您如何拒绝输入，提出要求并仅保留该值

Question

Hello Stack社区

我在询问用户输入时使用了if或语句，如果用户输入的值超出了我的if或语句，那么我会显示一条消息，提示“请输入1到500之间的有效数字”向用户的另一个输入请求。

所以我已经完成了目标的2/3。我已经拒绝了原始输入，但是已经重新提出了要求，但是在重新询问用户之后，我无法找到仅保留新值的方法。相反，我的程序保留了所需参数之外的原始用户输入，然后在重新设置后保留了新输入并将它们组合为一个值。这是在Python IDLE 3.6上。

确切的问题：如何从用户输入中删除原始输入值，并仅在重新设置后保留新值？

这是我的代码：

O(N^2)

以下是shell中发生的情况的屏幕截图：

Answer 1

您可以使用无限循环不断询问用户有效输入，直到用户输入一个有效输入为止。请注意，您还应该使用<div id="home" class="section"> <div class="cover"> <svg viewBox="0 0 1366 768" width="100%" height="100%" preserveAspectRatio="xMinYMin meet"> <defs> <path id="shape" d="M1039.5,770.5a125.77,125.77,0,0,0-17-32c-12.32-16.72-25.68-25.84-33-31-6-4.23-59.88-42.55-100-90-13.64-16.14-20.57-24.48-26-38-15-37.48-3.73-73.65,0-85,8.68-26.45,23-43.26,35-57,19-21.82,33.56-29.9,67-54,33.68-24.27,55.39-39.91,77-60,40.56-37.69,35.94-49.35,81-96,34.18-35.39,51.27-53.08,79-65,7.79-3.35,76-31.44,140,2a142,142,0,0,1,31,22l7.5,7.5V770.5Z" /> <clipPath id="art"> <use xlink:href="#shape" /> </clipPath> </defs> <image xlink:href="http://1.bp.blogspot.com/-xV_Q0sj_3HQ/T4NjscM0ibI/AAAAAAAAB_U/DKjSJvD0vls/s1600/Simple+unique+odd+weird+wallpapers+minimalistic+%2310.jpg" width="1366" height="768" preserveAspectRatio="xMidYMid slice" /> <use xlink:href="#shape" id="background" /> <image xlink:href="http://1.bp.blogspot.com/-xV_Q0sj_3HQ/T4NjscM0ibI/AAAAAAAAB_U/DKjSJvD0vls/s1600/Simple+unique+odd+weird+wallpapers+minimalistic+%2310.jpg" width="1366" height="768" preserveAspectRatio="xMidYMid slice" id="middleground" /> <use xlink:href="#shape" id="foreground" /> </svg> </div> </div>块来确保用户输入有效的整数：

try-except

由于您要针对座位的不同部分提出相同的问题，因此您可以将不同部分的价格存储在字典中，然后在计算总收入时遍历各个部分以提出问题：

while True:
    try:
        resultsA = int(input("lease enter how many seats you sold is section A:"))
        if 1 <= resultsA <= 300:
            break
        raise RuntimeError()
    except (ValueError, RuntimeError):
        print("Please enter a valid number from 1 to 300.")

Answer 2

您可以轻松地创建一个函数，通过其中的try / except块来完成您的任务：

def int_input(prompt, lower, higher, bad_input='bad input:'):
  while True:
    try:
      r = int(input(prompt))
      if r < lower or r > higher:
        raise ValueError('enter a number between {} and {}'.format(lower, higher))
    except ValueError as e:
      print(bad_input, e)
    else:
      return r

然后在您的代码中使用它：

resultsA = int_input(
    'enter no. of seats sold in section A',
    1,
    200,
    'invalid input'
)

Answer 3

尝试这个：

def main():

    resultsA = int(input("Please enter how many seats you sold is section A:"))
    while(resultsA < 1 or resultsA > 300):
        resultsA = int(input("Please enter a valid number from 1 to 300:"))


    resultsB = int(input("Please enter how many seats you sold is section B:"))
    while(resultsB < 1 or resultsB > 500):
        resultsB = int(input("Please enter a valid number from 1 to 500:"))


    resultsC = int(input("Please enter how many seats you sold is section C:"))
    while(resultsC < 1 or resultsC > 200):
        resultsC = int(input("Please enter a valid number from 1 to 200:"))


    finalResultA = 20*resultsA
    finalResultsB = 15*resultsB
    finalResultsC = 10*resultsC
    trulyFinal = finalResultA + finalResultsB + finalResultsC

    print ("Congratulations, here is your total revenue from tickets: $",trulyFinal)