Python Input Loops: Keep Asking Until It's Right

Keep Asking Until the Answer Is Valid

An input loop is a while loop that calls input() again and again. Its first job: keep asking until the user gives a valid answer, instead of stopping at the first mistake.

In chapter 8, an if could only complain about bad input once. A while can ask again:

12345marks = int(input())
while marks < 0 or marks > 100:
    print("Marks must be 0 to 100. Try again.")
    marks = int(input())
print("Saved:", marks)

The Input box has three lines, one for each input() call: 150 and -5 fail the check, and 80 passes. Change the lines and run it again.

Think of a shop assistant who asks your shoe size. Say "99", and they ask again, until you say a size they have.

The pattern

1. Ask once before the loop.
2. While the answer is bad: say what's wrong, and ask again inside the loop.
3. After the loop, the answer is sure to be good.

The input() inside the loop is the loop's change: it's the line that can make the condition False. Leave it out, and you get an endless loop, printing "Try again" forever.

Write the bad case in the condition

The condition describes a bad answer, because that's when the loop should keep going:

Common beginner doubt: "What if the user types abc?" Then int() stops the program with a ValueError, before the loop can help. Catching that needs try and except, from chapter 17.

Ask once, then while the answer is bad, explain and ask again.

Stop at a Word Like done (Sentinel Value)

How does a program know when the user has finished typing? You agree on a stop word, like done. A special value that means "stop" is called a sentinel value (a sentinel is a guard who keeps watch).

123456items = []
item = input()
while item != "done":
    items.append(item)
    item = input()
print("You need", len(items), "things:", items)

Think of a waiter taking your order: they keep writing until you say "That's all".

Two ways to write it

The loop above asks before the loop, and again at the end of each round. The same program with while True and break asks in just one place:

1234567items = []
while True:
    item = input()
    if item == "done":
        break
    items.append(item)
print("You need", len(items), "things:", items)

Both are fine. Pick the one you find easier to read.

The stop word is not data

Notice that done never ends up in the list. Check for the stop word before you use the answer.

Avoid: running out of input

Here in LivePython, every input() takes the next line of the Input box. If your loop asks for more lines than there are, Python stops with an error:

1234item = input()
while item != "done":
    print("Got", item)
    item = input()

EOF means "end of file": there was no more input to read. Add the stop word as the last line, and it works.

A stop word ends an input loop. Check for it before you use the answer.

Add Up Numbers Until the User Stops

Input loops shine when you add up values and don't know how many there will be. Combine the stop word with a running total and a counter, and you can work out sums and averages of any length.

12345678910total = 0
count = 0
n = int(input())
while n != 0:
    total += n
    count += 1
    n = int(input())
print("Total:", total)
if count > 0:
    print(f"Average: {total / count:.2f}")

Here the stop value is the number 0. It's a good sentinel when 0 can't be a real answer, like marks or prices here.

Think of a cashier: they scan items one by one, keeping a running total, until there's nothing left on the belt.

The parts

Both start before the loop. Start them inside, and they're reset to 0 every round.

Watch out for an empty list of numbers

What if the very first answer is 0? Then the loop never runs, count stays 0, and total / count would give a ZeroDivisionError. That's why the average sits inside if count > 0:, the check-first habit from chapter 8.

1234567891011total = 0
count = 0
n = int(input())
while n != 0:
    total += n
    count += 1
    n = int(input())
if count > 0:
    print(f"Average: {total / count:.2f}")
else:
    print("No numbers given")

Text or number stop?

If 0 could be a real answer (a temperature, a score in a game), use a word like done as in the last lesson: read the text, check for the stop word, and only then use int().

Start the total and the counter before the loop, add each round, and check count before you divide.

Give the User Only a Few Tries

A cash machine doesn't let you guess a PIN forever. To limit the tries, join two conditions with and (chapter 8): keep asking while the answer is wrong and there are tries left.

12345678910pin = input()
tries = 1
while pin != "1234" and tries < 3:
    print("Wrong PIN. Try again.")
    pin = input()
    tries += 1
if pin == "1234":
    print("Welcome, Lucky!")
else:
    print("Card blocked.")

Now try the Input lines 1, 2 and 3: two "Wrong PIN" messages, then Card blocked.

Think of a game with three lives: you keep playing while you haven't won and you still have lives left.

Two ways out

This loop can end for two reasons:

• the PIN is right, so pin != "1234" is False, or
• the tries are used up, so tries < 3 is False.

and needs both sides True to go on, so either one ends the loop.

Find out why it ended

After the loop, you don't know which reason stopped it. Ask with an if, as the example does: if the PIN is right, the user got in; otherwise the tries ran out.

Check the answer, not the counter. On the third try, tries is 3 whether the PIN was right or wrong, so if tries == 3: can't tell you if the user got in.

Count the tries carefully

tries starts at 1, because one answer is already read before the loop. Each new input() adds 1. That's the off-by-one thinking from topic 2: trace the edges, and make sure the user gets exactly 3 tries, not 2 or 4.

Limit the tries with and, then use if after the loop to see why it ended.

Practice: Input loops

Solve these to lock in what you learned in this topic. Each one runs in your browser.

1. 95.Python Program to Add Numbers Until the User Types 0EasySolve
2. 96.Python Program to Ask Again Until the Age Is ValidEasySolve
3. 97.Python Program to Find the Largest Number Until done Is TypedMediumSolve