Exit while loop by user hitting ENTER key

Question

I am a python newbie and have been asked to carry out some exercises using while and for loops. I have been asked to 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
else
    running == 1

I have tried: (as instructed in the exercise)

if User == <Carriage return>

and also

if User == <Return>

but this only results in invalid syntax. Please could you advise me on how to do this in the simplest way possible. Thanks

Answer 1

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")

Answer 2

The exact thing you want ;)

https://stackoverflow.com/a/22391379/3394391

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

Answer 3

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.

Answer 4

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()

Answer 5

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

Answer 6

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")

Answer 7

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.
'''

Answer 8

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."

Answer 9

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...')

Answer 10

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)

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

The following works from me:

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

Answer 15

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
else
    running == 1