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  • Programmatically specifying Django model attributes

    - by mojbro
    Hi! I would like to add attributes to a Django models programmatically, at run time. For instance, lets say I have a Car model class and want to add one price attribute (database column) per currency, given a list of currencies. What is the best way to do this? I had an approach that I thought would work, but it didn't exactly. This is how I tried doing it, using the car example above: from django.db import models class Car(models.Model): name = models.CharField(max_length=50) currencies = ['EUR', 'USD'] for currency in currencies: Car.add_to_class('price_%s' % currency.lower(), models.IntegerField()) This does seem to work pretty well at first sight: $ ./manage.py syncdb Creating table shop_car $ ./manage.py dbshell shop=# \d shop_car Table "public.shop_car" Column | Type | Modifiers -----------+-----------------------+------------------------------------------------------- id | integer | not null default nextval('shop_car_id_seq'::regclass) name | character varying(50) | not null price_eur | integer | not null price_usd | integer | not null Indexes: "shop_car_pkey" PRIMARY KEY, btree (id) But when I try to create a new Car, it doesn't really work anymore: >>> from shop.models import Car >>> mycar = Car(name='VW Jetta', price_eur=100, price_usd=130) >>> mycar <Car: Car object> >>> mycar.save() Traceback (most recent call last): File "<console>", line 1, in <module> File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/base.py", line 410, in save self.save_base(force_insert=force_insert, force_update=force_update) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/base.py", line 495, in save_base result = manager._insert(values, return_id=update_pk) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/manager.py", line 177, in _insert return insert_query(self.model, values, **kwargs) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/query.py", line 1087, in insert_query return query.execute_sql(return_id) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/sql/subqueries.py", line 320, in execute_sql cursor = super(InsertQuery, self).execute_sql(None) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/sql/query.py", line 2369, in execute_sql cursor.execute(sql, params) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/backends/util.py", line 19, in execute return self.cursor.execute(sql, params) ProgrammingError: column "price_eur" specified more than once LINE 1: ...NTO "shop_car" ("name", "price_eur", "price_usd", "price_eur... ^

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  • Django: CharField with fixed length, how?

    - by Giovanni Di Milia
    Hi everybody, I wold like to have in my model a CharField with fixed length. In other words I want that only a specified length is valid. I tried to do something like volumenumber = models.CharField('Volume Number', max_length=4, min_length=4) but it gives me an error (it seems that I can use both max_length and min_length at the same time). Is there another quick way? Thanks EDIT: Following the suggestions of some people I will be a bit more specific: My model is this: class Volume(models.Model): vid = models.AutoField(primary_key=True) jid = models.ForeignKey(Journals, db_column='jid', null=True, verbose_name = "Journal") volumenumber = models.CharField('Volume Number') date_publication = models.CharField('Date of Publication', max_length=6, blank=True) class Meta: db_table = u'volume' verbose_name = "Volume" ordering = ['jid', 'volumenumber'] unique_together = ('jid', 'volumenumber') def __unicode__(self): return (str(self.jid) + ' - ' + str(self.volumenumber)) What I want is that the volumenumber must be exactly 4 characters. I.E. if someone insert '4b' django gives an error because it expects a string of 4 characters. So I tried with volumenumber = models.CharField('Volume Number', max_length=4, min_length=4) but it gives me this error: Validating models... Unhandled exception in thread started by <function inner_run at 0x70feb0> Traceback (most recent call last): File "/Library/Python/2.5/site-packages/django/core/management/commands/runserver.py", line 48, in inner_run self.validate(display_num_errors=True) File "/Library/Python/2.5/site-packages/django/core/management/base.py", line 249, in validate num_errors = get_validation_errors(s, app) File "/Library/Python/2.5/site-packages/django/core/management/validation.py", line 28, in get_validation_errors for (app_name, error) in get_app_errors().items(): File "/Library/Python/2.5/site-packages/django/db/models/loading.py", line 131, in get_app_errors self._populate() File "/Library/Python/2.5/site-packages/django/db/models/loading.py", line 58, in _populate self.load_app(app_name, True) File "/Library/Python/2.5/site-packages/django/db/models/loading.py", line 74, in load_app models = import_module('.models', app_name) File "/Library/Python/2.5/site-packages/django/utils/importlib.py", line 35, in import_module __import__(name) File "/Users/Giovanni/src/djangoTestSite/../djangoTestSite/journaldb/models.py", line 120, in <module> class Volume(models.Model): File "/Users/Giovanni/src/djangoTestSite/../djangoTestSite/journaldb/models.py", line 123, in Volume volumenumber = models.CharField('Volume Number', max_length=4, min_length=4) TypeError: __init__() got an unexpected keyword argument 'min_length' That obviously doesn't appear if I use only "max_length" OR "min_length". I read the documentation on the django web site and it seems that I'm right (I cannot use both together) so I'm asking if there is another way to solve the problem. Thanks again

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  • python networkx

    - by krisdigitx
    hi, i am trying to use networkx with python, when i run this program, it get this error, is there anything missing? #!/usr/bin/env python import networkx as nx import matplotlib import matplotlib.pyplot import matplotlib.pyplot as plt G=nx.Graph() G.add_node(1) G.add_nodes_from([2,3,4,5,6,7,8,9,10]) #nx.draw_graphviz(G) #nx_write_dot(G, 'node.png') nx.draw(G) plt.savefig("/var/www/node.png") Traceback (most recent call last): File "graph.py", line 13, in <module> nx.draw(G) File "/usr/lib/pymodules/python2.5/networkx/drawing/nx_pylab.py", line 124, in draw cf=pylab.gcf() File "/usr/lib/pymodules/python2.5/matplotlib/pyplot.py", line 276, in gcf return figure() File "/usr/lib/pymodules/python2.5/matplotlib/pyplot.py", line 254, in figure **kwargs) File "/usr/lib/pymodules/python2.5/matplotlib/backends/backend_tkagg.py", line 90, in new_figure_manager window = Tk.Tk() File "/usr/lib/python2.5/lib-tk/Tkinter.py", line 1650, in __init__ self.tk = _tkinter.create(screenName, baseName, className, interactive, wantobjects, useTk, sync, use) _tkinter.TclError: no display name and no $DISPLAY environment variable

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  • Python: Define Classes in Packages

    - by rfkrocktk
    I'm learning Python and I have been playing around with packages. I wanted to know the best way to define classes in packages. It seems that the only way to define classes in a package is to define them in init.py of that package. Coming from Java, I'd kind of like to define individual files for my classes. Is this a recommended practice? I'd like to have my directory look somewhat like this: recursor/ __init__.py RecursionException.py RecursionResult.py Recursor.py So I could refer to my classes as "recursor.Recursor," "recursor.RecursionException," and "recursor.RecursionResult.py". Is this "do-able" or recommended in Python?

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  • Facebook publish HTTP Error 400 : bad request

    - by Abhishek
    Hey I am trying to publish a score to Facebook through python's urllib2 library. import urllib2,urllib url = "https://graph.facebook.com/USER_ID/scores" data = {} data['score']=SCORE data['access_token']='APP_ACCESS_TOKEN' data_encode = urllib.urlencode(data) request = urllib2.Request(url, data_encode) response = urllib2.urlopen(request) responseAsString = response.read() I am getting this error: response = urllib2.urlopen(request) File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 124, in urlopen return _opener.open(url, data, timeout) File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 389, in open response = meth(req, response) File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 502, in http_response 'http', request, response, code, msg, hdrs) File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 427, in error return self._call_chain(*args) File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 361, in _call_chain result = func(*args) File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 510, in http_error_default raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) urllib2.HTTPError: HTTP Error 400: Bad Request Not sure if this is relating to Facebook's Open Graph or improper urllib2 API use.

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  • How to get HTTP status message in (py)curl?

    - by mykhal
    spending some time studying pycurl and libcurl documentation, i still can't find a (simple) way, how to get HTTP status message (reason-phrase) in pycurl. status code is easy: import pycurl import cStringIO curl = pycurl.Curl() buff = cStringIO.StringIO() curl.setopt(pycurl.URL, 'http://example.org') curl.setopt(pycurl.WRITEFUNCTION, buff.write) curl.perform() print "status code: %s" % curl.getinfo(pycurl.HTTP_CODE) # -> 200 # print "status message: %s" % ??? # -> "OK"

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  • Which revision of html5lib is stable?

    - by Mat
    html5lib notes that it's latest release (0.11) is somewhat old. Using the Python portion, I have recursion problems as noted in Issue 70 and Issue 59 but can't find a recent Mercurial revision that is stable. The latest tip is no good, I got the following error from python setup.py install: byte-compiling build/bdist.linux-x86_64/egg/html5lib/treewalkers/_base.py to _base.pyc File "build/bdist.linux-x86_64/egg/html5lib/treewalkers/_base.py", line 40 "data": []} ^ SyntaxError: invalid syntax And I get the following errors at runtime: soup = parser.parse(page.read()) File "build/bdist.linux-x86_64/egg/html5lib/html5parser.py", line 165, in parse File "build/bdist.linux-x86_64/egg/html5lib/html5parser.py", line 144, in _parse File "build/bdist.linux-x86_64/egg/html5lib/html5parser.py", line 454, in processDoctype TypeError: insertDoctype() takes exactly 4 arguments (2 given) I'm using it on Python 2.5.2 with lxml and BeautifulSoup.

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  • Install TurboGears on windows xp

    - by coder
    I've been trying to get TurboGears installed on Windows by following this site. I've installed virtualenv but when I execute the command "virtualenv --no-site-packages testproj", I get the following message: New python executable in testproj\Scripts\python.exe Traceback (most recent call last): File "C:\Python26\Scripts\virtualenv-script.py", line 8, in load_entry_point('virtualenv==1.4.5', 'console_scripts', 'virtualenv')() File "C:\Python26\lib\site-packages\virtualenv-1.4.5-py2.6.egg\virtualenv.py", line 529, in main use_distribute=options.use_distribute) File "C:\Python26\lib\site-packages\virtualenv-1.4.5-py2.6.egg\virtualenv.py", line 612, in create_environment site_packages=site_packages, clear=clear)) File "C:\Python26\lib\site-packages\virtualenv-1.4.5-py2.6.egg\virtualenv.py", line 837, in install_python stdout=subprocess.PIPE) File "C:\Python26\lib\subprocess.py", line 621, in __init__ errread, errwrite) File "C:\Python26\lib\subprocess.py", line 830, in _execute_child startupinfo) WindowsError: [Error 14001] This application has failed to start because the application configuration is incorrect. Reinstalling the application may fix this problem Can someone help me debug this ? If any one knows a better tutorial to install turbogears, please let me know.

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  • I keep Getting KeyError: 'tried' Whenever I Tried to Run Django Dev Server from Remote Machine

    - by Spikie
    I am running django on python2.6.1, and did start the django web server like this manage.py runserver 192.0.0.1:8000 then tried to connect to the django dev web server on http://192.0.0.1:8000/ keep getting this message on the remote computer Traceback (most recent call last): File "C:\Python26\Lib\site-packages\django\core\servers\basehttp.py", line 279, in run self.result = application(self.environ, self.start_response) File "C:\Python26\Lib\site-packages\django\core\servers\basehttp.py", line 651, in call return self.application(environ, start_response) File "C:\Python26\lib\site-packages\django\core\handlers\wsgi.py", line 241, in call response = self.get_response(request) File "C:\Python26\lib\site-packages\django\core\handlers\base.py", line 115, in get_response return debug.technical_404_response(request, e) File "C:\Python26\Lib\site-packages\django\views\debug.py", line 247, in technical_404_response tried = exception.args[0]['tried'] KeyError: 'tried' what i am doing wrong ? it seen to work ok if i run http://192.0.0.1:8000/ on the computer that runs the Django web server and have that ip 192.0.0.1:8000

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  • Difference between URLLIB2 call in IDLE and from Django?

    - by danspants
    The following piece of code works as expected when running in a local install of django apache 2.2 fx = urllib2.Request(f); fx.add_header('User-Agent','Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US) AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.36 Safari/525.19'); url_opened = urllib2.urlopen(fx); However when I enter that code into IDLE on the same machine I get the following error: url_opened = urllib2.urlopen(fx); File "C:\Python25\lib\urllib2.py", line 124, in urlopen return _opener.open(url, data) File "C:\Python25\lib\urllib2.py", line 387, in open response = meth(req, response) File "C:\Python25\lib\urllib2.py", line 498, in http_response 'http', request, response, code, msg, hdrs) File "C:\Python25\lib\urllib2.py", line 425, in error return self._call_chain(*args) File "C:\Python25\lib\urllib2.py", line 360, in _call_chain result = func(*args) File "C:\Python25\lib\urllib2.py", line 506, in http_error_default raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) HTTPError: HTTP Error 407: Proxy Authentication Required Any ideas?

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  • In Python, urllib2 giving error

    - by pythBegin
    I tried running this, >>> urllib2.urlopen('http://tycho.usno.navy.mil/cgi-bin/timer.pl') But it is giving error like this, can anyone tell me a solution ? Traceback (most recent call last): File "<pyshell#11>", line 1, in <module> urllib2.urlopen('http://tycho.usno.navy.mil/cgi-bin/timer.pl') File "C:\Python26\lib\urllib2.py", line 126, in urlopen return _opener.open(url, data, timeout) File "C:\Python26\lib\urllib2.py", line 391, in open response = self._open(req, data) File "C:\Python26\lib\urllib2.py", line 409, in _open '_open', req) File "C:\Python26\lib\urllib2.py", line 369, in _call_chain result = func(*args) File "C:\Python26\lib\urllib2.py", line 1161, in http_open return self.do_open(httplib.HTTPConnection, req) File "C:\Python26\lib\urllib2.py", line 1136, in do_open raise URLError(err) URLError: <urlopen error [Errno 11001] getaddrinfo failed>

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  • How to do relative imports in Python?

    - by Joril
    Imagine this directory structure: app/ __init__.py sub1/ __init__.py mod1.py sub2/ __init__.py mod2.py I'm coding mod1, and I need to import something from mod2. How should I do it? I tried from ..sub2 import mod2 but I'm getting an "Attempted relative import in non-package". I googled around but found only "sys.path manipulation" hacks. Isn't there a clean way? Edit: all my __init__.py's are currently empty Edit2: I'm trying to do this because sub2 contains classes that are shared across sub packages (sub1, subX, etc.). Edit3: The behaviour I'm looking for is the same as described in PEP 366 (thanks John B)

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  • how to read a static file in .py file using django ..

    - by zjm1126
    this is my error code: text = open('/media/a.txt', 'rb').read() and my perplexed is: when i use this : text = open('a.txt', 'rb').read() it can be running but when i put the 'a.txt' to the 'media' folder, i can't running , why ? thanks IOError at / [Errno 13] file not accessible: '/media/a.txt'

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  • Import boto from local library

    - by ensnare
    I'm trying to use boto as a downloaded library, rather than installing it globally on my machine. I'm able to import boto, but when I run boto.connect_dynamodb() I get an error: ImportError: No module named dynamodb.layer2 Here's my file structure: project/ project/ __init__.py libraries/ __init__.py flask/ boto/ views/ .... modules/ __init__.py db.py .... templates/ .... static/ .... runserver.py And the contents of the relevant files as follows: project/project/modules/db.py from project.libraries import boto conn = boto.connect_dynamodb( aws_access_key_id='<YOUR_AWS_KEY_ID>', aws_secret_access_key='<YOUR_AWS_SECRET_KEY>') What am I doing wrong? Thanks in advance.

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  • Google App Engine with local Django 1.1 gets Intermittent Failures

    - by Jon Watte
    I'm using the Windows Launcher development environment for Google App Engine. I have downloaded Django 1.1.2 source, and un-tarrred the "django" subdirectory to live within my application directory (a peer of app.yaml) At the top of each .py source file, I do this: import settings import os os.environ["DJANGO_SETTINGS_MODULE"] = 'settings' In my file settings.py (which lives at the root of the app directory, as well), I do this: DEBUG = True TEMPLATE_DIRS = ('html') INSTALLED_APPS = ('filters') import os os.environ["DJANGO_SETTINGS_MODULE"] = 'settings' from google.appengine.dist import use_library use_library('django', '1.1') from django.template import loader Yes, this looks a bit like overkill, doesn't it? I only use django.template. I don't explicitly use any other part of django. However, intermittently I get one of two errors: 1) Django complains that DJANGO_SETTINGS_MODULE is not defined. 2) Django complains that common.html (a template I'm extending in other templates) doesn't exist. 95% of the time, these errors are not encountered, and they randomly just start happening. Once in that state, the local server seems "wedged" and re-booting it generally fixes it. What's causing this to happen, and what can I do about it? How can I even debug it? Here is the traceback from the error: Traceback (most recent call last): File "C:\code\kwbudget\edit_budget.py", line 34, in get self.response.out.write(t.render(template.Context(values))) File "C:\code\kwbudget\django\template\__init__.py", line 165, in render return self.nodelist.render(context) File "C:\code\kwbudget\django\template\__init__.py", line 784, in render bits.append(self.render_node(node, context)) File "C:\code\kwbudget\django\template\__init__.py", line 797, in render_node return node.render(context) File "C:\code\kwbudget\django\template\loader_tags.py", line 71, in render compiled_parent = self.get_parent(context) File "C:\code\kwbudget\django\template\loader_tags.py", line 66, in get_parent raise TemplateSyntaxError, "Template %r cannot be extended, because it doesn't exist" % parent TemplateSyntaxError: Template u'common.html' cannot be extended, because it doesn't exist And edit_budget.py starts with exactly the lines that I included up top. All templates live in a directory named "html" in my root directory, and "html/common.html" exists. I know the template engine finds them, because I start out with "html/edit_budget.html" which extends common.html. It looks as if the settings module somehow isn't applied (because that's what adds html to the search path for templates).

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  • Is Django's manage.py syncdb or South used to create the test database?

    - by Thierry Lam
    With Django 1.1.1 and South 0.62, running a test from the CLI usually have the following output: Creating table some_model Installing index for my_app.SomeModel model . ----- Ran 1 test in 1s OK After upgrading to South 0.7, the output is invoking South's migration: Creating table some_model Installing index for my_app.SomeModel model Migrating... Running migrations for my_app: - Migrating forwards to 0001_initial > my_app:0001_initial - Loading initial data for my_app Migrated: - my_app . ----- Ran 1 test in 1s OK To create the test DB, has the test always used South migration in the past(before South 0.7) even if the output is not explicitly being shown?

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  • Python: Importing a file from a parent folder

    - by Sascha
    ...Now I know this question has been asked many times & I have looked at these other threads. Nothing so far has worked, from using sys.path.append('.') to just import foo I have a python file that wishes to import a file (that is in its parent directory). Can you help me figure out how my child file can successfully import its a file in its parent directory. I am using python 2.7 The structure is like so (each directory also has the __init__.py file in it): StockTracker/ __Comp/ ____a.py ____SubComp/ ______b.py Inside b.py, I would like to import a.py: So I have tried each of the following but I still get an error inside b.py saying "There is no such module a" import a import .a import Comp.a import StockTracker.Comp.a import os import sys sys.path.append('.') import a sys.path.remove('.')

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  • yet another confusion with multiprocessing error, 'module' object has no attribute 'f'

    - by gatoatigrado
    I know this has been answered before, but it seems that executing the script directly "python filename.py" does not work. I have Python 2.6.2 on SuSE Linux. Code: #!/usr/bin/python # -*- coding: utf-8 -*- from multiprocessing import Pool p = Pool(1) def f(x): return x*x p.map(f, [1, 2, 3]) Command line: > python example.py Process PoolWorker-1: Traceback (most recent call last): File "/usr/lib/python2.6/multiprocessing/process.py", line 231, in _bootstrap self.run() File "/usr/lib/python2.6/multiprocessing/process.py", line 88, in run self._target(*self._args, **self._kwargs) File "/usr/lib/python2.6/multiprocessing/pool.py", line 57, in worker task = get() File "/usr/lib/python2.6/multiprocessing/queues.py", line 339, in get return recv() AttributeError: 'module' object has no attribute 'f'

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  • Python - Subprocess Popen and Thread error

    - by n0idea
    In both functions record and ftp, i have subprocess.Popen if __name__ == '__main__': try: t1 = threading.Thread(target = record) t1.daemon = True t1.start() t2 = threading.Thread(target = ftp) t2.daemon = True t2.start() except (KeyboardInterrupt, SystemExit): sys.exit() The error I'm receiving is: Exception in thread Thread-1 (most likely raised during interpreter shutdown): Traceback (most recent call last): File "/usr/lib/python2.7/threading.py", line 551, in __bootstrap_inner File "/usr/lib/python2.7/threading.py", line 504, in run File "./in.py", line 20, in recordaudio File "/usr/lib/python2.7/subprocess.py", line 493, in call File "/usr/lib/python2.7/subprocess.py", line 679, in __init__ File "/usr/lib/python2.7/subprocess.py", line 1237, in _execute_child <type 'exceptions.AttributeError'>: 'NoneType' object has no attribute 'close' What might the issue be ?

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  • Model in sub-directory via app_label?

    - by prometheus
    In order to place my models in sub-folders I tried to use the app_label Meta field as described here. My directory structure looks like this: project apps foo models _init_.py bar_model.py In bar_model.py I define my Model like this: from django.db import models class SomeModel(models.Model): field = models.TextField() class Meta: app_label = "foo" I can successfully import the model like so: from apps.foo.models.bar_model import SomeModel However, running: ./manage.py syncdb does not create the table for the model. In verbose mode I do see, however, that the app "foo" is properly recognized (it's in INSTALLED_APPS in settings.py). Moving the model to models.py under foo does work. Is there some specific convention not documented with app_label or with the whole mechanism that prevents this model structure from being recognized by syncdb?

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  • How do I make a project in Django? Beginner

    - by ggfan
    Okay I just started with Django and it's totally different from PHP. I installed Python 2.6 and Django. Both are located in my C drive. C: Django build django docs Python26 I am doing the django site tutorial and when they say to write django-admin.py startproject mysite from my Python command line, I keep getting: Syntax error: invalid syntax >>>django-admin.py startproject mysite FILE "<stdin>", line 1 django-admin.py startproject mysite ^ My django-admin.py is in the django/bin folder. I installed Python via python setup.py. Am I suppose to use my window's CP? When I do that, I get window's can't open a .py file. I thought I was just creating a folder? How do I create a project with django? Thanks :)

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  • Python: How can I override one module in a package with a modified version that lives outside the pa

    - by zlovelady
    I would like to update one module in a python package with my own version of the module, with the following conditions: I want my updated module to live outside of the original package (either because I don't have access to the package source, or because I want to keep my local modifications in a separate repo, etc). I want import statements that refer to original package/module to resolve to my local module Here's an example of what I'd like to do using specifics from django, because that's where this problem has arisen for me: Say this is my project structure django/ ... the original, unadulterated django package ... local_django/ conf/ settings.py myproject/ __init__.py myapp/ myfile.py And then in myfile.py # These imports should fetch modules from the original django package from django import models from django.core.urlresolvers import reverse # I would like this following import statement to grab a custom version of settings # that I define in local_django/conf/settings.py from django.conf import settings def foo(): return settings.some_setting Can I do some magic with the __import__ statement in myproject/__init__.py to accomplish this? Is there a more "pythonic" way to achieve this?

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