如何设置 raw_input 的时间限制

Question

在 python 中，有没有办法在等待用户输入时计算时间，以便在 30 秒后自动跳过raw_input()函数？

Answer 1

The signal.alarm function, on which @jer's recommended solution is based, is unfortunately Unix-only. If you need a cross-platform or Windows-specific solution, you can base it on threading.Timer instead, using thread.interrupt_main to send a KeyboardInterrupt to the main thread from the timer thread. Ie:

import thread
import threading

def raw_input_with_timeout(prompt, timeout=30.0):
    print(prompt, end=' ')    
    timer = threading.Timer(timeout, thread.interrupt_main)
    astring = None
    try:
        timer.start()
        astring = input(prompt)
    except KeyboardInterrupt:
        pass
    timer.cancel()
    return astring

this will return None whether the 30 seconds time out or the user explicitly decides to hit control-C to give up on inputting anything, but it seems OK to treat the two cases in the same way (if you need to distinguish, you could use for the timer a function of your own that, before interrupting the main thread, records somewhere the fact that a timeout has happened, and in your handler for KeyboardInterrupt access that "somewhere" to discriminate which of the two cases occurred).

Edit : I could have sworn this was working but I must have been wrong -- the code above omits the obviously-needed timer.start() , and even with it I can't make it work any more. select.select would be the obvious other thing to try but it won't work on a "normal file" (including stdin) in Windows -- in Unix it works on all files, in Windows, only on sockets.

So I don't know how to do a cross-platform "raw input with timeout". A windows-specific one can be constructed with a tight loop polling msvcrt.kbhit , performing a msvcrt.getche (and checking if it's a return to indicate the output's done, in which case it breaks out of the loop, otherwise accumulates and keeps waiting) and checking the time to time out if needed. I cannot test because I have no Windows machine (they're all Macs and Linux ones), but here the untested code I would suggest:

import msvcrt
import time

def raw_input_with_timeout(prompt, timeout=30.0):
    print(prompt, end=' ')    
    finishat = time.time() + timeout
    result = []
    while True:
        if msvcrt.kbhit():
            result.append(msvcrt.getche())
            if result[-1] == '\r':   # or \n, whatever Win returns;-)
                return ''.join(result)
            time.sleep(0.1)          # just to yield to other processes/threads
        else:
            if time.time() > finishat:
                return None

The OP in a comment says he does not want to return None upon timeout, but what's the alternative? Raising an exception? Returning a different default value? Whatever alternative he wants he can clearly put it in place of my return None ;-).

If you don't want to time out just because the user is typing slowly (as opposed to, not typing at all!-), you could recompute finishat after every successful character input.

Answer 2

I found a solution to this problem in a blog post . Here's the code from that blog post:

import signal

class AlarmException(Exception):
    pass

def alarmHandler(signum, frame):
    raise AlarmException

def nonBlockingRawInput(prompt='', timeout=20):
    signal.signal(signal.SIGALRM, alarmHandler)
    signal.alarm(timeout)
    try:
        text = raw_input(prompt)
        signal.alarm(0)
        return text
    except AlarmException:
        print '\nPrompt timeout. Continuing...'
    signal.signal(signal.SIGALRM, signal.SIG_IGN)
    return ''

Please note: this code will only work on *nix OSs .

Answer 3

from threading import Timer


def input_with_timeout(x):    

def time_up():
    answer= None
    print('time up...')

t = Timer(x,time_up) # x is amount of time in seconds
t.start()
try:
    answer = input("enter answer : ")
except Exception:
    print('pass\n')
    answer = None

if answer != True:   # it means if variable have somthing 
    t.cancel()       # time_up will not execute(so, no skip)

input_with_timeout(5) # try this for five seconds

As it is self defined... run it in command line prompt , I hope you will get the answer read this python doc you will be crystal clear what just happened in this code!!

Answer 4

The input() function is designed to wait for the user to enter something (at least the [Enter] key).

If you are not dead set to use input(), below is a much lighter solution using tkinter. In tkinter, dialog boxes (and any widget) can be destroyed after a given time.

Here is an example :

import tkinter as tk

def W_Input (label='Input dialog box', timeout=5000):
    w = tk.Tk()
    w.title(label)
    W_Input.data=''
    wFrame = tk.Frame(w, background="light yellow", padx=20, pady=20)
    wFrame.pack()
    wEntryBox = tk.Entry(wFrame, background="white", width=100)
    wEntryBox.focus_force()
    wEntryBox.pack()

    def fin():
        W_Input.data = str(wEntryBox.get())
        w.destroy()
    wSubmitButton = tk.Button(w, text='OK', command=fin, default='active')
    wSubmitButton.pack()

# --- optionnal extra code in order to have a stroke on "Return" equivalent to a mouse click on the OK button
    def fin_R(event):  fin()
    w.bind("<Return>", fin_R)
# --- END extra code --- 

    w.after(timeout, w.destroy) # This is the KEY INSTRUCTION that destroys the dialog box after the given timeout in millisecondsd
    w.mainloop()

W_Input() # can be called with 2 parameter, the window title (string), and the timeout duration in miliseconds

if W_Input.data : print('\nYou entered this : ', W_Input.data, end=2*'\n')

else : print('\nNothing was entered \n')

Answer 5

A curses example which takes for a timed math test

#!/usr/bin/env python3

import curses
import curses.ascii
import time

#stdscr = curses.initscr() - Using curses.wrapper instead
def main(stdscr):
    hd = 100 #Timeout in tenths of a second
    answer = ''

    stdscr.addstr('5+3=') #Your prompt text

    s = time.time() #Timing function to show that solution is working properly

    while True:
        #curses.echo(False)
        curses.halfdelay(hd)
        start = time.time()
        c = stdscr.getch()
        if c == curses.ascii.NL: #Enter Press
            break
        elif c == -1: #Return on timer complete
            break
        elif c == curses.ascii.DEL: #Backspace key for corrections. Could add additional hooks for cursor movement
            answer = answer[:-1]
            y, x = curses.getsyx()
            stdscr.delch(y, x-1)
        elif curses.ascii.isdigit(c): #Filter because I only wanted digits accepted
            answer += chr(c)
            stdscr.addstr(chr(c))
        hd -= int((time.time() - start) * 10) #Sets the new time on getch based on the time already used

    stdscr.addstr('\n')

    stdscr.addstr('Elapsed Time: %i\n'%(time.time() - s))
    stdscr.addstr('This is the answer: %s\n'%answer)
    #stdscr.refresh() ##implied with the call to getch
    stdscr.addstr('Press any key to exit...')
curses.wrapper(main)

Answer 6

under linux one could use curses and getch function, its non blocking. see getch()

https://docs.python.org/2/library/curses.html

function that waits for keyboard input for x seconds (you have to initialize a curses window (win1) first!

import time

def tastaturabfrage():

    inittime = int(time.time()) # time now
    waitingtime = 2.00          # time to wait in seconds

    while inittime+waitingtime>int(time.time()):

        key = win1.getch()      #check if keyboard entry or screen resize

        if key == curses.KEY_RESIZE:
            empty()
            resize()
            key=0
        if key == 118:
            p(4,'KEY V Pressed')
            yourfunction();
        if key == 107:
            p(4,'KEY K Pressed')
            yourfunction();
        if key == 99:
            p(4,'KEY c Pressed')
            yourfunction();
        if key == 120:
            p(4,'KEY x Pressed')
            yourfunction();

        else:
            yourfunction

        key=0

Answer 7

This is for newer python versions, but I believe it will still answer the question. What this does is it creates a message to the user that the time is up, then ends the code. I'm sure there's a way to make it skip the input rather than completely end the code, but either way, this should at least help...

import sys
import time
from threading import Thread
import pyautogui as pag
#imports the needed modules

xyz = 1 #for a reference call

choice1 = None #sets the starting status

def check():
    time.sleep(15)#the time limit set on the message
    global xyz
    if choice1 != None:  # if choice1 has input in it, than the time will not expire
        return
    if xyz == 1:  # if no input has been made within the time limit, then this message 
                  # will display
        pag.confirm(text = 'Time is up!', title = 'Time is up!!!!!!!!!')
        sys.exit()


Thread(target = check).start()#starts the timer
choice1 = input("Please Enter your choice: ")