Exit while loop by user hitting ENTER key

Question

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?

Barmen

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")

Juvenility

Simple, effective, and pythonic. – Averse

Raises SyntaxError: unexpected EOF while parsing in Python 2. Using raw_input may fix, but I haven't tried it. – Monkeypot

Answer 2

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

Pull

This is an excellent answer. Here is (I believe) the original source, which has further information about non-blocking stdin: repolinux.wordpress.com/2012/10/09/… – Newfoundland

os.system('cls' if os.name == 'nt' else 'clear') is not relevant to the question – Amling

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.

Tacklind

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()

Walz

Some very minor nitpicking: flag = 1 -> flag = True. while flag==1 -> while flag. if flag==False: -> if not flag:. And you only have to state the global if you want to write to a variable, so you can remove global flag from normal(). – Camshaft

Answer 5

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

Orvah

the print statement is blank so I have tried User == '' but still this line is highlighted as invalid syntax – Barmen

raw_input will not capture <enter> or <return> – Obscurant

You forgot repr(). Otherwise printing doesn't show anything. print(repr(User)) – Amling

Answer 6

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)

Yasmeen

I tried to use a print statement to do this and the variable is blank so I tried User == '' but this results in invalid syntax as does User == '\n' – Barmen

Why should you be doing User == "? " is invalid Syntax. I'll help you even a bit more (even though this is soooo obvious actually): print repr(raw_input()) and just hit enter. – Yasmeen

Answer 7

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.

Confluence

It's better to use only one input statement. See this answer under "The Redundant Use of Redundant input Statements" – Amling

Answer 8

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.

Picky

You need to provide some discussion explaining how your answer addresses the question. – Anderson

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...')

Salute

Answer 10

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

Ejectment

I edited your post to reduce the impression that you only want to comment. If you indeed want to comment instead of actually answering please delete this and wait for your commenting privilege. meta.stackexchange.com/questions/214173/… – Rolanderolando

Answer 11

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

Toul

this line is still being highlighted as invalid syntax – Barmen

repr? Why? You can just do User == ''... – Amling

Answer 12

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")

Disputation

Answer 13

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."

Mouthful

A bare except is bad practice. Instead, use the specific exception you're expecting like except ValueError for invalid int strings, or at least except Exception, which would also include KeyboardInterrupt and EOFError. – Amling

Answer 14

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

Strangles

I have attempted this but the 5th line is still highlighted as invalid syntax – Barmen

You also need a : after the else, and running == 1 is a boolean expression not an assignment. – Burget

Answer 15

The following works from me:

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

Rancidity

Please explain your code, and what more does it bring that other answers do not. – Salomie