I have a function called get_full_class_name(instance), which returns the full module-qualified class name of instance.
Example my_utils.py:
def get_full_class_name(instance):
return '.'.join([instance.__class__.__module__,
instance.__class__.__name__])
Unfortunately, this function fails when given a class that's defined in a currently running script.
Example my_module.py:
#! /usr/bin/env python
from my_utils import get_full_class_name
class MyClass(object):
pass
def main():
print get_full_class_name(MyClass())
if __name__ == '__main__':
main()
When I run the above script, instead of printing my_module.MyClass, it prints __main__.MyClass:
$ ./my_module.py
__main__.MyClass
I do get the desired behavior if I run the above main() from another script.
Example run_my_module.py:
#! /usr/bin/env python
from my_module import main
if __name__ == '__main__':
main()
Running the above script gets:
$ ./run_my_module.py
my_module.MyClass
Is there a way I could write the get_full_class_name() function such that it always returns my_module.MyClass regardless of whether my_module is being run as a script?
I propose handling the case __name__ == '__main__' using the techniques discussed in Find Path to File Being Run. This results in this new my_utils:
import sys
import os.path
def get_full_class_name(instance):
if instance.__class__.__module__ == '__main__':
return '.'.join([os.path.basename(sys.argv[0]),
instance.__class__.__name__])
else:
return '.'.join([instance.__class__.__module__,
instance.__class__.__name__])
This does not handle interactive sessions and other special cases (like reading from stdin). For this you may have to include techniques like discussed in detect python running interactively.
Following mkiever's answer, I ended up changing get_full_class_name() to what you see below.
If instance.__class__.__module__ is __main__, it doesn't use that as the module path. Instead, it uses the relative path from sys.argv[0] to the closest directory in sys.path.
One problem is that sys.path always includes the directory of sys.argv[0] itself, so this relative path ends up being just the filename part of sys.argv[0]. As a quick hack-around, the code below assumes that the sys.argv[0] directory is always the first element of sys.path, and disregards it. This seems unsafe, but safer options are too tedious for my personal code for now.
Any better solutions/suggestions would be greatly appreciated.
import os
import sys
from nose.tools import assert_equal, assert_not_equal
def get_full_class_name(instance):
'''
Returns the fully-qualified class name.
Handles the case where a class is declared in the currently-running script
(where instance.__class__.__module__ would be set to '__main__').
'''
def get_module_name(instance):
def get_path_relative_to_python_path(path):
path = os.path.abspath(path)
python_paths = [os.path.abspath(p) for p in sys.path]
assert_equal(python_paths[0],
os.path.split(os.path.abspath(sys.argv[0]))[0])
python_paths = python_paths[1:]
min_relpath_length = len(path)
result = None
for python_path in python_paths:
relpath = os.path.relpath(path, python_path)
if len(relpath) < min_relpath_length:
min_relpath_length = len(relpath)
result = os.path.join(os.path.split(python_path)[-1],
relpath)
if result is None:
raise ValueError("Path {} doesn't seem to be in the "
"PYTHONPATH.".format(path))
else:
return result
if instance.__class__.__module__ == '__main__':
script_path = os.path.abspath(sys.argv[0])
relative_path = get_path_relative_to_python_path(script_path)
relative_path = relative_path.split(os.sep)
assert_not_equal(relative_path[0], '')
assert_equal(os.path.splitext(relative_path[-1])[1], '.py')
return '.'.join(relative_path[1:-1])
else:
return instance.__class__.__module__
module_name = get_module_name(instance)
return '.'.join([module_name, instance.__class__.__name__])
Related
I am trying to write a logging module that I can use over and over again. The logging is not the problem. I want to call my logging module from any script and not have to pass the params for the logging file.
If I have a test script called mytest.py how would I return this name in the import logging module. I have tried this in the logging script itself but it returns the name of that file and not the file I am trying to log.
my_test.py:
from my_logger import logging
info.logging("something here")
print("something here")
if __name__ == "__main__":
logging()
I would expect the log file to be called my_test.log, but currently it is being named logging.log
Here is the part from the logging script:
def logging(filename=False, level=DEFAULT_LOG_LEVEL):
if filename is False:
file_ext = os.path.basename(__file__) # Need this to be my_test.py
filename = ("C:/Users/Logs/{0}").format(file_ext)
"Start logging with given filename and level."
#print(filename)
logging.basicConfig(filename=filename, level=LEVELS[level])
else:
"Start logging with given filename and level."
logging.basicConfig(filename=filename, level=LEVELS[level])
Assuming your compiler is CPython (thanks to Matthew Trevor);
You can use inspect.getouterframes to get the caller's frame, plus the filename and line number etc.
import inspect
def logging(filename=False, level=DEFAULT_LOG_LEVEL):
if filename is False:
caller_file = inspect.getouterframes(inspect.currentframe())[1][3]
# prints /home/foo/project/my_test.py
...
The __file__ binding you use in your logging function will hold the value of the file it's scoped in. You need to pass in the calling module's __file__ to get the behaviour you want:
# my_test.py
...
if __name__ == "__main__":
logging(__file__)
I'm surprised that your log file was called logging.log and not my_logger.log, though, as that's the name of the file in which the logging function is defined.
In python 2.7, I want to run:
$ ./script.py initparms.py
This is a trick to supply a parameter file to script.py, since initparms.py contains several python variables e.g.
Ldir = '/home/marzipan/jelly'
LMaps = True
# etc.
script.py contains:
X = __import__(sys.argv[1])
Ldir = X.Ldir
LMaps = X.Lmaps
# etc.
I want to do a bulk promotion of the variables in X so they are available to script.py, without spelling out each one in the code by hand.
Things like
import __import__(sys.argv[1])
or
from sys.argv[1] import *
don't work. Almost there perhaps... Any ideas? Thanks!
here's a one-liner:
globals().update(__import__(sys.argv[1]).__dict__)
You can use execfile:
execfile(sys.argv[1])
Of course, the usual warnings with exec or eval apply (Your script has no way of knowing whether it is running trusted or untrusted code).
My suggestion would be to not do what you're doing and instead use configparser and handling the configuration though there.
You could do something like this:
import os
import imp
import sys
try:
module_name = sys.argv[1]
module_info = imp.find_module(module_name, [os.path.abspath(os.path.dirname(__file__))] + sys.path)
module_properties = imp.load_module(module_name, *module_info)
except ImportError:
pass
else:
try:
attrlist = module_properties.__all__
except AttributeError:
attrlist = dir(module_properties)
for attr in attrlist:
if attr.startswith('__'):
continue
globals()[attr] = getattr(module_properties, attr)
Little complicated, but gets the job done.
I started using Python few days back and I think I have a very basic question where I am stuck. Maybe I am not doing it correctly in Python so wanted some advice from the experts:
I have a config.cfg & a class test in one package lib as follows:
myProj/lib/pkg1/config.cfg
[api_config]
url = https://someapi.com/v1/
username=sumitk
myProj/lib/pkg1/test.py
class test(object):
def __init__(self, **kwargs):
config = ConfigParser.ConfigParser()
config.read('config.cfg')
print config.get('api_config', 'username')
#just printing here but will be using this as a class variable
def some other foos()..
Now I want to create an object of test in some other module in a different package
myProj/example/useTest.py
from lib.pkg1.test import test
def temp(a, b, c):
var = test()
def main():
temp("","","")
if __name__ == '__main__':
main()
Running useTest.py is giving me error:
...
print config.get('api_config', 'username')
File "C:\Python27\lib\ConfigParser.py", line 607, in get
raise NoSectionError(section)
ConfigParser.NoSectionError: No section: 'api_config'
Now if I place thie useTest.py in the same package it runs perfectly fine:
myProj/lib/pkg1/useTest.py
myProj/lib/pkg1/test.py
myProj/lib/pkg1/config.cfg
I guess there is some very basic package access concept in Python that I am not aware of or is there something I am doing wrong here?
The issue here is that you have a different working directory depending on which module is your main script. You can check the working directory by adding the following lines to the top of each script:
import os
print os.getcwd()
Because you just provide 'config.cfg' as your file name, it will attempt to find that file inside of the working directory.
To fix this, give an absolute path to your config file.
You should be able to figure out the absolute path with the following method since you know that config.cfg and test.py are in the same directory:
# inside of test.py
import os
config_path = os.path.join(os.path.dirname(os.path.abspath(__file__)),
'config.cfg')
If I want the path of the current module, I'll use __file__.
Now let's say I want a function to return that. I can't do:
def get_path():
return __file__
Because it will return the path of the module the function has been declared in.
I need it to work even if the function is not called at the root of the module but at any level of nesting.
This is how I would do it:
import sys
def get_path():
namespace = sys._getframe(1).f_globals # caller's globals
return namespace.get('__file__')
Get it from the globals dict in that case:
def get_path():
return globals()['__file__']
Edit in response to the comment: given the following files:
# a.py
def get_path():
return 'Path from a.py: ' + globals()['__file__']
# b.py
import a
def get_path():
return 'Path from b.py: ' + globals()['__file__']
print get_path()
print a.get_path()
Running this will give me the following output:
C:\workspace>python b.py
Path from b.py: b.py
Path from a.py: C:\workspace\a.py
Next to the absolute/relative paths being different (for brevity, lets leave that out), it looks good to me.
I found a way to do it with the inspect module. I'm ok with this solution, but if somebody find a way to do it without dumping the whole stacktrace, it would be cleaner and I would accept his answer gratefully:
def get_path():
frame, filename, line_number, function_name, lines, index =\
inspect.getouterframes(inspect.currentframe())[1]
return filename
File "G:\Python25\Lib\site-packages\PyAMF-0.6b2-py2.5-win32.egg\pyamf\util\__init__.py", line 15, in <module>
ImportError: cannot import name python
How do I fix it?
If you need any info to know how to fix this problem, I can explain, just ask.
Thanks
Code:
from google.appengine.ext.webapp.util import run_wsgi_app
from google.appengine.ext import webapp
from TottysGateway import TottysGateway
import logging
def main():
services_root = 'services'
#services = ['users.login']
#gateway = TottysGateway(services, services_root, logger=logging, debug=True)
#app = webapp.WSGIApplication([('/', gateway)], debug=True)
#run_wsgi_app(app)
if __name__ == "__main__":
main()
Code:
from pyamf.remoting.gateway.google import WebAppGateway
import logging
class TottysGateway(WebAppGateway):
def __init__(self, services_available, root_path, not_found_service, logger, debug):
# override the contructor and then call the super
self.services_available = services_available
self.root_path = root_path
self.not_found_service = not_found_service
WebAppGateway.__init__(self, {}, logger=logging, debug=True)
def getServiceRequest(self, request, target):
# override the original getServiceRequest method
try:
# try looking for the service in the services list
return WebAppGateway.getServiceRequest(self, request, target)
except:
pass
try:
# don't know what it does but is an error for now
service_func = self.router(target)
except:
if(target in self.services_available):
# only if is an available service import it's module
# so it doesn't access services that should be hidden
try:
module_path = self.root_path + '.' + target
paths = target.rsplit('.')
func_name = paths[len(paths) - 1]
import_as = '_'.join(paths) + '_' + func_name
import_string = "from "+module_path+" import "+func_name+' as service_func'
exec import_string
except:
service_func = False
if(not service_func):
# if is not found load the default not found service
module_path = self.rootPath + '.' + self.not_found_service
import_string = "from "+module_path+" import "+func_name+' as service_func'
# add the service loaded above
assign_string = "self.addService(service_func, target)"
exec assign_string
return WebAppGateway.getServiceRequest(self, request, target)
You need to post your full traceback. What you show here isn't all that useful. I ended up digging up line 15 of pyamf/util/init.py. The code you should have posted is
from pyamf import python
This should not fail unless your local environment is messed up.
Can you 'import pyamf.util' and 'import pyamf.python' in a interactive Python shell? What about if you start Python while in /tmp (on the assumption that you might have a file named 'pyamf.py' in the current directory. Which is a bad thing.)
= (older comment below) =
Fix your question. I can't even tell where line 15 of util/__init__.py is supposed to be. Since I can't figure that out, I can't answer your question. Instead, I'll point out ways to improve your question and code.
First, use the markup language correctly, so that all the code is in a code block. Make sure you've titled the code, so we know it's from util/__init__.py and not some random file.
In your error message, include the full traceback, and not the last two lines.
Stop using parens in things like "if(not service_func):" and use a space instead, so its " if not service_func:". This is discussed in PEP 8.
Read the Python documentation and learn how to use the language. Something like "func_name = paths[len(paths) - 1]" should be "func_name = paths[-1]"
Learn about the import function and don't use "exec" for this case. Nor do you need the "exec assign_string" -- just do the "self.addService(service_func, target)"