python - 尝试/异常(exception)错误Python？

Question

我的代码的输出应该如下所示:

加密:

Enter 1 to encipher or 2 to decipher: 1
Enter text you wish to encipher: My dog has fleas.
Enter the number of characters to shift: 7
The encrypted text is: Fr whz atl yextl.

破译:

Enter 1 to encipher or 2 to decipher: 2
Enter text you wish to decipher: Fr whz atl yextl.
The most likely shift is: 7
My dog has fleas.

到目前为止，我已经得到了这个，但我不断收到无效的语法。我对如何在输出中输入答案感到困惑。它应该是带有 while 循环的 try/except 因为这是一项学校作业。

while True: 
    try: 
        num = int(raw_input('Enter 1 or 2:'))
        if num in [1,2]:
            break
        print "You have to enter 1 or 2, try again"

    if (num == 1):
        num = int(raw_input('Enter a number:'))
        num = int(raw_input('encipher'))
        print "Enter text to encipher"
        print "Enter the number of characters you want to shift" 


    elif (num == 2):
        num = int(raw_input('Enter a number:'))
        num = int(raw_input('decipher'))
        print "Enter text to decipher"
        print "Enter the number of characters you want to shift"

Answer 1

您没有正确使用try- except。这个想法是尝试一段代码；但如果遇到错误/异常，请执行其他操作。

while True: 
    try: 
        num = int(raw_input('Enter 1 or 2:'))
        if num in [1,2]:
            break
        print "You have to enter 1 or 2, try again"

在您的脚本中，您尝试获取数组中的 int，并且您正在处理它不在预定义选项列表中的情况。但是，如果它们是该代码块的异常(exception)，那么您就不会执行任何操作。

要利用try- except，请执行以下操作:

while True: 
    try: 
        num = int(raw_input('Enter 1 or 2:'))
        if num in [1,2]:
            break
        print "You have to enter 1 or 2, try again"
    except Exception, e:
        print e