我是 Python 新手，但已經撰寫了一些結構化文本和一點點 C ，現在我正在使用 if 陳述句解決一個問題，但它似乎不像我習慣的那樣作業
示例代碼：
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
我試過確保它是一個整數等等，只是繼續堅持完成？1 代表是 2 代表否，它總是讓我回過頭來輸入客人
在使用結構化文本時，我經常使用這種來回切換的方式，但我似乎無法在 python 中理解它，也許我應該使用 case/switch？無論如何，如果有人可以幫助我理解您是否可以在python中以這種方式使用IF陳述句，非常感謝
uj5u.com熱心網友回復：
你的代碼有一堆問題，這里已經修復：
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
請注意，您的大部分代碼并不是真正需要的，您正在做很多 Python 可以為您做的簿記，或者只是不需要：
# 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，這只是：
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，這會做同樣的事情：
more_guests = True
guests = []
while more_guests:
guests.append(input("type in guest "))
more_guests = input("done? 1 for yes") != '1'
print(guests)
uj5u.com熱心網友回復：
我認為你寫錯了運算子。ferdig == 1 只是比較值運算子。所以它回傳真（1）或假（0）。
如果您想更改 ferdig 的值，您應該在 if 陳述句中寫入 ferdig = 3。
uj5u.com熱心網友回復：
我已經清理了你的代碼以獲得我認為你想要的東西。
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")
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