Question

我尝试在Python 2.7上编写简单的加密程序时遇到了极其令人沮丧的错误。基本上它通过说我的语法无效而拒绝if / else语句，我不知道我做错了什么。：在if输入==＆＃34; 1＆＃34;：语句中，Python突出显示为不正确。

    #Main menu function, directs user to an operation
    def menu():
        input = raw_input("Type 1 for encryption, 2 for decryption, or 3 for information."
        if input == "1":
            print "Encryption selected..."
            encrypt()
        elif input == "2":
            print "Decryption selected..."
            decrypt()
        elif input == "3":
            print "Loading info..."
            info()
        else:
            print "ERROR: INVALID INPUT"
            menu()

Answer 1

您在raw_input()行的末尾错过了结束括号？

更改

input = raw_input("Type 1 for encryption, 2 for decryption, or 3 for information."

到

input = raw_input("Type 1 for encryption, 2 for decryption, or 3 for information.")

我在if / else语句中的语法有什么问题？

1 个答案: