我最近遇到了这个问题，并希望将此解决方案留在这里以备将来参考。这些解决方案从终端清除待处理的raw_input（readline）文本，打印新文本，然后将raw_input缓冲区中的内容重新打印到终端。
第一个程序非常简单，但是仅在只有一行文本等待raw_input时才能正确运行：
#!/usr/bin/python
import time,readline,thread,sys
def noisy_thread():
while True:
time.sleep(3)
sys.stdout.write('\r'+' '*(len(readline.get_line_buffer())+2)+'\r')
print 'Interrupting text!'
sys.stdout.write('> ' + readline.get_line_buffer())
sys.stdout.flush()
thread.start_new_thread(noisy_thread, ())
while True:
s = raw_input('> ')
输出：
$ ./threads_input.py
Interrupting text!
Interrupting text!
Interrupting text!
> WELL, PRINCE, Genoa and Lucca are Now no more than private estates of the Bo
Interrupting text!
> WELL, PRINCE, Genoa and Lucca are Now no more than private estates of the Bo
naparte family. No, I warn you, that if you do not tell me we are at war,
第二个可以正确处理2条或更多的缓冲行，但是具有更多（标准）模块依赖性，并且需要一点点终端黑客：
#!/usr/bin/python
import time,readline,thread
import sys,struct,fcntl,termios
def blank_current_readline():
# Next line said to be reasonably portable for varIoUs Unixes
(rows,cols) = struct.unpack('hh', fcntl.ioctl(sys.stdout, termios.TIOCGWINSZ,'1234'))
text_len = len(readline.get_line_buffer())+2
# ANSI escape sequences (All VT100 except ESC[0G)
sys.stdout.write('\x1b[2K') # Clear current line
sys.stdout.write('\x1b[1A\x1b[2K'*(text_len/cols)) # Move cursor up and clear line
sys.stdout.write('\x1b[0G') # Move to start of line
def noisy_thread():
while True:
time.sleep(3)
blank_current_readline()
print 'Interrupting text!'
sys.stdout.write('> ' + readline.get_line_buffer())
sys.stdout.flush() # Needed or text doesn't show until a key is pressed
if __name__ == '__main__':
thread.start_new_thread(noisy_thread, ())
while True:
s = raw_input('> ')
输出。先前的readline行已正确清除：
$ ./threads_input2.py
Interrupting text!
Interrupting text!
Interrupting text!
Interrupting text!
> WELL, PRINCE, Genoa and Lucca are Now no more than private estates of the Bo
naparte family. No, I warn you, that if you do not tell me we are at war,
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