Python-pexpect - TIMEOUT fällt in die traceback-und Ausgänge

Ich bin neu in der python-pexpect. In Tcl/erwarten, wenn ich traf einen timeout - ich würde Antworten, mit der Meldung und beenden Sie die Funktion. Ich habe versucht, zu Experimentieren mit ähnlichen Antwort mit Beispiel-code geschrieben
http://pexpect.svn.sourceforge.net/viewvc/pexpect/trunk/pexpect/examples/sshls.py?revision=489&view=markup

Ich basierend auf diesem code oben - wenn ich ein falsche Passwort, ich würde erwarten, dass dies nur timeout, print "FEHLER!", und Programm beenden. Aber wenn ich es laufen - geht in einen 'Traceback-Ausgabe (siehe unten), kann mir jemand helfen um das Programm zu drucken, "FEHLERMELDUNG" und das Programm beenden würde.

test@ubuntu:~/scripts$ ./tmout.py 
Hostname: 192.168.26.84
User: root
Password: 
Timeout exceeded in read_nonblocking().
<pexpect.spawn object at 0xb77309cc>
version: 2.3 ($Revision: 399 $)
command: /usr/bin/ssh
args: ['/usr/bin/ssh', '-l', 'root', '192.168.26.84', '/bin/ls', '-l']
searcher: searcher_re:
    0: EOF
buffer (last 100 chars): 
Permission denied, please try again.
root@192.168.26.84's password: 
before (last 100 chars): 
Permission denied, please try again.
[email protected]'s password: 
after: <class 'pexpect.TIMEOUT'>
match: None
match_index: None
exitstatus: None
flag_eof: False
pid: 14997
child_fd: 3
closed: False
timeout: 30
delimiter: <class 'pexpect.EOF'>
logfile: None
logfile_read: None
logfile_send: None
maxread: 2000
ignorecase: False
searchwindowsize: None
delaybeforesend: 0.05
delayafterclose: 0.1
delayafterterminate: 0.1
Traceback (most recent call last):
  File "./tmout.py", line 54, in <module>
    traceback.print_exc()
NameError: name 'traceback' is not defined
test@ubuntu:~/scripts$ 

Source Code:

#!/usr/bin/env python

"""This runs 'ls -l' on a remote host using SSH. At the prompts enter hostname,
user, and password.

$Id$
"""

import pexpect
import getpass, os

def ssh_command (user, host, password, command):

    """This runs a command on the remote host. This could also be done with the
pxssh class, but this demonstrates what that class does at a simpler level.
This returns a pexpect.spawn object. This handles the case when you try to
connect to a new host and ssh asks you if you want to accept the public key
fingerprint and continue connecting. """

    ssh_newkey = 'Are you sure you want to continue connecting'
    child = pexpect.spawn('ssh -l %s %s %s'%(user, host, command))
    i = child.expect([pexpect.TIMEOUT, ssh_newkey, 'password: '])
    if i == 0: # Timeout
        print 'ERROR!'
        print 'SSH could not login. Here is what SSH said:'
        print child.before, child.after
        return None
    if i == 1: # SSH does not have the public key. Just accept it.
        child.sendline ('yes')
        child.expect ('password: ')
        i = child.expect([pexpect.TIMEOUT, 'password: '])
        if i == 0: # Timeout
            print 'ERROR!'
            print 'SSH could not login. Here is what SSH said:'
            print child.before, child.after
            return None       
    child.sendline(password)
    return child

def main ():

    host = raw_input('Hostname: ')
    user = raw_input('User: ')
    password = getpass.getpass('Password: ')
    child = ssh_command (user, host, password, '/bin/ls -l')
    child.expect(pexpect.EOF)
    print child.before

if __name__ == '__main__':
    try:
        main()
    except Exception, e:
        print str(e)
        traceback.print_exc()
        os._exit(1)
können Sie wieder Bearbeiten Ihren text? setzen Sie in: >>> print '\n\x20\x20\x20\x20'.join(your_code.split('\n'))

InformationsquelleAutor Tester315 | 2012-05-05

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