How to re-prompt loop user input after invalid input range-CodePudding

User input range for num of rooms is [1-8]. Invalid input for this should display 'invalid input' and keep reprompt until receive valid input for executing the rest of the program. I've made it so invalid input for services returns and 'invalid service', the same should be done for input of num of rooms.

def main():
    small = 60
    medium = 120
    large = 145
    servOne = 40
    servTwo = 80


    numRooms = int(input("Number of rooms in the house?: "))

    while True:
        restart = int(input("Sorry Invalid number of rooms,try again? "))
        if int(numRooms) < 1 or int(numRooms) > 8 in restart:
            continue
            (???)

       




    servType = input("Type of cleaning service requested? (carpet cleaning or window cleaning): ")


    if int(numRooms) <= 2:
        print("Calculating fees for small house...")

        if servType == "carpet cleaning":
            print("Total cost of service:$",(small servOne))

        elif servType == "window cleaning":
            print("Total cost of service:$",(small servTwo))

        else:
            print("Sorry! Invalid service input")



    if int(numRooms) >= 3 and int(numRooms) <= 5:
        print("Calculating fees for medium house...")

        if servType == "carpet cleaning":
            print("Total cost of service:$",(medium servOne))

        elif servType == "window cleaning":
            print("Total cost of service:$",(medium servTwo))

        else:
            print("Sorry! Invalid service input")



    if int(numRooms) >= 6:
        print("Calculating fees for large house...")

        if servType == "carpet cleaning":
            print("Total cost of service:$",(large servOne))

        elif servType == "window cleaning":
            print("Total cost of service:$",(large servTwo))

        else:
            print("Sorry! Invalid service input")

main()

CodePudding user response：

One option (If I assume what you are trying to do) is to have a recursive function instead of a while loop, and call the function again on invalid input:


def Main():
  validnums = [1,2,3]
  i = Input("Please Enter input")
  if int(i) not in validnums:
    Main()
  else:
    ##Do stuff
  Main()

By having Main() call Main(), it functions identical to a while loop but you can restart it anytime. I hope this helps.

CodePudding user response：

just wrap input in a while loop. if valid input is recieved, you end the loop

valid_input=False 
while not valid_input:
    i=input("Number of rooms in the house?:")
    if int(1) in range(1,9):  #(9 is not included)
        valid_input=True 
... 
... remaining code 
...