Question

我一直在尝试简化此代码，在上一个问题中，我问用户一个可能的解决方案，但我无法使用列表列表，下面是我正在尝试简化的代码：

option = int(input("Table to serve "))
while  option != 0:
    menu()
    if option == 1:
        option2 = int(input("Put the code of the food (0 to END): "))
        while option2 != 0:
            menu()
            x = option2 - 1
            table1.append((names[x],prices[x],types[x]))
            option2 = int(input("Put the code of the food (0 to END): "))

如您所见，它是为单个选项定义的，但是我有6个选项，如果我将其复制并粘贴6次，代码将很长，我如何在考虑到以下情况的情况下简化函数呢？ / p>

table1.append((names[x],prices[x],types[x]))

表的名称取决于变量'option'

谢谢。

编辑：

def files():
    try: 
        file = open("menu.txt","r",1,"utf-8")
    except:
        print("file not found")

    line = file.readline().strip()    
    while line != "":
        parts = line.split(";")
        name = parts[0]
        price = int(parts[1])
        type = parts[2]
        names.append(name)
        prices.append(price)
        types.append(type)

        line = file.readline().strip()

def menu():
    for i in range(len(names)):
        print(i+1," ",names[i], " ", prices[i], " ",types[i]," \n")

试图简化代码，附加问题

0 个答案: