
     try: 
      x = (int(input(msg))) 
      if x <=0: 
       print("Positive numbers only please.") 
       continue 
      if x not in range(20): 
       print("Enter a number between 1 - 20.") 
       continue 
      return x 
     except ValueError: 
      print("Integer numbers only please.") 

將接受20

Answer 2

問題是這部分

except: 
    if x <=0: 

空except條款上還沒有被抓到還沒有任何錯誤引發，但負數不會引發異常。你想要這樣的東西。請注意，在try子句中，我們可以像x已經是int一樣繼續，因爲我們可以假設沒有ValueError被拋出。

def get_plays(msg): 
    while True: 
     try: 
      x = (int(input(msg))) 
      if x <=0: 
       print ("Positive numbers only please.") 
      else: 
       return x 

     except ValueError: 
      print ("Integer numbers only please.")