Search Results

Search found 5007 results on 201 pages for 'django inheritance'.

Page 98/201 | < Previous Page | 94 95 96 97 98 99 100 101 102 103 104 105  | Next Page >

  • Correct CSS inheritance behavior for properties that aren't inherited?

    - by Chris
    So say we you have a CSS property that is not inherited by default. We'll call it "foo" and its default value is "black". Then we make the following html. <div id="div1" style="foo: red;"> <div id="div2"> <div id="div3" style="foo: inherit;"> </div> </div> </div> Since this property does not inherit by default, you'd think that in div2, "foo" must be "black" - the default value because it does not inherit by default. But ... in div3 should the value for "foo" inherit "black" from its parent that did not inherit foo, or should it inherit "red" from its grandparent because its parent did not specify foo? I need to know because I'm trying to implement something exactly to the spec.

    Read the article

  • Is it possible to use inheritance in this situation? (Java)

    - by they changed my name
    I have ClassA and ClassB, with ClassA being the superclass. ClassA uses NodeA, ClassB uses NodeB. First problem: method parameters. ClassB needs NodeB types, but I can't cast from the subclass to the superclass. That means I can't set properties which are unique to NodeB's. Second problem: When I need to add nodes toClassB, I have to instantiate a new NodeB. But, I can't do this in the superclass, so I'd have to rewrite the insertion to use NodeB. Is there a way around it or am I gonna have to rewrite the whole thing?

    Read the article

  • Objective-C inheritance; calling overriden method from superclass?

    - by anshuchimala
    Hello, I have an Objective-C class that has a method that is meant to be overridden, which is uses in a different method. Something like this: @interface BaseClass - (id)overrideMe; - (void)doAwesomeThings; @end @implementation BaseClass - (id)overrideMe { [self doesNotRecognizeSelector:_cmd]; return nil; } - (void)doAwesomeThings { id stuff = [self overrideMe]; /* do stuff */ } @end @interface SubClass : BaseClass @end @implementation SubClass - (id)overrideMe { /* Actually do things */ return <something>; } @end However, when I create a SubClass and try to use it, it still calls overrideMe on the BaseClass and crashes due to doesNotRecognizeSelector:. (I'm not doing a [super overrideMe] or anything stupid like that). Is there a way to get BaseClass to call the overridden overrideMe?

    Read the article

  • Go for Zend framework or Django for a modular web application?

    - by dr. squid
    I am using both Zend framework and Django, and they both have they strengths and weakness, but they are both good framworks in their own way. I do want to create a highly modular web application, like this example: modules: Admin cms articles sections ... ... ... I also want all modules to be self contained with all confid and template files. I have been looking into a way to solve this is zend the last days, but adding one omer level to the module setup doesn't feel right. I am sure this could be done, but should I? I have also included Doctrine to my zend application that could give me even more problems in my module setup! When we are talking about Django this is easy to implement (Easy as in concept, not in implementation time or whatever) and a great way to create web apps. But one of the downsides of Django is the web hosing part. There are some web hosts offering Django support, but not that many.. So then I guess the question is what have the most value; rapid modular development versus hosting options! Well, comments are welcome! Thanks

    Read the article

  • Django: How can I identify the calling view from a template?

    - by bryan
    Short version: Is there a simple, built-in way to identify the calling view in a Django template, without passing extra context variables? Long (original) version: One of my Django apps has several different views, each with its own named URL pattern, that all render the same template. There's a very small amount of template code that needs to change depending on the called view, too small to be worth the overhead of setting up separate templates for each view, so ideally I need to find a way to identify the calling view in the template. I've tried setting up the views to pass in extra context variables (e.g. "view_name") to identify the calling view, and I've also tried using {% ifequal request.path "/some/path/" %} comparisons, but neither of these solutions seems particularly elegant. Is there a better way to identify the calling view from the template? Is there a way to access to the view's name, or the name of the URL pattern? Update 1: Regarding the comment that this is simply a case of me misunderstanding MVC, I understand MVC, but Django's not really an MVC framework. I believe the way my app is set up is consistent with Django's take on MVC: the views describe which data is presented, and the templates describe how the data is presented. It just happens that I have a number of views that prepare different data, but that all use the same template because the data is presented the same way for all the views. I'm just looking for a simple way to identify the calling view from the template, if this exists. Update 2: Thanks for all the answers. I think the question is being overthought -- as mentioned in my original question, I've already considered and tried all of the suggested solutions -- so I've distilled it down to a "short version" now at the top of the question. And right now it seems that if someone were to simply post "No", it'd be the most correct answer :) Update 3: Carl Meyer posted "No" :) Thanks again, everyone.

    Read the article

  • What's the next steps for moving from appengine to full django?

    - by tomcritchlow
    Hey guys, I'm super new to programming and I've been using appengine to help me learn python and general coding. I'm getting better quickly and I'm loving it all the way :) Appengine was awesome for allowing me to just dive into writing my app and getting something live that works (see http://www.7bks.com/). But I'm realising that the longer I continue to learn on appengine the more I'm constraining myself and locking myself into a single system. I'd like to move to developing on full django (since django looks super cool!). What are my next steps? To give you a feel for my level of knowledge: I'm not a unix user I'm not familiar with command line controls (I still use appengine/python completely via the appengine SDK) I've never programmed in anything other than python, anywhere other than appengine I know the word SQL, but don't know what MySQL is really or how to use it. So, specifically: What are the skills I need to learn to get up and running with full django/python? If I'm going to host somewhere else I suppose I'll need to learn some sysadmin type skills (maybe even unix?). Is there anywhere that offers easy hosting (like appengine) but that supports django? I hear such great things about heroku I'm considering switching to RoR and going there I appreciate that I'm likely not quite ready to move away from appengine just yet but I'm a fiercely passionate learner (http://www.7bks.com/blog/179001) and would love it if I knew all the steps I needed to learn so I could set about learning them. At the moment, I don't even know what the steps are I need to learn! Thank you very much. Sorry this isn't a specific programming question but I've looked around and haven't found a good how-to for someone of my level of experience and I think others would appreciate a good roadmap for the things we need to learn to get up and running. Thanks, Tom PS - if anyone is in London and fancies showing me the ropes in person that would be super awesome :)

    Read the article

  • How can I get sessions to work if I'm using Google App Engine + Django 1.1?

    - by user341642
    Is there a way for me to get sessions working? I know Django has built in session management, and GAE has some tools for it if you're using their watered down version of Django 0.96, but is there a way to get sessions to work if you're trying to use GAE w/ Django 1.1 (i.e. use_library() call). I assume using a db-backed session doesn't work, and a file system backed one won't work b/c we don't have access to the filesystem if we deploy to the Google production servers. This kinda worked (as in didn't crap out) when I used SessionMiddleware backed by a local-memory backed cache and a non-persistent cache (i.e. setting SESSION_ENGINE to django.contrib.sessions.backends.cache). But the session never seems to persist in this case, no matter how I set the timeouts. A new session key is generated on every page reload. Maybe this is b/c the GAE assumes complete statelessness with each request and blows away my local cache? Apologies in advance, I'm pretty new to Python. Any suggestions would be greatly appreciated.

    Read the article

  • In practice, what are the key differences between Heroku and webfaction? [closed]

    - by jdotjdot
    I've been building and hosting webapps, mainly in Django and Flask, for some time now. Mainly, I've been hosting them on Heroku, because of the free tier and the ease of git-enabled application updating. I have seen that a lot of Django users prefer Webfaction. I looked through their offerings, and they seem to me like a standard web hosting service. Questions: Why might be webfaction considered a good hosting service for Django apps? If Heroku is generally called a "Platform-as-a-Service," what does that make Webfaction? Does it have any important similiarities/distinctions from Heroku that I might somehow be missing?

    Read the article

  • Inherit one instance variable from the global scope

    - by Julian
    I'm using Curses to create a command line GUI with Ruby. Everything's going well, but I have hit a slight snag. I don't think Curses knowledge (esoteric to be fair) is required to answer this question, just Ruby concepts such as objects and inheritance. I'm going to explain my problem now, but if I'm banging on, just look at the example below. Basically, every Window instance needs to have .close called on it in order to close it. Some Window instances have other Windows associated with it. When closing a Window instance, I want to be able to close all of the other Window instances associated with it at the same time. Because associated Windows are generated in a logical fashion, (I append the name with a number: instance_variable_set(self + integer, Window.new(10,10,10,10)) ), it's easy to target generated windows, because methods can anticipate what assosiated windows will be called, (I can recreate the instance variable name from scratch, and almost query it: instance_variable_get(self + integer). I have a delete method that handles this. If the delete method is just a normal, global method (called like this: delete_window(@win543) then everything works perfectly. However, if the delete method is an instance method, which it needs to be in-order to use the self keyword, it doesn't work for a very clear reason; it can 'query' the correct instance variable perfectly well (instance_variable_get(self + integer)), however, because it's an instance method, the global instances aren't scoped to it! Now, one way around this would obviously be to simply make a global method like this: delete_window(@win543). But I have attributes associated with my window instances, and it all works very elegantly. This is very simplified, but it literally translates the problem exactly: class Dog def speak woof end end def woof if @dog_generic == nil puts "@dog_generic isn't scoped when .woof is called from a class method!\n" else puts "@dog_generic is scoped when .woof is called from the global scope. See:\n" + @dog_generic end end @dog_generic = "Woof!" lassie = Dog.new lassie.speak #=> @dog_generic isn't scoped when .woof is called from an instance method!\n woof #=> @dog_generic is scoped when .woof is called from the global scope. See:\nWoof! TL/DR: I need lassie.speak to return this string: "@dog_generic is scoped when .woof is called from the global scope. See:\nWoof!" @dog_generic must remain as an insance variable. The use of Globals or Constants is not acceptable. Could woof inherit from the Global scope? Maybe some sort of keyword: def woof < global # This 'code' is just to conceptualise what I want to do, don't take offence! end Is there some way the .woof method could 'pull in' @dog_generic from the global scope? Will @dog_generic have to be passed in as a parameter?

    Read the article

  • How hard is it to modify the Django Models?

    - by alex
    I am doing geolocation, and Django does not have a PointField. So, I am forced to writing in RAW SQL. GeoDjango, the Django library, does not support the following query for MYSQL databases (can someone verify that for me?) cursor.execute("SELECT id FROM l_tag WHERE\ (GLength(LineStringFromWKB(LineString(asbinary(utm),asbinary(PointFromWKB(point(%s, %s)))))) < %s + accuracy + %s)\ I don't nkow why GeoDjango library cannot do this in MYSQL database. I hate writing RAW SQL for calculating distances between two points. Is there a way I can create my own library for Django that can handle this? If so, how hard is it?

    Read the article

  • After extending the User model in django, how do you create a ModelForm?

    - by mlissner
    I extended the User model in django to include several other variables, such as location, and employer. Now I'm trying to create a form that has the following fields: First name (from User) Last name (from User) Location (from UserProfile, which extends User via a foreign key) Employer (also from UserProfile) I have created a modelform: from django.forms import ModelForm from django.contrib import auth from alert.userHandling.models import UserProfile class ProfileForm(ModelForm): class Meta: # model = auth.models.User # this gives me the User fields model = UserProfile # this gives me the UserProfile fields So, my question is, how can I create a ModelForm that has access to all of the fields, whether they are from the User model or the UserProfile model? Hope this makes sense. I'll be happy to clarify if there are any questions.

    Read the article

  • How to create an exception folder in a django site?

    - by ninja123
    There are a few folders where I house my django site that I want to be rendered as it would on any other non-django site. Namely, forum (vbulletin) and cpanel. I currently run the site with fastcgi. My .htaccess looks like this: AddHandler application/x-httpd-php5 .htm AddHandler application/x-httpd-php5 .html AddHandler fastcgi-script .fcgi Options +FollowSymLinks RewriteEngine On RewriteBase / AddHandler application/x-httpd-php5 .htm RewriteCond %{REQUEST_URI} !(mysite.fcgi) RewriteRule ^(.*)$ mysite.fcgi/$1 [QSA,L] What are lines I can add so www.mysite.com/forum can not be picked up by django url and be rendered as it would do normally. Thanks.

    Read the article

  • How do I configure the Python logging module in Django?

    - by mipadi
    I'm trying to configure logging for a Django app using the Python logging module. I have placed the following bit of configuration code in my Django project's settings.py file: import logging import logging.handlers import os date_fmt = '%m/%d/%Y %H:%M:%S' log_formatter = logging.Formatter(u'[%(asctime)s] %(levelname)-7s: %(message)s (%(filename)s:%(lineno)d)', datefmt=date_fmt) log_dir = os.path.join(PROJECT_DIR, "var", "log", "my_app") log_name = os.path.join(log_dir, "nyrb.log") bytes = 1024 * 1024 # 1 MB if not os.path.exists(log_dir): os.makedirs(log_dir) handler = logging.handlers.RotatingFileHandler(log_name, maxBytes=bytes, backupCount=7) handler.setFormatter(log_formatter) handler.setLevel(logging.DEBUG) logging.getLogger().setLevel(logging.DEBUG) logging.getLogger().addHandler(handler) logging.getLogger(__name__).info("Initialized logging subsystem") At startup, I get a couple Django-related messages, as well as the "Initialized logging subsystem", in the log files, but then all the log messages end up going to the web server logs (/var/log/apache2/error.log, since I'm using Apache), and use the standard log format (not the formatter I designated). Am I configuring logging incorrectly?

    Read the article

  • django-lfs upload Image doesn't work on some environment ?

    - by vernomcrp
    yesterday, after I complete setup django-lfs without buildout. Happlily create categories and products but while I upload image to product after I push upload button its stay always 'pendings'. I use fedora django==1.1.2,PIL==1.1.7. but its work on osx. Now I try on Ubuntu9.10 with completely PIL==1.1.7 and Django==1.1.2 and its won't work. Anyone hav some good solution for this ? (i may think of flash version because upload part looklike its come from flash)

    Read the article

  • How to layout class definition when inheriting from multiple interfaces

    - by gabr
    Given two interface definitions ... IOmniWorkItem = interface ['{3CE2762F-B7A3-4490-BF22-2109C042EAD1}'] function GetData: TOmniValue; function GetResult: TOmniValue; function GetUniqueID: int64; procedure SetResult(const value: TOmniValue); // procedure Cancel; function DetachException: Exception; function FatalException: Exception; function IsCanceled: boolean; function IsExceptional: boolean; property Data: TOmniValue read GetData; property Result: TOmniValue read GetResult write SetResult; property UniqueID: int64 read GetUniqueID; end; IOmniWorkItemEx = interface ['{3B48D012-CF1C-4B47-A4A0-3072A9067A3E}'] function GetOnWorkItemDone: TOmniWorkItemDoneDelegate; function GetOnWorkItemDone_Asy: TOmniWorkItemDoneDelegate; procedure SetOnWorkItemDone(const Value: TOmniWorkItemDoneDelegate); procedure SetOnWorkItemDone_Asy(const Value: TOmniWorkItemDoneDelegate); // property OnWorkItemDone: TOmniWorkItemDoneDelegate read GetOnWorkItemDone write SetOnWorkItemDone; property OnWorkItemDone_Asy: TOmniWorkItemDoneDelegate read GetOnWorkItemDone_Asy write SetOnWorkItemDone_Asy; end; ... what are your ideas of laying out class declaration that inherits from both of them? My current idea (but I don't know if I'm happy with it): TOmniWorkItem = class(TInterfacedObject, IOmniWorkItem, IOmniWorkItemEx) strict private FData : TOmniValue; FOnWorkItemDone : TOmniWorkItemDoneDelegate; FOnWorkItemDone_Asy: TOmniWorkItemDoneDelegate; FResult : TOmniValue; FUniqueID : int64; strict protected procedure FreeException; protected //IOmniWorkItem function GetData: TOmniValue; function GetResult: TOmniValue; function GetUniqueID: int64; procedure SetResult(const value: TOmniValue); protected //IOmniWorkItemEx function GetOnWorkItemDone: TOmniWorkItemDoneDelegate; function GetOnWorkItemDone_Asy: TOmniWorkItemDoneDelegate; procedure SetOnWorkItemDone(const Value: TOmniWorkItemDoneDelegate); procedure SetOnWorkItemDone_Asy(const Value: TOmniWorkItemDoneDelegate); public constructor Create(const data: TOmniValue; uniqueID: int64); destructor Destroy; override; public //IOmniWorkItem procedure Cancel; function DetachException: Exception; function FatalException: Exception; function IsCanceled: boolean; function IsExceptional: boolean; property Data: TOmniValue read GetData; property Result: TOmniValue read GetResult write SetResult; property UniqueID: int64 read GetUniqueID; public //IOmniWorkItemEx property OnWorkItemDone: TOmniWorkItemDoneDelegate read GetOnWorkItemDone write SetOnWorkItemDone; property OnWorkItemDone_Asy: TOmniWorkItemDoneDelegate read GetOnWorkItemDone_Asy write SetOnWorkItemDone_Asy; end; As noted in answers, composition is a good approach for this example but I'm not sure it applies in all cases. Sometimes I'm using multiple inheritance just to split read and write access to some property into public (typically read-only) and private (typically write-only) part. Does composition still apply here? I'm not really sure as I would have to move the property in question out from the main class and I'm not sure that's the correct way to do it. Example: // public part of the interface interface IOmniWorkItemConfig = interface function OnExecute(const aTask: TOmniBackgroundWorkerDelegate): IOmniWorkItemConfig; function OnRequestDone(const aTask: TOmniWorkItemDoneDelegate): IOmniWorkItemConfig; function OnRequestDone_Asy(const aTask: TOmniWorkItemDoneDelegate): IOmniWorkItemConfig; end; // private part of the interface IOmniWorkItemConfigEx = interface ['{42CEC5CB-404F-4868-AE81-6A13AD7E3C6B}'] function GetOnExecute: TOmniBackgroundWorkerDelegate; function GetOnRequestDone: TOmniWorkItemDoneDelegate; function GetOnRequestDone_Asy: TOmniWorkItemDoneDelegate; end; // implementing class TOmniWorkItemConfig = class(TInterfacedObject, IOmniWorkItemConfig, IOmniWorkItemConfigEx) strict private FOnExecute : TOmniBackgroundWorkerDelegate; FOnRequestDone : TOmniWorkItemDoneDelegate; FOnRequestDone_Asy: TOmniWorkItemDoneDelegate; public constructor Create(defaults: IOmniWorkItemConfig = nil); public //IOmniWorkItemConfig function OnExecute(const aTask: TOmniBackgroundWorkerDelegate): IOmniWorkItemConfig; function OnRequestDone(const aTask: TOmniWorkItemDoneDelegate): IOmniWorkItemConfig; function OnRequestDone_Asy(const aTask: TOmniWorkItemDoneDelegate): IOmniWorkItemConfig; public //IOmniWorkItemConfigEx function GetOnExecute: TOmniBackgroundWorkerDelegate; function GetOnRequestDone: TOmniWorkItemDoneDelegate; function GetOnRequestDone_Asy: TOmniWorkItemDoneDelegate; end;

    Read the article

  • Does Django tests run slower on the mac compared to linux?

    - by Thierry Lam
    I'm currently developing my Django projects on both: Mac OS X 10.5, 32 bit Ubuntu Server 9.10 64 bits (1 CPU, 512MB RAM) Both of the above OS are using: Python 2.6.4 Django 1.1.1 MySQL 5.1 Running 12 tests for one of my application take: Mac: 57.513s Linux: 30.935s EDIT: Mac Hardware Spec: MacBook Pro 2.2 GHz Intel Core 2 Duo 3GB RAM I'm running the Ubuntu OS on the same mac above through VMware Fusion 2.0.6. You might argue that Ubuntu Server 64 bits is faster but I have observed a similar speed difference on Ubuntu 8.10 32 bits desktop edition. Even if I turn off my linux VM and other mac applications, I still experience the slowness. Has anyone else experienced this Django test speed difference across those two OS?

    Read the article

  • Threaded Django task doesn't automatically handle transactions or db connections?

    - by Gabriel Hurley
    I've got Django set up to run some recurring tasks in their own threads, and I noticed that they were always leaving behind unfinished database connection processes (pgsql "Idle In Transaction"). I looked through the Postgres logs and found that the transactions weren't being completed (no ROLLBACK). I tried using the various transaction decorators on my functions, no luck. I switched to manual transaction management and did the rollback manually, that worked, but still left the processes as "Idle". So then I called connection.close(), and all is well. But I'm left wondering, why doesn't Django's typical transaction and connection management work for these threaded tasks that are being spawned from the main Django thread?

    Read the article

  • Does Python Django support custom SQL and denormalized databases with no Foreign Key relationships?

    - by Jay
    I've just started learning Python Django and have a lot of experience building high traffic websites using PHP and MySQL. What worries me so far is Python's overly optimistic approach that you will never need to write custom SQL and that it automatically creates all these Foreign Key relationships in your database. The one thing I've learned in the last few years of building Chess.com is that its impossible to NOT write custom SQL when you're dealing with something like MySQL that frequently needs to be told what indexes it should use (or avoid), and that Foreign Keys are a death sentence. Percona's strongest recommendation was for us to remove all FKs for optimal performance. Is there a way in Django to do this in the models file? create relationships without creating actual DB FKs? Or is there a way to start at the database level, design/create my database, and then have Django reverse engineer the models file?

    Read the article

  • 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?

    Read the article

  • Server-side access to Client Browser's Latitude/Longitude using Django.

    - by ZenGyro
    Hello, So i am writing a little app that compares a user's position against a database on web-based server written using Django and performs some functions with it. Accessing the browser's geolocation data (in supported browsers ) is fairly trivial using JavaScript. But what is the best way to allow the Django server to access the longitude and latitude variables? Is it best to wrap them up as a JSON object and send to the server via POST? Or is there some easier (Geo)Django-based way to access the Navigator.geolocation browser object. Please forgive a newbie a question like this, but my Google-Fuing only seems to find ways to insert variables into JavaScript via template tag, whereas I need it to work the other way! Any advice or code snippets greatly appreciated. Feel free to talk to me like I am an idiot.

    Read the article

  • How to use external static files with Django (serving external files once again)?

    - by Tomas Novotny
    Hi, even after Googling and reading all relevant posts at StackOverflow, I still can't get static files working in my Django application. Here is how my files look: settings.py MEDIA_ROOT = os.path.join(SITE_ROOT, 'static') MEDIA_URL = '/static/' urs.py from DjangoBandCreatorSite.settings import DEBUG if DEBUG: urlpatterns += patterns('', ( r'^static/(?P<path>.*)$', 'django.views.static.serve', {'document_root': 'static'} )) template: <script type="text/javascript" src="/static/jquery.js"></script> <script type="text/javascript"> I am trying to use jquery.js stored in directory "static". I am using: Windows XP Python 2.6.4 Django 1.2.3 Thank you very much for any help

    Read the article

< Previous Page | 94 95 96 97 98 99 100 101 102 103 104 105  | Next Page >