[英]What am I doing wrong in my code python that it won't print the totall cost?
此代码最多需要允许3个输入用于所需物料,并打印所有物料的总成本。 我对这一切非常陌生,需要我能得到的所有建议。 我无法打印总数。
pie = 2.75
coffee = 1.50
icecream = 2.00
while choice:
choice01 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
if choice01 == "nothing":
break
choice02 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
if choice02 == "nothing":
break
choice03 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
if choice03 == "nothing":
break
cost = choice01+choice02+choice03
print "Your total is: ${0}" .format(cost)
让我们专注于代码在做什么。
choice01 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
当用户回答这个问题时,他们的答案就是一个字符串。 现在处于选择choice01
。 对于此示例,假设它们键入“ pie”(不带引号)。
我们使用choice02 =
line重复此操作,然后用户选择“咖啡”。
让我们看一下这两个选择的cost =
行。
cost = choice01 + choice02
我们只是确定choice01
是字符串值“ pie”,而choice02
是字符串值“ coffee”。因此, cost = "piecoffee"
您想在顶部使用这些变量。 一种方法是创建字典:
prices = {"pie": 2.75,
"coffee": 1.50,
"icecream": 2.00,
"nothing": 0.00
}
...
cost = prices[choice01]+prices[choice02]+prices[choice03]
在字典中,我设置了4个可能的值和相关价格。 “无”的值为0.00,因为您在成本计算中使用了该值。 它使数学变得简单并且有效,因为您假设始终会做出3个选择。
重要的是要注意,使用这种方法,如果用户输入答案(即“ cofee”而不是“ coffee”),它将抛出异常。 这是确定您如何处理此类错误的活动。 您可以在很多方面添加这样的检查。
您还需要解决其他一些问题:
while choice:
将无法按照您的代码立场行事。 您没有定义choice
。 另一种选择是简单地while True
,然后在循环结束时中断。 例:
prices = {"pie": 2.75,
"coffee": 1.50,
"icecream": 2.00
}
cost = 0.00
while True:
choice = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
if choice == "nothing":
break
cost = cost + prices[choice]
print "Your total is: ${0}" .format(cost)
输出:
What would you like to buy? pie, coffee, icecream, or nothing? pie
What would you like to buy? pie, coffee, icecream, or nothing? nothing
Your total is: $2.75
请注意,我们只有一次用户输入问题,并且在循环开始之前定义了cost
。 然后,我们每次循环都要添加。 我还从字典中删除了“ nothing”键,因为您在将选择添加到成本之前就跳出了循环。
price_dict = dict( pie=2.75, coffee=1.50, icecream=2.00, nothing=None)
max_choices = 3
cost = 0
while max_choices:
choice = raw_input(
"What would you like to buy? pie, coffee, icecream, or nothing?")
# please check the input
if choice not in price_dict:
print 'Your input is invalid, please enter again...'
continue
if choice == 'nothing':
break
max_choices -= 1
cost += price_dict[choice]
print "Your total is: ${0}" .format(cost)
您在顶部定义的是派,咖啡和冰淇淋的变量名。
从raw_input得到的是文本字符串。
它们不仅仅因为相同而匹配,您还必须以某种方式使其匹配。 我建议更像:
pie = 2.75
coffee = 1.50
icecream = 2.00
cost = 0
while True:
choice = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
if choice == "nothing":
break
if choice == "pie":
cost = cost + pie
if choice = "coffee":
cost = cost + coffee
#... etc.
print "Your total is: ${0}".format(cost)
而且,如果您想避免使用大量if
语句,请查看@Evert建议使用包含价格的字典。
看来您根本不需要三个选择。 考虑以下代码
pie = 2.75
coffee = 1.50
icecream = 2.00
cost = 0
while True:
choice01 = 0
choice01 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
if choice01 == "nothing":
break
elif choice01=="coffee":
choice01 = coffee
elif choice01=="pie":
choice01 = pie
elif choice01=="icecream":
choice01==icecream
else:
choice01=0
cost+=choice01
print "Your total is: ${0}" .format(cost)
确保用户仅输入pie而不输入“ pie”,因为Python会将“ pie”作为字符串。
考虑使用字典: choices = {'pie' : 2.75, 'coffee':1.50, 'icecream': 2.00, 'nothing':0}
,然后使用for循环。 您也可以使用文件循环,但是如果不需要,建议使用更简单的循环。
还要在启动循环之前声明cost= 0
。 这是循环:-
for x in range(0, 3): choosed = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?") cost = cost + choices[choosed]
然后最后(外部循环)用户print "Your total is: ${0}" .format(cost)
以获取总计。
将所有这些放在一起:
choices = {'pie' :2.75, 'coffee':1.50, 'icecream': 2.00, 'nothing': 0}
cost = 0
for x in range(0, 3):
choosed = input("What would you like to buy? pie, coffee, icecream, or nothing?")
cost = cost + choices[choosed]
print "Your total is: ${0}" .format(cost)
PS:-我建议您不要询问3次输入,而是在用户输入EOF后停止询问。 为此,您可以考虑if choosed == "nothing": break
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