我正在编写一个程序,根据用户输入的日期和时间,通知用户哪些动物将在参观动物园时醒来并喂食。 我有一个类,一个函数,它读取一个文件的信息,包括名字,醒着的时间,他们什么时候被喂,还有一些其他的东西,所有这些都可以工作
class Animal:
def __init__(self, name, sleep, diet, awake, feed, number):
self.name = name
self.sleep = sleep
self.diet = diet
self.awake = awake
self.feed = feed
self.number = number
def __repr__(self):
return self.name + " " + self.sleep + " " + self.diet + " " + str(self.awake) + " " + str(self.feed) + " " + str(self.number)
def readInfo():
infile = open("zoo.txt", "r", encoding="UTF-8")
animals = []
lines = infile.readlines()
infile.close
for line in lines:
lineparts = line.split(" / ")
name = lineparts[0]
sleep = lineparts[1]
diet = lineparts[2]
awake = lineparts[3]
feed = lineparts[4]
number = lineparts[5]
animals.append(Animal(name, sleep, diet, awake, feed, number))
return animals
def Awake(x):
awakeanimals = x
print("\nYou can see ")
for object in awakeanimals:
print(object)
def Feed(x):
matadjur = x
print("\nand you can feed: ")
for object in feedanimals:
print(object)
以下是我的代码:
def open():
animals = readInfo()
awake = list()
feed = list()
time = int(input("Enter a time interval, eg 07-16")).split("-")
if 9 <= time <= 20:
awakeanimals.append(animals[0].name)
if 12 <= time <= 14:
awakeanimals.append(animals[1].name)
if 21 >= time >= 05:
awakeanimals.append(animals[2].name)
#same for the rest of the animals
if time <= 12 <= time:
feedanimals.append(animals[0].name)
if time <= 13 <= time:
feedanimals.append(animals[0].name)
#same for the rest of the animals
Awake(awakeanimals)
Feed(feedanimals)
之后,我有一个简单的菜单,根据用户输入的日期,调用函数open()
,然后进入时间部分
我不知道如何在if
条件下获得正确的输入
另外,由于ValueError: invalid literal for int() with base 10
,time = input().split("-")
不起作用,所以我考虑使用两个时间输入time1 = input()
和time2 = input()
。然而,在if
条件中,这似乎更复杂
我不太清楚你想要什么逻辑。假设它是一个范围之间的检查,下面是
split("-")
失败,因为您正在intger值上调用它split
只能应用于字符串。当07-16
不是有效数字时,会出现base 10错误这是一段经过编辑的代码
这将主要用于输入问题和比较
相关问题 更多 >
编程相关推荐