if语句，来回

Question

我是 python 的新手，但已经编写了一些结构化文本和一点点 C++ 现在我正在使用 if 语句解决一个问题，它似乎不像我习惯的那样工作

示例代码：

more_guests = 0
loop = bool; loop = False
ferdig = int; ferdig = 1
while loop == False:
    guests = []
    more_guests = 0
    if int(ferdig) == 1:
        guest = input("type in guest ")
        more_guests = int(input("done? 1 for yes 2 for no "))
        if int(more_guests) == 1:
          guests.append(guest)
          ferdig == 3
        elif int(more_guests) == 2:
            guests.append(guest)
            ferdig == 2
        else:
            print("unvalid answer, please use 1 or 2")
            ferdig == 1
    elif int(ferdig) == 2:
        ferdig = 1
    elif int(ferdig) == 3:
        print(guests)
    else:
        loop = True

我试过确保它是一个整数等等，只是继续坚持完成吗？ 1 表示是 2 表示否，它总是让我返回输入 guest

在使用结构化文本时，我经常使用这种来回的方式，但在 python 中我似乎无法理解它，也许我更应该使用 case/switch？无论如何，如果有人可以帮助我理解您是否可以在 python 中以这种方式使用 IF 语句，我们将不胜感激

Answer 1

我想你写错了运算符。 ferdig == 1 只是比较值运算符。所以它是 return 真 (1) 或假 (0)。

如果你想改变 ferdig 的值，你应该在 if 语句中写 ferdig = 3。

Answer 2

我已经稍微清理了你的代码以获得我认为你想要的。

guests = []
while True:
    guest = input("type in guest ")
    guests.append(guest)
    response = int(input("done? 1 for yes 2 for no "))
    if response == 1:
        print(guests)
        break
    elif response == 2:
        continue
    elif response == 3:
        print(guests)
    else:
        print("invalid answer, please use 1 or 2")

Answer 3

你的代码有很多问题，这里已经解决了：

more_guests = 0
# you don't need to declare types, but if you want to, this is how
loop: bool = False
# however, Python can just infer the type itself like this
ferdig = 1
# you don't want to reset guests every time around the loop
guests = []
while loop == False:
    more_guests = 0
    if int(ferdig) == 1:
        guest = input("type in guest ")
        # you always want to append a guest, including if someone types a 1 after
        guests.append(guest)
        more_guests = int(input("done? 1 for yes 2 for no "))
        if int(more_guests) == 1:
            ferdig = 3
        elif int(more_guests) == 2:
            ferdig = 2
        else:
            print("invalid answer, please use 1 or 2")
            ferdig = 1
    elif int(ferdig) == 2:
        ferdig = 1
    elif int(ferdig) == 3:
        # after printing, you're done, you don't want to print forever
        print(guests)
        loop = True

请注意，您的大部分代码并不是真正需要的，您正在做很多 Python 可以为您做的簿记工作，或者根本不需要：

# this isn't needed, because you set that at the start of the loop anyway
# more_guests = 0
# this isn't needed, because you can tell when to stop from ferdig
# loop: bool = False
# starting at 2, since that means you want to keep going
ferdig = 2
guests = []
while ferdig != 1:
    # this isn't needed, you can just read ferdig
    # more_guests = 0
    # this isn't needed, you want a new guest on every loop
    #if int(ferdig) == 1:
    guest = input("type in guest ")
    guests.append(guest)
    ferdig = int(input("done? 1 for yes 2 for no "))
    # none of this is needed, all you need to know is if ferdig is 1 or 2
    # if int(more_guests) == 1:
    #     ferdig = 3
    # elif int(more_guests) == 2:
    #     ferdig = 2
    # else:
    if ferdig not in (1, 2):
        print("invalid answer, please use 1 or 2")
        ferdig = 1
    # this is also not needed, at this point ferdig will be 1 or 2
    # elif int(ferdig) == 2:
    #     ferdig = 1
    # elif int(ferdig) == 3:
# put the print outside the loop and it only prints once
print(guests)
# got rid of this
# loop = True

所以，这只是：

ferdig = 2
guests = []
while ferdig != 1:
    guest = input("type in guest ")
    guests.append(guest)
    ferdig = int(input("done? 1 for yes 2 for no "))
    if ferdig not in (1, 2):
        print("invalid answer, please use 1 or 2")
        ferdig = 1
print(guests)

最后，这样做也是一样的：

more_guests = True
guests = []
while more_guests:
    guests.append(input("type in guest "))
    more_guests = input("done? 1 for yes") != '1'
print(guests)

Answer 4

这里有另一种方法可以使用函数来完成您想要做的事情。

guests = []

def add_guest():

    guest = input("type in guest ")
    guests.append(guest)
    add_more()

def add_more():

    more_guests = int(input("done? 1 for yes 2 for no ")) 

    #if they answer 1 we print out the final list
    if more_guests == 1: 
        print("the guest list is {}".format(guests))
        print('have a nice day')  

    #if they answer 2 we go to the add_guest function    
    elif more_guests == 2:  
        add_guest()

    #if they answer anything other than 1 or 2 we re-call the add_more function.  
    else: 
        print("unvalid answer, please use 1 or 2") 
        add_more()

add_guest()

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