pexpect output not showing - python

Our simple pexpect script has this:
import pexpect
import sys
test = pexpect.spawn('ftp www.today.com')
test.logfile = sys.stdout
test.expect('Name.*')
However, on the shell the script was invoked, there's no output shown.
Instead it seems to hang but we could see the process ftp ... is spawned.
How to have the output shown on the shell the script is invoked ?
thanks

Should this line:
test = pexpect.spawn('ftp www.today.com')
not be:
test = pexpect.spawn('ftp ftp.today.com')
because normally if you want ftp, you'll have to use ftp.something.com.

test.logfile will only contain the output of the command, the command line itself is not logged in the logfile attribute.
So as long as the command is spawned and that there is no output, nothing will be displayed in the shell when invoking your script.
There will be a display when for example the ftp connection timout has been reached.

You might need to use logfile_read. Here is the code:
import pexpect
import sys
test = pexpect.spawn('ftp www.today.com')
test.logfile_read = sys.stdout
test.expect('Name.*')

Related

Calling a shell script from python

I have a python script that calls a shell scrips, that in turn calls a .exe called iv4_console. I need to print the stdout of iv4_console for debugging purposes. I used this:
Python:
import sys
import subprocess
var="rW015005000000"
proc = subprocess.Popen(["c.sh", var], shell=True, stdout=subprocess.PIPE)
output = ''
for line in iter(proc.stdout.readline, ""):
print line
output += line
Shell:
start_dir=$PWD
release=$1
echo Release inside shell: $release
echo Directory: $start_dir
cd $start_dir
cd ../../iv_system4/ports/visualC12/Debug
echo Debug dir: $PWD
./iv4_console.exe ../embedded/LUA/analysis/verbose-udp-toxml.lua ../../../../../logs/$release/VASP_DUN722_20160307_Krk_Krk_113048_092_1_$release.dvl &>../../../../FCW/ObjectDetectionTest/VASP_DUN722_20160307_Krk_Krk_113048_092_1_$release.xml
./iv4_console.exe ../embedded/LUA/analysis/verbose-udp-toxml.lua ../../../../../logs/$release/VASP_FL140_20170104_C60_Checkout_afterIC_162557_001_$release.dvl &>../../../../FCW/ObjectDetectionTest/VASP_FL140_20170104_C60_Checkout_afterIC_162557_001_$release.xml
exit
But this didn't work, it prints nothing. What do you think?
See my comment, best approach (i.m.o) would be to just use python only.
However, in answer of your question, try:
import sys
import subprocess
var="rW015005000000"
proc = subprocess.Popen(["/bin/bash", "/full/path/to/c.sh"], stdout=subprocess.PIPE)
# Best to always avoid shell=True because of security vulnerabilities.
proc.wait() # To make sure the shell script does not continue running indefinitely in the background
output, errors = proc.communicate()
print(output.decode())
# Since subprocess.communicate() returns a bytes-string, you can use .decode() to print the actual output as a string.
You can use
import subprocess
subprocess.call(['./c.sh'])
to call the shell script in python file
or
import subprocess
import shlex
subprocess.call(shlex.split('./c.sh var'))

running a python script on server

i have a python script on the server
#!/usr/bin/env python
import cgi
import cgitb; #cgitb.enable()
import sys, os
from subprocess import call
import time
import subprocess
form = cgi.FieldStorage()
component = form.getvalue('component')
command = form.getvalue('command')
success = True
print """Content-Type: text/html\n"""
if component=="Engine" and command=="Start":
try:
process = subprocess.Popen(['/usr/sbin/telepath','engine','start'], shell=False, stdout=subprocess.PIPE)
print "{ans:12}"
except Exception, e:
success = False
print "{ans:0}"
When I run this script and add the component and command parameters to be "Engine" and "Start" respectively - it starts the process and prints to the shell
"""Content-Type: text/html\n"""
{ans:12}
but most importantly - it starts the process!
however, when I run the script by POSTing to it, it returns {ans:12} but does not run the process which was the whole intention in the first place. Any logical explanation?
I suspect it's one of two things, firstly your process is probably running but your python code doesn't handle the output so do:
process = subprocess.Popen(['/usr/sbin/telepath','engine','start'], shell=False, stdout=subprocess.PIPE)
print process.stdout.read()
This is the most likely and explains why you see the output from the command line and not the browser, or secondly because the script is run through the browsers as the user apache and not with your userid check the permission for /usr/sbin/telepath.

How to execute a UNIX command in Python script

#!/usr/bin/python
import os
import shutil
import commands
import time
import copy
name = 'test'
echo name
I have a simple python scripts like the above. When I attempt to execute it I get a syntax error when trying to output the name variable.
You cannot use UNIX commands in your Python script as if they were Python code, echo name is causing a syntax error because echo is not a built-in statement or function in Python. Instead, use print name.
To run UNIX commands you will need to create a subprocess that runs the command. The simplest way to do this is using os.system(), but the subprocess module is preferable.
you can also use subprocess module.
import subprocess
proc = subprocess.Popen(['echo', name],
stdin = subprocess.PIPE,
stdout = subprocess.PIPE,
stderr = subprocess.PIPE
)
(out, err) = proc.communicate()
print out
Read: http://www.doughellmann.com/PyMOTW/subprocess/

How can I tell whether screen is running?

I am trying to run a Python program to see if the screen program is running. If it is, then the program should not run the rest of the code. This is what I have and it's not working:
#!/usr/bin/python
import os
var1 = os.system ('screen -r > /root/screenlog/screen.log')
fd = open("/root/screenlog/screen.log")
content = fd.readline()
while content:
if content == "There is no screen to be resumed.":
os.system ('/etc/init.d/tunnel.sh')
print "The tunnel is now active."
else:
print "The tunnel is running."
fd.close()
I know there are probably several things here that don't need to be and quite a few that I'm missing. I will be running this program in cron.
from subprocess import Popen, PIPE
def screen_is_running():
out = Popen("screen -list",shell=True,stdout=PIPE).communicate()[0]
return not out.startswith("This room is empty")
Maybe the error message that you redirect on the first os.system call is written on the standard error instead of the standard output. You should try replacing this line with:
var1 = os.system ('screen -r 2> /root/screenlog/screen.log')
Note the 2> to redirect standard error to your file.

How to redirect stderr in Python?

I would like to log all the output of a Python script. I tried:
import sys
log = []
class writer(object):
def write(self, data):
log.append(data)
sys.stdout = writer()
sys.stderr = writer()
Now, if I "print 'something' " it gets logged. But if I make for instance some syntax error, say "print 'something# ", it wont get logged - it will go into the console instead.
How do I capture also the errors from Python interpreter?
I saw a possible solution here:
http://www.velocityreviews.com/forums/showpost.php?p=1868822&postcount=3
but the second example logs into /dev/null - this is not what I want. I would like to log it into a list like my example above or StringIO or such...
Also, preferably I don't want to create a subprocess (and read its stdout and stderr in separate thread).
I have a piece of software I wrote for work that captures stderr to a file like so:
import sys
sys.stderr = open('C:\\err.txt', 'w')
so it's definitely possible.
I believe your problem is that you are creating two instances of writer.
Maybe something more like:
import sys
class writer(object):
log = []
def write(self, data):
self.log.append(data)
logger = writer()
sys.stdout = logger
sys.stderr = logger
You can't do anything in Python code that can capture errors during the compilation of that same code. How could it? If the compiler can't finish compiling the code, it won't run the code, so your redirection hasn't even taken effect yet.
That's where your (undesired) subprocess comes in. You can write Python code that redirects the stdout, then invokes the Python interpreter to compile some other piece of code.
I can't think of an easy way. The python process's standard error is living on a lower level than a python file object (C vs. python).
You could wrap the python script in a second python script and use subprocess.Popen. It's also possible you could pull some magic like this in a single script:
import os
import subprocess
import sys
cat = subprocess.Popen("/bin/cat", stdin=subprocess.PIPE, stdout=subprocess.PIPE)
os.close(sys.stderr.fileno())
os.dup2(cat.stdin.fileno(), sys.stderr.fileno())
And then use select.poll() to check cat.stdout regularly to find output.
Yes, that seems to work.
The problem I foresee is that most of the time, something printed to stderr by python indicates it's about to exit. The more usual way to handle this would be via exceptions.
---------Edit
Somehow I missed the os.pipe() function.
import os, sys
r, w = os.pipe()
os.close(sys.stderr.fileno())
os.dup2(w, sys.stderr.fileno())
Then read from r
To route the output and errors from Windows, you can use the following code outside of your Python file:
python a.py 1> a.out 2>&1
Source: https://support.microsoft.com/en-us/help/110930/redirecting-error-messages-from-command-prompt-stderr-stdout
Since python 3.5 you can use contextlib.redirect_stderr
with open('help.txt', 'w') as f:
with redirect_stdout(f):
help(pow)
For such a request, usually it would be much easier to do it in the OS instead of in Python.
For example, if you're going to run "a.py" and record all the messages it will generate into file "a.out", it would just be
python a.py 2>&1 > a.out
The first part 2>&1 redirects stderr to stdout (0: stdin, 1:stdout, 2:stderr), and the second redirects that to a file called a.out.
And as far as I know, this command works in Windows, Linux or MacOS! For other file redirection techniques, just search the os plus "file redirection"
I found this approach to redirecting stderr particularly helpful. Essentially, it is necessary to understand if your output is stdout or stderr. The difference? Stdout is any output posted by a shell command (think an 'ls' list) while sterr is any error output.
It may be that you want to take a shell commands output and redirect it to a log file only if it is normal output. Using ls as an example here, with an all files flag:
# Imports
import sys
import subprocess
# Open file
log = open("output.txt", "w+")
# Declare command
cmd = 'ls -a'
# Run shell command piping to stdout
result = subprocess.run(cmd, stdout=subprocess.PIPE, shell=True)
# Assuming utf-8 encoding
txt = result.stdout.decode('utf-8')
# Write and close file
log.write(txt)
log.close()
If you wanted to make this an error log, you could do the same with stderr. It's exactly the same code as stdout with stderr in its place. This pipes an error messages that get sent to the console to the log. Doing so actually keeps it from flooding your terminal window as well!
Saw this was a post from a while ago, but figured this could save someone some time :)
import sys
import tkinter
# ********************************************
def mklistenconsswitch(*printf: callable) -> callable:
def wrapper(*fcs: callable) -> callable:
def newf(data):
[prf(data) for prf in fcs]
return newf
stdoutw, stderrw = sys.stdout.write, sys.stderr.write
funcs = [(wrapper(sys.stdout.write, *printf), wrapper(sys.stderr.write, *printf)), (stdoutw, stderrw)]
def switch():
sys.stdout.write, sys.stderr.write = dummy = funcs[0]
funcs[0] = funcs[1]
funcs[1] = dummy
return switch
# ********************************************
def datasupplier():
i = 5.5
while i > 0:
yield i
i -= .5
def testloop():
print(supplier.__next__())
svvitch()
root.after(500, testloop)
root = tkinter.Tk()
cons = tkinter.Text(root)
cons.pack(fill='both', expand=True)
supplier = datasupplier()
svvitch = mklistenconsswitch(lambda text: cons.insert('end', text))
testloop()
root.mainloop()
Python will not execute your code if there is an error. But you can import your script in another script an catch exceptions. Example:
Script.py
print 'something#
FinalScript.py
from importlib.machinery import SourceFileLoader
try:
SourceFileLoader("main", "<SCRIPT PATH>").load_module()
except Exception as e:
# Handle the exception here
To add to Ned's answer, it is difficult to capture the errors on the fly during the compilation.
You can write several print statements in your script and you can stdout to a file, it will stop writing to the file when the error occurs. To debug the code you could check the last logged output and check your script after that point.
Something like this:
# Add to the beginning of the script execution(eg: if __name__ == "__main__":).
from datetime import datetime
dt = datetime.now()
script_dir = os.path.dirname(os.path.abspath(__file__)) # gets the path of the script
stdout_file = script_dir+r'\logs\log'+('').join(str(dt.date()).split("-"))+r'.log'
sys.stdout = open(stdout_file, 'w')
This will create a log file and stream the print statements to the file.
Note: Watch out for escape characters in your filepath while concatenating with script_dir in the second line from the last in the code. You might want something similar to raw string. You can check this thread for this.

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