I'm trying to 12-factor my django settings by passing environment variables from ansible. My problem is that I don't know how to pass the variables host1 and host2 from an ansible yaml file, so that MYSETTING is a python list, not a string.
I'm trying to get:
MYSETTING = ['host1', 'host2'] # this should be a list, not a string
I have tried various options in ansible, but they both seem to result in MYSETTING being a string:
# settings.py
MYSETTING = os.environ.get(ANSIBLE_VALUE, [])
# ansible vars file
ANSIBLE_VALUE: "['host1', 'host2']"
or
# settings.py
MYSETTING = [os.environ.get(ANSIBLE_VALUE, None)]
# ansible vars file
ANSIBLE_VALUE: "'host1', 'host2'"
but neither option seems to work. I'm sure there is something simple, but I can't figure it out.
Your environment variable is a string type. You can change string into a list with ast library
import ast
MYSETTING = ast.literal_eval(os.environ.get(ANSIBLE_VALUE, '[]'))
# In settings.py
ADMINS = ['ws.kwak', 'jm0926.kim', 'viet.long', 'nam.nv12', 'phuong.nv2']
#In other files like views.py
from django.conf import settings
settings.ADMINS
Related
I have situation like this. I am creating a Python package. That Python package needs to use Redis, so I want to allow the user of the package to define the Redis url.
Here's how I attempted to do it:
bin/main.py
from my_package.main import run
from my_package.config import config
basicConfig(filename='logs.log', level=DEBUG)
# the user defines the redis url
config['redis_url'] = 'redis://localhost:6379/0'
run()
my_package/config.py
config = {
"redis_url": None
}
my_package/main.py
from .config import config
def run():
print(config["redis_url"]) # prints None instead of what I want
Unfortunately, it doesn't work. In main.py the value of config["redis_url"] is None instead of the url defined in bin/main.py file. Why is that? How can I make it work?
I could pass the config to the run() function, but then if I run some other function I will need to pass the config to that function as well. I'd like to pass it one time ideally.
replace "Variables config" with "Variables /path/CLOUD234/__init__.py" in robot framework .Cloud instance is defined at run time .In each run the value changes ,so I have created a python file initpath.py as follows with fun() keyword .It will return the required path .How can I call it in Variables section of robot framework ? Thank you in advance.
import socket
import re
import os
def fun():
name = socket.gethostname()
pattern = ".*CLOUD[0-9]*"
hname = re.findall(pattern,name)
cloud_instance = hname[0].replace("-","_")
init_file = "/path/{}/__init__.py".format(cloud_instance)
return init_file
Variables section do not execute any code.
I suggest you run the python under testcase/suite up and use Set Test Variable to set the variable.
Try following: Here, you are not required to use any variables/section to call py file.
Under *** Settings *** section add
Library initpath.py
Building from Reading config file as dictionary in Flask I'm attempting to define a custom configuration file in order to customize my Flask app.
Running code :
from flask import Flask
app = Flask(__name__)
import os
my_config = app.config.from_pyfile(os.path.join('.', 'config/app.conf'), silent=False)
#app.route('/')
def hello_world():
with app.open_instance_resource('app.cfg') as f:
print(type(f.read()))
return 'Hello World! {}'.format(app.config.get('LOGGING'))
if __name__ == '__main__':
app.run()
In config/app.conf these properties are added :
myvar1="tester"
myvar2="tester"
myvar3="tester"
Invoking the service for hello_world() function returns :
Hello World! None
Shouldn't the contents of the file be returned instead of None ?
I attempt to access a property using app.config.get('myvar1') but None is also returned in this case.
From the official Flask documentation:
The configuration files themselves are actual Python files. Only values in uppercase are actually stored in the config object later on. So make sure to use uppercase letters for your config keys.
Therefore, set structure your ./config/app.conf variables as:
VARIABLE_NAME = "variable value"
Load as:
app.config.from_pyfile(os.path.join(".", "config/app.conf"), silent=False)
NOTE: If you're not using instance specific configuration you don't actually need to build the path as it will take the Flask app path as its root.
And access the values later as:
app.config.get("VARIABLE_NAME")
I have a Pyramid web application managed with zc.buildout. In it, I need to read a file on disk, which is located in a sub-directory of buildout directory.
The problem is with determining the path to the file - I do not want to hard-code the absolute path and just providing a relative path does not work when serving the app in production (supposedly because the working directory is different).
So the promising "hooks" I am thinking about are:
the "root" buildout directory, which I can address in buildout.cfg as ${buildout:directory} - however, I can't figure out how can I "export" it so it can be accessed by the Python code
the location of the Paster's .ini file which starts the app
Like #MartijnPieters suggests in a comment on your own answer, I'd use collective.recipe.template to generate an entry in the .ini. I wondered myself how I could then access that data in my project, so I worked it out :-)
Let's work our way backwards to what you need. First in your view code where you want the buildout directory:
def your_view(request):
buildout_dir = request.registry.settings['buildout_dir']
....
request.registry.settings (see documentation) is a "dictonary-like deployment settings object". See deployment settings, that's the **settings that gets passed into your main method like def main(global_config, **settings)
Those settings are what's in the [app:main] part of your deployment.ini or production.ini file. So add the buildout directory there:
[app:main]
use = egg:your_app
buildout_dir = /home/you/wherever/it/is
pyramid.reload_templates = true
pyramid.debug_authorization = false
...
But, and this is the last step, you don't want to have that hardcoded path in there. So generate the .ini with a template. The template development.ini.in uses a ${partname:variable} expansion language. in your case you need${buildout:directory}:
[app:main]
use = egg:your_app
buildout_dir = ${buildout:dir}
# ^^^^^^^^^^^^^^^
pyramid.reload_templates = true
pyramid.debug_authorization = false
...
Add a buildout part in buildout.cfg to generate development.ini from development.ini.in:
[buildout]
...
parts =
...
inifile
...
[inifile]
recipe = collective.recipe.template
input = ${buildout:directory}/development.ini.in
output = ${buildout:directory}/development.ini
Note that you can do all sorts of cool stuff with collective.recipe.template. ${serverconfig:portnumber} to generate a matching port number in your production.ini and in your your_site_name.nginx.conf, for instance. Have fun!
If the path to the file relative to the buildout root or location of paster.ini is always the same, which it seems it is from your question, you could set it in paster.ini:
[app:main]
...
config_file = %(here)s/path/to/file.txt
Then access it from the registry as in Reinout's answer:
def your_view(request):
config_file = request.registry.settings['config_file']
Here's a rather clumsy solution I've devised:
In buildout.cfg I used extra-paths option of zc.recipe.egg to add the buildout directory to sys.path:
....
[webserver]
recipe = zc.recipe.egg:scripts
eggs = ${buildout:eggs}
extra-paths = ${buildout:directory}
then I put a file called app_config.py into the buildout directory:
# This remembers the root of the installation (similar to {buildout:directory}
# so we can import it and use where we need access to the filesystem.
# Note: we could use os.getcwd() for that but it feels kinda wonky
# This is not directly related to Celery, we may want to move it somewhere
import os.path
INSTALLATION_ROOT = os.path.dirname(__file__)
Now we can import it in our Python code:
from app_config import INSTALLATION_ROOT
filename = os.path.join(INSTALLATION_ROOT, "somefile.ext")
do_stuff_with_file(filename)
If anyone knows a nicer solution you're welcome :)
I have the following structure
sites
- SiteA
- Settings.py
- SiteA
- Settings.py
- SiteA
- Settings.py
all the settings.py file has a variable
DOMAIN_KEY = ''
Is there a way to programmically import this structure and look through this variable in all the setting files and print it?
EDIT (Updated code):
with cd(os.path.join(APPLICATION_DIR, 'config', 'sites')):
output = run ("ls -l | grep \"^d\"|awk {'print $8'}") # run will execute and return list of folders
for file in output.split():
__module = __import__('%s' % file)
print __module
print __module.TIME_ZONE
print __module.SITE_ID
The import is not working properly, am not sure how I can import the setting files without knowing the folder name.
Supposing you do not have duplicate file names you can access these modules and its variables like this:
import SiteA.Settings as A, SiteB.Settings as B, SiteC.Settings as C
print(A.DOMAIN_KEY, B.DOMAIN_KEY, C.DOMAIN_KEY)
EDIT: Another option would be the __import__() statement, that lets you load a module with its name:
for char in "ABC":
mod = __import__("Site%c.Settings")
print(mod.DOMAIN_KEY)
EDIT2: according to the edit in the OP.
Why don't you just append the "Settings" module name after the folder you got from the script?
for file in output.split():
__module = __import__('%s.Settings' % file)
print __module.Settings.DOMAIN_KEY