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Python

Posted by Freddy Venegas
Freddy Venegas
1,226 Points

Shopping list take 3 - double input

When I add the code to check for a ValueError I get double inputs of the same item. For example, I have this

while True:
    new_stuff = raw_input("> ")

    if new_stuff == 'DONE':
        print("\nHere's your list: ")
        show_list()
        break
    elif new_stuff == 'HELP':
        show_help()
        continue
    elif new_stuff == 'SHOW':
        show_list()
        continue
    else:
        new_list = new_stuff.split(",")
        index = raw_input("Add this at a certain spot? Press enter for the end of the list, " "or give me a number. Currently {} items in the list. ".format(len(shopping_list)))
    try:
        if index:
            spot = int(index) -1
            for item in new_list:
                shopping_list.insert(spot, item.strip())
                spot += 1
    except ValueError:
        print("That is not a number.")
    else:
        for item in new_list:
            shopping_list.append(item.strip())

if I enter bananas > pick any spot that's not enter > then enter SHOW I see the same item I just entered twice. And yes, even entering 0 adds the item twice. Like this:

1. bananas
2. bananas

Any ideas?

1 Answer

Freddy Venegas
Freddy Venegas
1,226 Points
Freddy Venegas
Freddy Venegas
1,226 Points

Seems like this did the trick, by moving try: & except: inside if index: - please respond if this is the best solution.

if index:
    try:
        spot = int(index) -1
        for item in new_list:
            shopping_list.insert(spot, item.strip())
            spot += 1
    except ValueError:
        print("This is not a number.")
else:
    for item in new_list:
        shopping_list.append(item.strip())