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Problems with sys.stdout.write() with time.sleep() in a function

开发者 https://www.devze.com 2022-12-29 17:19 出处:网络
What I wanted is printing out 5 dots that a dot printed per a second using time.sleep(), but the result was 5 dots were printed at once after 5 seconds delay.

What I wanted is printing out 5 dots that a dot printed per a second using time.sleep(), but the result was 5 dots were printed at once after 5 seconds delay.

Tried both print and sys.stdout.write, same result.

Thanks for any advices.

import time
import sys

def wait_for(n):
    """Wait for {n} seconds. {n} should be an integer greater than 0."""
    if not isinstance(n, int):
        print 'n in wait_for(n) should be an integer.'
        return
    elif n < 1:
        print 'n in wait_for(n) should be greater than 0.'
        return
    for i in range(0, n):
        sys.stdout.write('.')
        time.sleep(1)
    sys.stdout.write('\n')

def main():
    wait_for(5)    # FIXME: doesn't work as expected

if __name__ == '__main__':
    try:
  开发者_如何学运维      main()
    except KeyboardInterrupt:
        print '\nAborted.'


You need to flush after writing.

sys.stdout.write('foo')
sys.stdout.flush()
wastetime()
sys.stdout.write('bar')
sys.stdout.flush()


You should use sys.stderr.write for progress bars; stderr has the (not at all coincidental) advantage of not being buffered, so no sys.stderr.flush calls are needed.

See also this answer.

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