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  • How to create instances of related models in Django

    - by sevennineteen
    I'm working on a CMSy app for which I've implemented a set of models which allow for creation of custom Template instances, made up of a number of Fields and tied to a specific Customer. The end-goal is that one or more templates with a set of custom fields can be defined through the Admin interface and associated to a customer, so that customer can then create content objects in the format prescribed by the template. I seem to have gotten this hooked up such that I can create any number of Template objects, but I'm struggling with how to create instances - actual content objects - in those templates. For example, I can define a template "Basic Page" for customer "Acme" which has the fields "Title" and "Body", but I haven't figured out how to create Basic Page instances where these fields can be filled in. Here are my (somewhat elided) models... class Customer(models.Model): ... class Field(models.Model): ... class Template(models.Model): label = models.CharField(max_length=255) clients = models.ManyToManyField(Customer, blank=True) fields = models.ManyToManyField(Field, blank=True) class ContentObject(models.Model): label = models.CharField(max_length=255) template = models.ForeignKey(Template) author = models.ForeignKey(User) customer = models.ForeignKey(Customer) mod_date = models.DateTimeField('Modified Date', editable=False) def __unicode__(self): return '%s (%s)' % (self.label, self.template) def save(self): self.mod_date = datetime.datetime.now() super(ContentObject, self).save() Thanks in advance for any advice!

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  • Designing a Tag table that tells how many times it's used

    - by Satoru.Logic
    Hi, all. I am trying to design a tagging system with a model like this: Tag: content = CharField creator = ForeignKey used = IntergerField It is a many-to-many relationship between tags and what's been tagged. Everytime I insert a record into the assotication table, Tag.used is incremented by one, and decremented by one in case of deletion. Tag.used is maintained because I want to speed up answering the question 'How many times this tag is used?'. However, this seems to slow insertion down obviously. Please tell me how to improve this design. Thanks in advance.

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  • How to turn a list of tuples into a string?

    - by matt
    I have a list of tuples that I'm trying to incorporate into a SQL query but I can't figure out how to join them together without adding slashes. My like this: list = [('val', 'val'), ('val', 'val'), ('val', 'val')] If I turn each tuple into a string and try to join them with a a comma I'll get something like ' (\'val\, \'val\'), ... ' What's the right way to do this, so I can get the list (without brackets) as a string?

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  • Why is win32com so much slower than xlrd?

    - by Josh
    I have the same code, written using win32com and xlrd. xlrd preforms the algorithm in less than a second, while win32com takes minutes. Here is the win32com: def makeDict(ws): """makes dict with key as header name, value as tuple of column begin and column end (inclusive)""" wsHeaders = {} # key is header name, value is column begin and end inclusive for cnum in xrange(9, find_last_col(ws)): if ws.Cells(7, cnum).Value: wsHeaders[str(ws.Cells(7, cnum).Value)] = (cnum, find_last_col(ws)) for cend in xrange(cnum + 1, find_last_col(ws)): #finds end column if ws.Cells(7, cend).Value: wsHeaders[str(ws.Cells(7, cnum).Value)] = (cnum, cend - 1) break return wsHeaders And the xlrd def makeDict(ws): """makes dict with key as header name, value as tuple of column begin and column end (inclusive)""" wsHeaders = {} # key is header name, value is column begin and end inclusive for cnum in xrange(8, ws.ncols): if ws.cell_value(6, cnum): wsHeaders[str(ws.cell_value(6, cnum))] = (cnum, ws.ncols) for cend in xrange(cnum + 1, ws.ncols):#finds end column if ws.cell_value(6, cend): wsHeaders[str(ws.cell_value(6, cnum))] = (cnum, cend - 1) break return wsHeaders

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  • Loading url with cyrillic symbols

    - by Ockonal
    Hi guys, I have to load some url with cyrillic symbols. My script should work with this: http://wincode.org/%D0%BF%D1%80%D0%BE%D0%B3%D1%80%D0%B0%D0%BC%D0%BC%D0%B8%D1%80%D0%BE%D0%B2%D0%B0%D0%BD%D0%B8%D0%B5/ If I'll use this in browser it would replaced into normal symbols, but urllib code fails with 404 error. How to decode correctly this url? When I'm using that url directly in code, like address = 'that address', it works perfect. But I used parsing page for getting this url. I have a list of urls which contents cyrillic. Maybe they have uncorrect encoding? Here is more code: requestData = urllib2.Request( %SOME_ADDRESS%, None, {"User-Agent": user_agent}) requestHandler = pageHandler.open(requestData) pageData = requestHandler.read().decode('utf-8') soupHandler = BeautifulSoup(pageData) topicLinks = [] for postBlock in soupHandler.findAll('a', href=re.compile('%SOME_REGEXP%')): topicLinks.append(postBlock['href']) postAddress = choice(topicLinks) postRequestData = urllib2.Request(postAddress, None, {"User-Agent": user_agent}) postHandler = pageHandler.open(postRequestData) postData = postHandler.read() File "/usr/lib/python2.6/urllib2.py", line 518, in http_error_default raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) urllib2.HTTPError: HTTP Error 404: Not Found

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  • Error -3 while decompressing data: incorrect header check

    - by Rahul99
    I have .zip file which contain csv data. I am reading .zip file using <input type = "file" name = "select_file"/> I want to decompress that .zip file and read csv data. file_data = self.request.get('select_file') file_str = zlib.decompress(file_data) #file_data_list = file_str.split('\n') #file_Reader = csv.reader(file_data_list,quoting=csv.QUOTE_NONE ) I am expecting csv data in file_str but I am getting error. error :: Error -3 while decompressing data: incorrect header check What I have to use?

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  • Artificial Intelligence in online game using Google App Engine

    - by Hortinstein
    I am currently in the planning stages of a game for google app engine, but cannot wrap my head around how I am going to handle AI. I intend to have persistant NPCs that will move about the map, but short of writing a program that generates the same XML requests I use to control player actions, than run it on another server I am stuck on how to do it. I have looked at the Task Queue feature, but due to long running processes not being an option on the App engine, I am a little stuck. I intend to run multiple server instances with 200+ persistant NPC entities that I will need to update. Most action is slowly roaming around based on player movements/concentrations, and attacking close range players...(you can probably guess the type of game im developing)

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  • How do I get javascript results using selenium?

    - by Seth
    I have the following code: from selenium import selenium selenium = selenium("localhost", 4444, "*chrome", "http://some_site.com/") selenium.start() sel = selenium sel.open("/") sel.type("ctl00_ContentPlaceHolder1_SuburbTownTextBox", "Adelaide,SA,5000") sel.click("ctl00_ContentPlaceHolder1_SearchImageButton") #text = sel.get_body_text() text = sel.get_html_source() print(text) The click executes a javascript file which then produces results on the same page. Obviously print(text) will only print the orignal html source. How do I get to the results of the javascript?

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  • How can I detect whether an image is a PNG or APNG format?

    - by perlit
    APNG is backwards compatible with PNG. I opened up an apng and png file in a hex editor and the first few bytes look identical. So if a user uploads either of these formats, how do I detect what the format really is? I've seen this done on some sites that block apng. I'm guessing the ImageMagick library makes this easy, but what if I were to do the detect without the use of an image processing library (for learning purposes)? Can I look for specific bytes that tell me if the file is apng? Solutions in any language is welcome.

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  • How do I store multiple copies of the same field in Django?

    - by Alistair
    I'm storing OLAC metadata which describes linguistic resources. Many of the elements of the metadata are repeatable -- for example, a resource can have two languages, three authors and four dates associated with it. Is there any way of storing this in one model? It seems like overkill to define a model for each repeatable metadata element -- especially since the models will only have one field: it's value.

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  • How can I display multiple django modelformset forms together?

    - by JT
    I have a problem with needing to provide multiple model backed forms on the same page. I understand how to do this with single forms, i.e. just create both the forms call them something different then use the appropriate names in the template. Now how exactly do you expand that solution to work with modelformsets? The wrinkle, of course, is that each 'form' must be rendered together in the appropriate fieldset. For example I want my template to produce something like this: <fieldset> <label for="id_base-0-desc">Home Base Description:</label> <input id="id_base-0-desc" type="text" name="base-0-desc" maxlength="100" /> <label for="id_likes-0-icecream">Want ice cream?</label> <input type="checkbox" name="likes-0-icecream" id="id_likes-0-icecream" /> </fieldset> <fieldset> <label for="id_base-1-desc">Home Base Description:</label> <input id="id_base-1-desc" type="text" name="base-1-desc" maxlength="100" /> <label for="id_likes-1-icecream">Want ice cream?</label> <input type="checkbox" name="likes-1-icecream" id="id_likes-1-icecream" /> </fieldset> I am using a loop like this to process the results for base_form, likes_form in map(None, base_forms, likes_forms): which works as I'd expect (I'm using map because the # of forms can be different). The problem is that I can't figure out a way to do the same thing with the templating engine. The system does work if I layout all the base models together then all the likes models after wards, but it doesn't meet the layout requirements.

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  • Simple XML over http web service

    - by Mark
    I have a simple html service, developed in django. You enter your name - it posts this, and returns a value (male/female). I need to ofer this as a web service. I have no idea where to start. I want to accept a xml request, and provide an xml response - thats it. Can anyone give ma any pointers - Googling it is difficult when you dont know what your searching for.

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  • App only spawns one thread

    - by tipu
    I have what I thought was a thread-friendly app, and after doing some output I've concluded that of the 15 threads I am attempting to run, only one does. I have if __name__ == "__main__": fhf = FileHandlerFactory() tweet_manager = TweetManager("C:/Documents and Settings/Administrator/My Documents/My Dropbox/workspace/trie/Tweet Search Engine/data/partitioned_raw_tweets/raw_tweets.txt.001") start = time.time() for i in range(15): Indexer(tweet_manager, fhf).start() Then in my thread-entry point, I do def run(self): print(threading.current_thread()) self.index() That results in this: <Indexer(Thread-3, started 1168)> So of 15 threads that I thought were running, I'm only running one. Any idea as to why? Edit: code

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  • Can SQLAlchemy DateTime Objects Only Be Naive?

    - by Sean M
    I am working with SQLAlchemy, and I'm not yet sure which database I'll use under it, so I want to remain as DB-agnostic as possible. How can I store a timezone-aware datetime object in the DB without tying myself to a specific database? Right now, I'm making sure that times are UTC before I store them in the DB, and converting to localized at display-time, but that feels inelegant and brittle. Is there a DB-agnostic way to get a timezone-aware datetime out of SQLAlchemy instead of getting naive datatime objects out of the DB?

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  • How do I code this relationship in SQLAlchemy?

    - by Martin Del Vecchio
    I am new to SQLAlchemy (and SQL, for that matter). I can't figure out how to code the idea I have in my head. I am creating a database of performance-test results. A test run consists of a test type and a number (this is class TestRun below) A test suite consists the version string of the software being tested, and one or more TestRun objects (this is class TestSuite below). A test version consists of all test suites with the given version name. Here is my code, as simple as I can make it: from sqlalchemy import * from sqlalchemy.ext.declarative import declarative_base from sqlalchemy.orm import relationship, backref, sessionmaker Base = declarative_base() class TestVersion (Base): __tablename__ = 'versions' id = Column (Integer, primary_key=True) version_name = Column (String) def __init__ (self, version_name): self.version_name = version_name class TestRun (Base): __tablename__ = 'runs' id = Column (Integer, primary_key=True) suite_directory = Column (String, ForeignKey ('suites.directory')) suite = relationship ('TestSuite', backref=backref ('runs', order_by=id)) test_type = Column (String) rate = Column (Integer) def __init__ (self, test_type, rate): self.test_type = test_type self.rate = rate class TestSuite (Base): __tablename__ = 'suites' directory = Column (String, primary_key=True) version_id = Column (Integer, ForeignKey ('versions.id')) version_ref = relationship ('TestVersion', backref=backref ('suites', order_by=directory)) version_name = Column (String) def __init__ (self, directory, version_name): self.directory = directory self.version_name = version_name # Create a v1.0 suite suite1 = TestSuite ('dir1', 'v1.0') suite1.runs.append (TestRun ('test1', 100)) suite1.runs.append (TestRun ('test2', 200)) # Create a another v1.0 suite suite2 = TestSuite ('dir2', 'v1.0') suite2.runs.append (TestRun ('test1', 101)) suite2.runs.append (TestRun ('test2', 201)) # Create another suite suite3 = TestSuite ('dir3', 'v2.0') suite3.runs.append (TestRun ('test1', 102)) suite3.runs.append (TestRun ('test2', 202)) # Create the in-memory database engine = create_engine ('sqlite://') Session = sessionmaker (bind=engine) session = Session() Base.metadata.create_all (engine) # Add the suites in version1 = TestVersion (suite1.version_name) version1.suites.append (suite1) session.add (suite1) version2 = TestVersion (suite2.version_name) version2.suites.append (suite2) session.add (suite2) version3 = TestVersion (suite3.version_name) version3.suites.append (suite3) session.add (suite3) session.commit() # Query the suites for suite in session.query (TestSuite).order_by (TestSuite.directory): print "\nSuite directory %s, version %s has %d test runs:" % (suite.directory, suite.version_name, len (suite.runs)) for run in suite.runs: print " Test '%s', result %d" % (run.test_type, run.rate) # Query the versions for version in session.query (TestVersion).order_by (TestVersion.version_name): print "\nVersion %s has %d test suites:" % (version.version_name, len (version.suites)) for suite in version.suites: print " Suite directory %s, version %s has %d test runs:" % (suite.directory, suite.version_name, len (suite.runs)) for run in suite.runs: print " Test '%s', result %d" % (run.test_type, run.rate) The output of this program: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 Version v1.0 has 1 test suites: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Version v1.0 has 1 test suites: Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Version v2.0 has 1 test suites: Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 This is not correct, since there are two TestVersion objects with the name 'v1.0'. I hacked my way around this by adding a private list of TestVersion objects, and a function to find a matching one: versions = [] def find_or_create_version (version_name): # Find existing for version in versions: if version.version_name == version_name: return (version) # Create new version = TestVersion (version_name) versions.append (version) return (version) Then I modified my code that adds the records to use it: # Add the suites in version1 = find_or_create_version (suite1.version_name) version1.suites.append (suite1) session.add (suite1) version2 = find_or_create_version (suite2.version_name) version2.suites.append (suite2) session.add (suite2) version3 = find_or_create_version (suite3.version_name) version3.suites.append (suite3) session.add (suite3) Now the output is what I want: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 Version v1.0 has 2 test suites: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Version v2.0 has 1 test suites: Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 This feels wrong to me; it doesn't feel right that I am manually keeping track of the unique version names, and manually adding the suites to the appropriate TestVersion objects. Is this code even close to being correct? And what happens when I'm not building the entire database from scratch, as in this example. If the database already exists, do I have to query the database's TestVersion table to discover the unique version names? Thanks in advance. I know this is a lot of code to wade through, and I appreciate the help.

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  • Scrapy Not Returning Additonal Info from Scraped Link in Item via Request Callback

    - by zoonosis
    Basically the code below scrapes the first 5 items of a table. One of the fields is another href and clicking on that href provides more info which I want to collect and add to the original item. So parse is supposed to pass the semi populated item to parse_next_page which then scrapes the next bit and should return the completed item back to parse Running the code below only returns the info collected in parse If I change the return items to return request I get a completed item with all 3 "things" but I only get 1 of the rows, not all 5. Im sure its something simple, I just can't see it. class ThingSpider(BaseSpider): name = "thing" allowed_domains = ["somepage.com"] start_urls = [ "http://www.somepage.com" ] def parse(self, response): hxs = HtmlXPathSelector(response) items = [] for x in range (1,6): item = ScrapyItem() str_selector = '//tr[@name="row{0}"]'.format(x) item['thing1'] = hxs.select(str_selector")]/a/text()').extract() item['thing2'] = hxs.select(str_selector")]/a/@href').extract() print 'hello' request = Request("www.nextpage.com", callback=self.parse_next_page,meta={'item':item}) print 'hello2' request.meta['item'] = item items.append(item) return items def parse_next_page(self, response): print 'stuff' hxs = HtmlXPathSelector(response) item = response.meta['item'] item['thing3'] = hxs.select('//div/ul/li[1]/span[2]/text()').extract() return item

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  • Avoiding nesting two for loops

    - by chavanak
    Hi, Please have a look at the code below: import string from collections import defaultdict first_complex=open( "residue_a_chain_a_b_backup.txt", "r" ) first_complex_lines=first_complex.readlines() first_complex_lines=map( string.strip, first_complex_lines ) first_complex.close() second_complex=open( "residue_a_chain_a_c_backup.txt", "r" ) second_complex_lines=second_complex.readlines() second_complex_lines=map( string.strip, second_complex_lines ) second_complex.close() list_1=[] list_2=[] for x in first_complex_lines: if x[0]!="d": list_1.append( x ) for y in second_complex_lines: if y[0]!="d": list_2.append( y ) j=0 list_3=[] list_4=[] for a in list_1: pass for b in list_2: pass if a==b: list_3.append( a ) kvmap=defaultdict( int ) for k in list_3: kvmap[k]+=1 print kvmap Normally I use izip or izip_longest to club two for loops, but this time the length of the files are different. I don't want a None entry. If I use the above method, the run time becomes incremental and useless. How am I supposed to get the two for loops going? Cheers, Chavanak

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  • SQLAlchemy - select for update example

    - by Mark
    I'm looking for a complete example of using select for update in SQLAlchemy, but haven't found one googling. I need to lock a single row and update a column, the following code doesn't work (blocks forever): s = table.select(table.c.user=="test",for_update=True) u = table.update().where(table.c.user=="test") u.execute(email="foo") Do I need a commit? How do I do that? As far as I know you need to: begin transaction select ... for update update commit

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  • How do I use django settings in my logging.ini file?

    - by slypete
    I have a BASE_DIR setting in my settings.py file: BASE_DIR = os.path.dirname(os.path.abspath(__file__)) I need to use this variable in my logging.ini file to setup my file handler paths. The initialization of logging happens in the same file, the settings.py file, below my BASE_DIR variable: LOG_INIT_DONE=False if not LOG_INIT_DONE: logging.config.fileConfig(LOGGING_INI) LOG_INIT_DONE=True Thanks, Pete

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  • Find subset with K elements that are closest to eachother

    - by Nima
    Given an array of integers size N, how can you efficiently find a subset of size K with elements that are closest to each other? Let the closeness for a subset (x1,x2,x3,..xk) be defined as: 2 <= N <= 10^5 2 <= K <= N constraints: Array may contain duplicates and is not guaranteed to be sorted. My brute force solution is very slow for large N, and it doesn't check if there's more than 1 solution: N = input() K = input() assert 2 <= N <= 10**5 assert 2 <= K <= N a = [] for i in xrange(0, N): a.append(input()) a.sort() minimum = sys.maxint startindex = 0 for i in xrange(0,N-K+1): last = i + K tmp = 0 for j in xrange(i, last): for l in xrange(j+1, last): tmp += abs(a[j]-a[l]) if(tmp > minimum): break if(tmp < minimum): minimum = tmp startindex = i #end index = startindex + K? Examples: N = 7 K = 3 array = [10,100,300,200,1000,20,30] result = [10,20,30] N = 10 K = 4 array = [1,2,3,4,10,20,30,40,100,200] result = [1,2,3,4]

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  • How do I add a trailing slash for Django MPTT-based categorization app?

    - by Patrick Beeson
    I'm using Django-MPTT to develop a categorization app for my Django project. But I can't seem to get the regex pattern for adding a trailing slash that doesn't also break on child categories. Here's an example URL: http://mydjangoapp.com/categories/parentcat/childcat/ I'd like to be able to use http://mydjangoapp.com/categories/parentcat and have it redirect to the trailing slash version. The same should apply to http://mydjangoapp.com/categories/parentcat/childcat (it should redirect to http://mydjangoapp.com/categories/parentcat/childcat/). Here's my urls.py: from django.conf.urls.defaults import patterns, include, url from django.views.decorators.cache import cache_page from storefront.categories.models import Category from storefront.categories.views import SimpleCategoryView urlpatterns = patterns('', url(r'^(?P<full_slug>[-\w/]+)', cache_page(SimpleCategoryView.as_view(), 60 * 15), name='category_view'), ) And here is my view: from django.core.exceptions import ImproperlyConfigured from django.core.urlresolvers import reverse from django.views.generic import TemplateView, DetailView from django.views.generic.detail import SingleObjectTemplateResponseMixin, SingleObjectMixin from django.utils.translation import ugettext as _ from django.contrib.syndication.views import Feed from storefront.categories.models import Category class SimpleCategoryView(TemplateView): def get_category(self): return Category.objects.get(full_slug=self.kwargs['full_slug']) def get_context_data(self, **kwargs): context = super(SimpleCategoryView, self).get_context_data(**kwargs) context["category"] = self.get_category() return context def get_template_names(self): if self.get_category().template_name: return [self.get_category().template_name] else: return ['categories/category_detail.html'] And finally, my model: from django.db import models from mptt.models import MPTTModel from mptt.fields import TreeForeignKey class CategoryManager(models.Manager): def get(self, **kwargs): defaults = {} defaults.update(kwargs) if 'full_slug' in defaults: if defaults['full_slug'] and defaults['full_slug'][-1] != "/": defaults['full_slug'] += "/" return super(CategoryManager, self).get(**defaults) class Category(MPTTModel): title = models.CharField(max_length=255) description = models.TextField(blank=True, help_text='Please use <a href="http://daringfireball.net/projects/markdown/syntax">Markdown syntax</a> for all text-formatting and links. No HTML is allowed.') slug = models.SlugField(help_text='Prepopulates from title field.') full_slug = models.CharField(max_length=255, blank=True) template_name = models.CharField(max_length=70, blank=True, help_text="Example: 'categories/category_parent.html'. If this isn't provided, the system will use 'categories/category_detail.html'. Use 'categories/category_parent.html' for all parent categories and 'categories/category_child.html' for all child categories.") parent = TreeForeignKey('self', null=True, blank=True, related_name='children') objects = CategoryManager() class Meta: verbose_name = 'category' verbose_name_plural = 'categories' def save(self, *args, **kwargs): orig_full_slug = self.full_slug if self.parent: self.full_slug = "%s%s/" % (self.parent.full_slug, self.slug) else: self.full_slug = "%s/" % self.slug obj = super(Category, self).save(*args, **kwargs) if orig_full_slug != self.full_slug: for child in self.get_children(): child.save() return obj def available_product_set(self): """ Returns available, prioritized products for a category """ from storefront.apparel.models import Product return self.product_set.filter(is_available=True).order_by('-priority') def __unicode__(self): return "%s (%s)" % (self.title, self.full_slug) def get_absolute_url(self): return '/categories/%s' % (self.full_slug)

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  • verbose_name for a model's method

    - by mawimawi
    How can I set a verbose_name for a model's method, so that it might be displayed in the admin's change_view form? example: class Article(models.Model): title = models.CharField(max_length=64) created_date = models.DateTimeField(....) def created_weekday(self): return self.created_date.strftime("%A") in admin.py: class ArticleAdmin(admin.ModelAdmin): readonly_fields = ('created_weekday',) fields = ('title', 'created_weekday') Now the label for created_weekday is "Created Weekday", but I'd like it to have a different label which should be i18nable using ugettext_lazy as well. I've tried created_weekday.verbose_name=... after the method, but that did not show any result. Is there a decorator or something I can use, so I could make my own "verbose_name" / "label" / whateverthename is?

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