程序循环非常快速且无限期？无法弄清楚为什么

I'm working on a Tweet Manager program in python for my programming class. For the assignment, I'm supposed to create a Tweet class that stores the author of a tweet, the tweet itself, and the time the tweet was created. Then, I'm supposed to create a Twitter program that gives users a menu of options to choose from.

我正在python中为我的编程类开发一个Tweet Manager程序。对于作业,我应该创建一个Tweet类来存储推文的作者,推文本身以及创建推文的时间。然后,我应该创建一个Twitter程序,为用户提供一个选项菜单供选择。

When I try to run my Twitter program, it opens without syntax errors, but prints the Menu over and over and over again really rapidly without stopping. I can't figure out what in my code is causing this problem.

当我尝试运行我的Twitter程序时,它会打开而不会出现语法错误,但会在不停止的情况下一遍又一遍地打印菜单。我无法弄清楚我的代码中是什么导致了这个问题。

Here is my Twitter code:

这是我的Twitter代码:

    import Tweet
    import pickle
    def main():
        try:
            load_file = open('tweets.dat', 'rb')
            tweets = pickle.load('tweets.dat')
            load_file.close()
        except:
            tweet_list = []
        while (True):
            choice = display_menu()

            #Make a Tweet
            if (choice == 1):
                tweet_author = input("\nWhat is your name? ")
                tweet_text = input("What would you like to tweet? ")
                print()
            if len(tweet_text) > 140:
                print("Tweets can only be 140 characters!\n")
            else:
                print(tweet_author, ", your Tweet has been saved.")

            age = 0
            tweets = tweet.Tweet(tweet_author, tweet_text)

            tweet_list.append(tweets)

            try:
                output_file = open('tweets.dat', 'wb')
                pickle.dump(tweets, output_file)
                output_file.close()
            except:
                print("Your tweets could not be saved!")

            #View Recent Tweets
            elif (choice == 2):
                print("Recent Tweets")
                print("--------------")
                if len(tweet_list) == 0:
                    print("There are no recent tweets. \n")
                for tweets in tweet_list[-5]:
                    print(tweets.get_author(), "-", tweets.get_age())
                    print(tweets.get_text(), "\n")

            #Search Tweets
            elif (choice == 3):
                match = 0
                tweet_list.reverse()

                if tweet_list == []:
                    print("There are no tweets to search. \n")

                search = input("What would you like to search for? ")
                for tweets in tweet_list:
                    if (search in tweets.get_text()):
                        match = 1
                    if match = 1:
                        print("Search Results")
                        print("--------------")
                        print(tweets.get_author(), "-", tweets.get_age())
                        print(tweets.get_text(), "\n")
                    elif match == 0:
                        print("No tweets contained", search, "\n")

            #Quit
            elif (choice == 4):
                print("Thank you for using the Tweet Manager!")
                exit()

    def display_menu():
        print("Tweet Menu")
        print("------------")
        print()
        print("1. Make a Tweet")
        print("2. View Recent Tweets")
        print("3. Search Tweets")
        print("4. Quit")
        print()

    main()

4 个解决方案

#1

There's no problem, the code is doing precisely what you told it to.

没有问题,代码正是按照你告诉它做的。

display_menu does exactly what the name implies: displays the menu. The main function calls that function inside a loop.

display_menu正如名称所暗示的那样:显示菜单。 main函数在循环内调用该函数。

At no point do you actually ask for any input corresponding to the menu options.

在任何时候你都不会要求输入与菜单选项相对应的任何输入。

#2

display_menu always returns None - that's the default if you don't return anything.

display_menu总是返回None - 如果你没有返回任何东西,这是默认值。

#3

Display menu does not retrieve any input into choice... Just displays the menu. Thus, no if is matched and you loop indefinitely.

显示菜单不会检索任何输入选择...只显示菜单。因此,如果匹配则无关,并且无限循环。

#4

Your display_menu function doesn't actually give a value.

您的display_menu函数实际上没有给出值。

Change the last line of that function (print()) to

将该函数的最后一行(print())更改为

return input()

or if you are using python 2.x

或者如果您使用的是python 2.x.

return raw_input()

#1