檢查輸入的變量是否在兩個數字之間，然后打印出一條消息

Question

name = input("Hello user! What is your name?")
print("Welcome " + name + " to Rock, Scissor, Paper, Lizard, Spock!")
while True:
    try:
        roundsplayed = int(input("Choose how many rounds you want to play from 1 to 5!"))
    except ValueError:
        print("Sorry, Not a interger!")
        continue
    else:
        break
if 1 < roundsplayed < 6:
    print("You have chosen " + roundsplayed " rounds to play!")

所以我有了這個，到現在為止一切似乎都在起作用。

if 1 < roundsplayed < 6:
print("You have chosen " + roundsplayed " rounds to play!")

它帶有無效的語法，但不知道為什么，因為“ print”命令正確對齊（至少我認為是這樣）

有人有解決方案嗎？

Answer 1

我編譯了這段代碼 ，現在可以正常使用了

name = input("Hello user! What is your name?")
print("Welcome " + name + " to Rock, Scissor, Paper, Lizard, Spock!")
while True:
    try:
        roundsplayed = int(input("Choose how many rounds you want to play from 1 to 5!"))
    except ValueError:
        print("Sorry, Not a interger!")
        continue
    else:
        break
if 1 < roundsplayed < 6:
  print("You have chosen " + str(roundsplayed)+ " rounds to play!")

問題在於最后一行和str（roundsplayed）的縮進。 另外，您在同一行中缺少+號。

Answer 2

您無法在python中使用+運算符連接字符串和數字。 您要么需要將-運算符用作-

print("print("You have chosen",roundsplayed,"rounds to play!")

或者您可以將數字轉換為字符串值-

 print("You have chosen " + str(roundsplayed) " rounds to play!")

如果您使用+運算符而不將數字值轉換為字符串，則會收到無效的運算符錯誤。

Answer 3

1. 您可以使用range函數來檢查數字范圍內的成員資格。 鏈接

    >>> range(10) [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] >>> range(2,6) [2, 3, 4, 5] 

2. 使用字符串格式而不是串聯鏈接 。 這樣可以避免不必要的類型轉換

    name = input("Hello user! What is your name?") print("Welcome " + name + " to Rock, Scissor, Paper, Lizard, Spock!") while True: try: roundsplayed = int(input("Choose how many rounds you want to play from 1 to 5!")) except ValueError: print("Sorry, Not a interger!") continue else: break if roundsplayed in range(2, 6): print("You have chosen {} round to play".format(roundsplayed))