I've wrote a nice app for myself using the Twisted framework. I launch it using a command like:
twistd -y myapp.py --pidfile=/var/run/myapp.pid --logfile=/var/run/myapp.log
It works great =)
To launch my app I wrote a script with this command because I'm lazy^^
But since I launch my app with the same twistd option, and I tink the script shell solution is ugly, how I can do the same but inside my app? I'd like to launch my app by just doing ./myapp and without a shell work around.
I've tried to search about it in twisted documentation and by reading twisted source but I don't understand it since it's my first app in Python (wonderful language btw!)
Thanks in advance for anyhelp.
You need to import the twistd script as a module from Twisted and invoke it. The simplest solution for this, using your existing command-line, would be to import the sys module to replace the argv command line to look like how you want twistd to run, and then run it.
Here's a simple example script that will take your existing command-line and run it with a Python script instead of a shell script:
#!/usr/bin/python
from twisted.scripts.twistd import run
from sys import argv
argv[1:] = [
'-y', 'myapp.py',
'--pidfile', '/var/run/myapp.pid',
'--logfile', '/var/run/myapp.log'
]
run()
If you want to bundle this up nicely into a package rather than hard-coding paths, you can determine the path to myapp.py by looking at the special __file__ variable set by Python in each module. Adding this to the example looks like so:
#!/usr/bin/python
from twisted.scripts.twistd import run
from my.application import some_module
from os.path import join, dirname
from sys import argv
argv[1:] = [
'-y', join(dirname(some_module.__file__), "myapp.py"),
'--pidfile', '/var/run/myapp.pid',
'--logfile', '/var/run/myapp.log'
]
run()
and you could obviously do similar things to compute appropriate pidfile and logfile paths.
A more comprehensive solution is to write a plugin for twistd. The axiomatic command-line program from the Axiom object-database project serves as a tested, production-worthy example of how to do similar command-line manipulation of twistd to what is described above, but with more comprehensive handling of command-line options, different non-twistd-running utility functionality, and so on.
You can also create the options / config for a twisted command and pass it to the twisted runner.
#!/usr/bin/env python3
import twisted.scripts.twistd
import twisted.scripts._twistd_unix
config = twisted.scripts._twistd_unix.ServerOptions()
config.parseOptions([
"--nodaemon",
"web",
"--listen=tcp:80",
"--path=/some/path"
])
twisted.scripts._twistd_unix.UnixApplicationRunner(config).run()
Related
I want to create terminal commands in my python script.
for example,
$ cd my_project_folder
$ --help (I use a command in this python folder)
outputs
$ this is the help command. You've activated it using --help. This is only possible because you cd'd to the folder and inputted this command.
I am looking for user defined commands ( ones that I've already defined in my python function.)
I am not looking for commands like 'ls' and 'pwd'.
You can use os.system, as below:
import os
os.system('your command')
Use os module if you wanna execute command that is specific to bash or shell that you use
import os
os.system("string command")
As Leemosh sugested you can use click module to make your own commands that are related to scripts
And you can also use sys module to read args that you put affter your script
example
$ python3 name_of_script.py arg_n
import sys
if sys.argv[1] == "commnd":
do something
What you are looking for is creating a setup.py file and defining some entry points. I'd recommend watching https://www.youtube.com/watch?v=GIF3LaRqgXo&ab_channel=CodingTech or https://www.youtube.com/watch?v=4fzAMdLKC5ks&ab_channel=NextDayVideo to better understand how to deal with setup.py.
from setuptools import setup
setup(
name="Name_of_your_app",
version="0.0.1",
desciption="some description",
py_modules=["app"], # I think you don't need that
package_dir={"": "src"}, # and probably you don't need this as well
entry_points={"console_scripts": {"THIS_IS_THE_DEFINED_COMMAND=app:main"}},
)
app is name of the py file, main is function you wanna call.
app.py
def main():
print("hello")
I'm trying to start the django development server from another module in my package. My module can import manage.py, and I want to execute the equivalent of manage.py runserver without using subprocess or anything of that sort (why? see below).
Currently the best solution I could come up with is to use subprocess:
def run_with_default_settings():
import inspect
import subprocess
currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
subprocess.Popen(['python', 'manage.py', 'runserver'], cwd=currentdir)
However this solution seems to me rather overcomplicated, and more importantly it is not platform independent (for example if someone has both python 2 and python 3 and python is defined as python 3; or if python is not defined in the environment PATH... etc.).
I couldn't find any solutions online, and every way I tried to run execute_from_command_line() failed miserably.
Any ideas?
Yes. Just do what's in the manage.py:
import os
from django.core.management import execute_from_command_line
os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'web.settings')
execute_from_command_line(list_of_args)
This should work fine. Just remember that execute_from_command_line accepts originally sys.argv as argument, so the command runserver is on the index 1:
list_of_args = ['', 'runserver']
I have a fabric script called fwp.py that I run without calling it throug fab by using:
if __name__ == '__main__':
# imports for standalone mode only
import sys
import fabric.main
fabric.main.main(fabfile_locations=[__file__])
The thing is then have to call the script by calling fwp.py. I'd like to rename it as fwp to be able to call it as fwp. But doing that would result in
Fatal error: Couldn't find any fabfiles!
Is there a way to make Python/Fabric import this file, despite the lack of a ".py" extension?
To reiterate and clarify:
I'm not using the "fab" utility (e.g. as fab task task:parameter); just calling my script as fwp.py task task:parameter, and would like to be able to call it as fwp task task:parameter.
Update
It's not a duplicate of this question. The question is not "How to run a stand-alone fabric script?", but "How to do so, while having a script without a .py" extension.
EDIT: Original answer corrected
The fabric.main.main() function automatically adds .py to the end of supplied fabfile locations (see https://github.com/fabric/fabric/blob/master/fabric/main.py#L93). Unfortunately that function also uses Python's import machinery to load the file so it has to look like a module or package. Without reimplementing much of the fabric.main module I don't think it will be possible. You could try monkey-patching both fabric.main.find_fabfiles and fabric.main.load_fabfiles to make it work.
Origininal answer (wrong)
I can get this to work unaltered on a freshly installed fabric package. The following will execute with a filename fwp and executable permission on version 1.10.1, Python2.7. I would just try upgrading fabric.
#!/usr/bin/env python
from fabric.api import *
import fabric.main
def do():
local('echo "Hello World"')
if __name__ == '__main__':
fabric.main.main(fabfile_locations=[__file__])
Output:
$ ./fwp do
Hello World
Done
I wish to write a python script that allows me to navigate and git pull multiple repositories. Basically the script should type the following on the command-line:
cd
cd ~/Desktop/Git_Repo
git pull Git_Repo
I am not sure if there is a python library already out there that can perform such a task.
Use subprocess, os, and shlex. This should work, although you might require some minor tweaking:
import subprocess
import shlex
import os
# relative dir seems to work for me, no /'s or ~'s in front though
dir = 'Desktop/Git_Repo'
# I did get fetch (but not pull) to work
cmd = shlex.split('git pull Git_Repo')
# you need to give it a path to find git, this lets you do that.
env = os.environ
subprocess.Popen(cmd, cwd=dir, env=env)
Also, you'll need your login preconfigured.
I am trying to execute a program from a directory
import os
os.chdir("/home/user/a/b")
with cd("/home/user/a/b"):
run ("./program")
i get cd is not defined...
any help appreciated cheers
I'm not sure what instructions you're following to get what you showed. There is no built-in function called cd or run in Python.
You can call a program in a specific directory using the subprocess module:
import subprocess
subprocess.call("./program", cwd="/home/user/a/b")
The cwd argument causes the call function to automatically switch to that directory before starting the program named in the first argument.
It looks like you are trying to use functionalities of fabric. Make sure fabric is installed, and cd and run are imported from fabric. Something like,
from fabric.context_managers import cd
from fabric.operations import run
import os
os.chdir("/home/user/a/b")
with cd("/home/user/a/b"):
run ("./program")
Save your file as fabfile.py,and from the same directory run it as:
fab -H localhost
For more information on the fabric, checkout: fabric