Candace
Candace

Reputation: 131

Exit while loop by user hitting ENTER key

I am a python newbie and have been asked to carry out an exercise: make a program loop until exit is requested by the user hitting <Return> only. So far I have:

User = raw_input('Enter <Carriage return> only to exit: ')
running = 1
while running == 1:
    ...  # Run my program
    if User == # Not sure what to put here
        break

I have tried: (as instructed in the exercise)

if User == <Carriage return>

and also

if User == <Return>

but this only results in invalid syntax.

How do I do this in the simplest way possible?

Upvotes: 13

Views: 138961

Answers (15)

bstpierre
bstpierre

Reputation: 31186

Here's a solution (resembling the original) that works:

User = raw_input('Enter <Carriage return> only to exit: ')
while True:
    #Run my program
    print 'In the loop, User=%r' % (User, )

    # Check if the user asked to terminate the loop.
    if User == '':
        break

    # Give the user another chance to exit.
    User = raw_input('Enter <Carriage return> only to exit: ')

Note that the code in the original question has several issues:

  1. The if/else is outside the while loop, so the loop will run forever.
  2. The else is missing a colon.
  3. In the else clause, there's a double-equal instead of equal. This doesn't perform an assignment, it is a useless comparison expression.
  4. It doesn't need the running variable, since the if clause performs a break.

Upvotes: 0

Serban Razvan
Serban Razvan

Reputation: 4490

If you want your user to press enter, then the raw_input() will return "", so compare the User with "":

User = raw_input('Press enter to exit...')
running = 1
while running == 1:
    ...  # Run your program
    if User == "":
        break

Upvotes: -2

MIGHTY BOMBER
MIGHTY BOMBER

Reputation: 1

user_input = input("ENTER SOME POSITIVE INTEGER : ")
if (not user_input) or (int(user_input) <= 0):
   print("ENTER SOME POSITIVE INTEGER GREATER THAN ZERO")  # print some info
   import sys  # import
   sys.exit(0)  # exit program

not user_input checks if user has pressed enter key without entering number.

int(user_input) <= 0 checks if user has entered any number less than or equal to zero.

Upvotes: 0

user3394391
user3394391

Reputation: 487

The exact thing you want ;)

from answer by limasxgoesto0 on "Exiting while loop by pressing enter without blocking. How can I improve this method?"

import sys, select, os

i = 0
while True:
    os.system('cls' if os.name == 'nt' else 'clear')
    print "I'm doing stuff. Press Enter to stop me!"
    print i
    if sys.stdin in select.select([sys.stdin], [], [], 0)[0]:
        line = raw_input()
        break
    i += 1

Upvotes: 17

Pitmaster
Pitmaster

Reputation: 1

I recommend to use u\000D. It is the CR in unicode.

Upvotes: 0

gtcoder
gtcoder

Reputation: 151

This works for python 3.5 using parallel threading. You could easily adapt this to be sensitive to only a specific keystroke.

import time
import threading


# set global variable flag
flag = 1

def normal():
    global flag
    while flag==1:
        print('normal stuff')
        time.sleep(2)
        if flag==False:
            print('The while loop is now closing')

def get_input():
    global flag
    keystrk=input('Press a key \n')
    # thread doesn't continue until key is pressed
    print('You pressed: ', keystrk)
    flag=False
    print('flag is now:', flag)

n=threading.Thread(target=normal)
i=threading.Thread(target=get_input)
n.start()
i.start()

Upvotes: 2

Anonymous
Anonymous

Reputation: 1

The following works from me:

i = '0'
while len(i) != 0:
    i = list(map(int, input(),split()))

Upvotes: -3

DaNNuN
DaNNuN

Reputation: 358

You are nearly there. the easiest way to get this done would be to search for an empty variable, which is what you get when pressing enter at an input request. My code below is 3.5

running = 1
while running == 1:

    user = input(str('Enter <Carriage return> only to exit: '))

    if user == '':
        running = 0
    else:
        print('You had one job...')

Upvotes: 0

Oybek
Oybek

Reputation: 1

Here is the best and simplest answer. Use try and except calls.

x = randint(1,9)
guess = -1

print "Guess the number below 10:"
while guess != x:
    try:
        guess = int(raw_input("Guess: "))

        if guess < x:
            print "Guess higher."
        elif guess > x:
            print "Guess lower."
        else:
            print "Correct."
    except:
        print "You did not put any number."

Upvotes: -1

crackedshiva
crackedshiva

Reputation: 1

a very simple solution would be, and I see you have said that you would like to see the simplest solution possible. A prompt for the user to continue after halting a loop Etc.

raw_input("Press<enter> to continue")

Upvotes: -1

ptay
ptay

Reputation: 795

I ran into this page while (no pun) looking for something else. Here is what I use:

while True:
    i = input("Enter text (or Enter to quit): ")
    if not i:
        break
    print("Your input:", i)
print("While loop has exited")

Upvotes: 22

hymloth
hymloth

Reputation: 7035

if repr(User) == repr(''):
    break

Upvotes: -1

Surya Kasturi
Surya Kasturi

Reputation: 4828

Actually, I suppose you are looking for a code that runs a loop until a key is pressed from the keyboard. Of course, the program shouldn't wait for the user all the time to enter it.

  1. If you use raw_input() in python 2.7 or input() in python 3.0, The program waits for the user to press a key.
  2. If you don't want the program to wait for the user to press a key but still want to run the code, then you got to do a little more complex thing where you need to use kbhit() function in msvcrt module.

Actually, there is a recipe in ActiveState where they addressed this issue. Please follow this link

I think the following links would also help you to understand in much better way.

  1. python cross platform listening for keypresses

  2. How do I get a single keypress at a time

  3. Useful routines from the MS VC++ runtime

I hope this helps you to get your job done.

Upvotes: 5

naeg
naeg

Reputation: 4002

You need to find out what the variable User would look like when you just press Enter. I won't give you the full answer, but a tip: Fire an interpreter and try it out. It's not that hard ;) Notice that print's sep is '\n' by default (was that too much :o)

Upvotes: 0

Tom Zych
Tom Zych

Reputation: 13576

Use a print statement to see what raw_input returns when you hit enter. Then change your test to compare to that.

Upvotes: 1

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