为什么没有一个函数调用像在这个程序中那样执行？

Question

标题几乎说明了一切。每个小功能都可以独立工作（我在单独的文件中测试了它们），但是在放入主功能时它们不起作用。我究竟做错了什么？ （代码块很长，对此感到抱歉） 基本问题是它在选择选项时它不会移动到执行功能，它只是结束。

    from sys import exit

def readFrom():
    file = input('What file do you want to read from? (Please include extension)')
    try:
        infile = open(file, 'r')

        line = infile.readline()
        cleanLine = line.strip()

        while line != '':
            print(cleanLine)
            line = infile.readline()
            cleanLine = line.strip()
        infile.close()
        menu()
    except IOError:
        print('Sorry, that is not a valid file. It might not exist or might not be in the correct directory.')

def writeTest():
    file = input('What file do you want to write to?')
    outfile = open(file,'w')

    addition = input('What would you like to write to the file?')
    try:
        addition = int(addition)
    except:
        print('Sorry, that is invalid.')
        writeTest()
    outfile.write(str(addition))
    outfile.close()
    menu()

def fileAppend():
    outfile = open('example.txt','a')

    addition = input('What would you like to add to the file?')
    try:
        addition = int(addition)
    except:
        print('Sorry, that is invalid.')
        fileAppend()
    outfile.write(str(addition))
    outfile.close()
    menu()

def menu():
    print('Hello! Welcome to the file editing helper!')
    answer = input('Selection Menu:\n'
                '0. Exit\n'
                '1. Read from file\n'
                '2. Write integers to a file\n'
                '3. Append integers to a file\n'
                'Which would you like to do? ')

    if answer == 1:
        readFrom()
    if answer == 2:
        writeTest()
    if answer == 3:
        fileAppend()
    if answer == 0:
        exit()

menu()

我已经尝试了所有我能想到的方法，我已经将完整的函数放入了我本来要调用函数的部分中，如果我不必按照要求将它们放在函数中我已经得到了，我不会在函数中使用它们。任何提示都是有帮助的，因为我真的不擅长这个。

Answer 1

好的，在你的 menu() 中你要求输入，然后你将该输入与一个整数进行比较，但你的输入被认为是一个字符串。

示例：

user_input = input('Enter a number') # 7

user_input 将是 '7'，而不是 7，所以你不能说

if user_input == 7:
    #do this

因为它永远不会。所以这里的菜单是固定的。

def menu():
    print('Hello! Welcome to the file editing helper!')
    answer = int(input('Selection Menu:\n'
                '0. Exit\n'
                '1. Read from file\n'
                '2. Write integers to a file\n'
                '3. Append integers to a file\n'
                'Which would you like to do? '))

    if answer == 1:
        readFrom()
    if answer == 2:
        writeTest()
    if answer == 3:
        fileAppend()
    if answer == 0:
        exit()

menu()

我将输入函数包装在一个 int() 方法中，现在应该可以使用了。 1 不等于 '1'，这是您要执行的操作。