Search Results

Search found 16680 results on 668 pages for 'python datetime'.

Page 365/668 | < Previous Page | 361 362 363 364 365 366 367 368 369 370 371 372  | Next Page >

  • Extended Django base-class with multiple instances

    - by Gijs
    I'm modeling a simple movie database using Django. models.py defines a base model Person. I extend Person into Actor and Director, which works as I imagined. Persons must be unique. When (in the Admin) I create an instance of Actor, and this person is also a Director, it won't save because of the unique = True. Any ideas how to solve this problem? (generic foreign keys?) Thx

    Read the article

  • How to display a QGraphicsScene?

    - by Chris
    I've got the following code and I'm not sure how to add the QGraphicsScene to my layout.. class MainForm(QDialog): def __init__(self, parent=None): super(MainForm, self).__init__(parent) self.scene = QGraphicsScene(self) self.scene.setSceneRect(0, 0, 500, 500) self.view = QGraphicsView() self.view.setRenderHint(QPainter.Antialiasing) self.view.setScene(self.scene) self.view.setFocusPolicy(Qt.NoFocus) zoomSlider = QSlider(Qt.Horizontal) zoomSlider.setRange(5, 200) zoomSlider.setValue(100) self.pauseButton = QPushButton("Pause") quitButton = QPushButton("Quit") layout = QVBoxLayout() layout.addWidget(zoomSlider) self.setLayout(layout) self.startTimer(10) How can I get my QGraphicsScene running? I'm new to Qt. Am I even supposed to be adding a QGraphicsScene to a layout/

    Read the article

  • Django: reverse lookup URL of feeds?

    - by Santa
    I am having trouble doing a reverse URL lookup for Django-generated feeds. I have the following setup in urls.py: feeds = { 'latest': LatestEntries, } urlpatterns = patterns('', # ... # enable feeds (RSS) url(r'^feeds/(?P<url>.*)/$', 'django.contrib.syndication.views.feed', {'feed_dict': feeds}, name='feeds_view'), ) I have tried using the following template tag: <a href="{% url feeds_view latest %}">RSS feeds</a> But the resulting link is not what want (http://my.domain.com/feeds//). It should be http://my.domain.com/feeds/latest/. For now, I am using a hack to generate the URL for the template: <a href="http://{{ request.META.HTTP_HOST }}/feeds/latest">RSS feeds</a> But, as you can see, it clearly is not DRY. Is there something I am missing?

    Read the article

  • How to keep text inside a circle using Cairo?

    - by miguelrios
    I a drawing a graph using Cairo (pycairo specifically) and I need to know how can I draw text inside a circle without overlapping it, by keeping it inside the bounds of the circle. I have this simple code snippet that draws a letter "a" inside the circle: ''' Created on May 8, 2010 @author: mrios ''' import cairo, math WIDTH, HEIGHT = 1000, 1000 #surface = cairo.PDFSurface ("/Users/mrios/Desktop/exampleplaces.pdf", WIDTH, HEIGHT) surface = cairo.ImageSurface (cairo.FORMAT_ARGB32, WIDTH, HEIGHT) ctx = cairo.Context (surface) ctx.scale (WIDTH/1.0, HEIGHT/1.0) # Normalizing the canvas ctx.rectangle(0, 0, 1, 1) # Rectangle(x0, y0, x1, y1) ctx.set_source_rgb(255,255,255) ctx.fill() ctx.arc(0.5, 0.5, .4, 0, 2*math.pi) ctx.set_source_rgb(0,0,0) ctx.set_line_width(0.03) ctx.stroke() ctx.arc(0.5, 0.5, .4, 0, 2*math.pi) ctx.set_source_rgb(0,0,0) ctx.set_line_width(0.01) ctx.set_source_rgb(255,0,255) ctx.fill() ctx.set_source_rgb(0,0,0) ctx.select_font_face("Georgia", cairo.FONT_SLANT_NORMAL, cairo.FONT_WEIGHT_BOLD) ctx.set_font_size(1.0) x_bearing, y_bearing, width, height = ctx.text_extents("a")[:4] print ctx.text_extents("a")[:4] ctx.move_to(0.5 - width / 2 - x_bearing, 0.5 - height / 2 - y_bearing) ctx.show_text("a") surface.write_to_png ("/Users/mrios/Desktop/node.png") # Output to PNG The problem is that my labels have variable amount of characters (with a limit of 20) and I need to set the size of the font dynamically. It must fit inside the circle, no matter the size of the circle nor the size of the label. Also, every label has one line of text, no spaces, no line breaks. Any suggestion?

    Read the article

  • How do I pass a lot of parameters to views in Django?

    - by Mark
    I'm very new to Django and I'm trying to build an application to present my data in tables and charts. Till now my learning process went very smooth, but now I'm a bit stuck. My pageview retrieves large amounts of data from a database and puts it in the context. The template then generates different html-tables. So far so good. Now I want to add different charts to the template. I manage to do this by defining <img src=".../> tags. The Matplotlib chart is generate in my chartview an returned via: response=HttpResponse(content_type='image/png') canvas.print_png(response) return response Now I have different questions: the data is retrieved twice from the database. Once in the pageview to render the tables, and again in the chartview for making the charts. What is the best way to pass the data, already in the context of the page to the chartview? I need a lot of charts, each with different datasets. I could make a chartview for each chart, but probably there is a better way. How do I pass the different dataset names to the chartview? Some charts have 20 datasets, so I don't think that passing these dataset parameters via the url (like: <imgm src="chart/dataset1/dataset2/.../dataset20/chart.png />) is the right way. Any advice?

    Read the article

  • Django Admin Page missing CSS

    - by super9
    I saw this question and recommendation from Django Projects here but still can't get this to work. My Django Admin pages are not displaying the CSS at all. This is my current configuration. settings.py ADMIN_MEDIA_PREFIX = '/media/admin/' httpd.conf <VirtualHost *:80> DocumentRoot /home/django/sgel ServerName ec2-***-**-***-***.ap-**********-1.compute.amazonaws.com ErrorLog /home/django/sgel/logs/apache_error.log CustomLog /home/django/sgel/logs/apache_access.log combined WSGIScriptAlias / /home/django/sgel/apache/django.wsgi <Directory /home/django/sgel/media> Order deny,allow Allow from all </Directory> <Directory /home/django/sgel/apache> Order deny,allow Allow from all </Directory> LogLevel warn Alias /media/ /home/django/sgel/media/ </VirtualHost> <VirtualHost *:80> ServerName sgel.com Redirect permanent / http://www.sgel.com/ </VirtualHost> In addition, I also ran the following to create (I think) the symbolic link ln -s /home/djangotest/sgel/media/admin/ /usr/lib/python2.6/site-packages/django/contrib/admin/media/ UPDATE In my httpd.conf file, User django Group django When I run ls -l in my /media directory drwxr-xr-x 2 root root 4096 Apr 4 11:03 admin -rw-r--r-- 1 root root 9 Apr 8 09:02 test.txt Should that root user be django instead? UPDATE 2 When I enter ls -la in my /media/admin folder total 12 drwxr-xr-x 2 root root 4096 Apr 13 03:33 . drwxr-xr-x 3 root root 4096 Apr 8 09:02 .. lrwxrwxrwx 1 root root 60 Apr 13 03:33 media -> /usr/lib/python2.6/site-packages/django/contrib/admin/media/ The thing is, when I navigate to /usr/lib/python2.6/site-packages/django/contrib/admin/media/, the folder was empty. So I copied the CSS, IMG and JS folders from my Django installation into /usr/lib/python2.6/site-packages/django/contrib/admin/media/ and it still didn't work

    Read the article

  • Apps not showing in Django admin site

    - by jack
    I have a Django project with about 10 apps in it. But the admin interface only shows Auth and Site models which are part of Django distribution. Yes, the admin interface is up and working but none of my self-written apps shows there. INSTALLED_APPS INSTALLED_APPS = ( 'django.contrib.auth', 'django.contrib.sites', 'django.contrib.contenttypes', 'django.contrib.humanize', 'django.contrib.sessions', 'django.contrib.admin', 'django.contrib.admindocs', 'project.app1', ... app1/admin.py from django.contrib import admin from project.app1.models import * admin.site.register(model1) admin.site.register(model2) admin.site.register(model3) What could be wrong in this case? Looks like everything is configured as what document says. Thank you in advance.

    Read the article

  • django admin - adding fields on the fly

    - by Thomas
    Basically I am writing a simple shopping cart. Each item can have multiple prices. (i.e. shirts where each size is priced differently). I would like to have a single price field in my admin panel, where when the first price is entered, an additional price field pops up. However I am kind of at a loss as to how to do this. What would be the best way to do this?

    Read the article

  • Django - provide additional information in template

    - by Ninefingers
    Hi all, I am building an app to learn Django and have started with a Contact system that currently stores Contacts and Addresses. C's are a many to many relationship with A's, but rather than use Django's models.ManyToManyField() I've created my own link-table providing additional information about the link, such as what the address type is to the that contact (home, work etc). What I'm trying to do is pass this information out to a view, so in my full view of a contact I can do this: def contact_view_full(request, contact_id): c = get_object_or_404(Contact, id=contact_id) a = [] links = ContactAddressLink.objects.filter(ContactID=c.id) for link in links: b = Address.objects.get(id=link.AddressID_id) a.append(b) return render_to_response('contact_full.html', {'contact_item': c, 'addresses' : a }, context_instance=RequestContext(request)) And so I can do the equivalent of c.Addresses.all() or however the ManyToManyField works. What I'm interested to know is how can I pass out information about the link in the link object with the 'addresses' : a information, so that when my template does this: {% for address in addresses %} <!-- ... --> {% endfor %} and properly associate the correct link object data with the address. So what's the best way to achieve this? I'm thinking a union of two objects might be an idea but I haven't enough experience with Django to know if that's considered the best way of doing it. Suggestions? Thanks in advance. Nf

    Read the article

  • Extending Django Flatpages to accept template tags

    - by Tristan
    I use django flatpages for a lot of content on our site, I'd like to extend it to accept django template tags in the content as well. I found this snippet but after much larking about I couldn't get it to work. Am I correct in assuming that you would need too "subclass" the django flatpages app to get this to work? Is this best way of doing it? I'm not quite sure how to structure it, as I don't really want to directly modify the django distribution.

    Read the article

  • Modifying Bresenham's line algorithm

    - by sphennings
    I'm trying to use Bresenham's line algorithm to compute Field of View on a grid. The code I'm using calculates the lines without a problem but I'm having problems getting it to always return the line running from start point to endpoint. What do I need to do so that all lines returned run from (x0,y0) to (x1,y1) def bresenham_line(self, x0, y0, x1, y1): steep = abs(y1 - y0) > abs(x1 - x0) if steep: x0, y0 = y0, x0 x1, y1 = y1, x1 if x0 > x1: x0, x1 = x1, x0 y0, y1 = y1, y0 if y0 < y1: ystep = 1 else: ystep = -1 deltax = x1 - x0 deltay = abs(y1 - y0) error = -deltax / 2 y = y0 line = [] for x in range(x0, x1 + 1): if steep: line.append((y,x)) else: line.append((x,y)) error = error + deltay if error > 0: y = y + ystep error = error - deltax return line

    Read the article

  • Copying contents of a module

    - by Hulk
    If there exists an old data of a model say , query=Emp.objects.filter(pk=profile.id) Is there a easier way to copy the same values into the same model again.. Now that the id will be different so.. I have this requirement. Thanks..

    Read the article

  • Rewriting An URL With Regular Expression Substitution in Routes

    - by Sean M
    In my Pylons app, some content is located at URLs that look like http://mysite/data/31415. Users can go to that URL directly, or search for "31415" via the search page. My constraints, however, mean that http://mysite/data/000031415 should go to the same page as the above, as should searches for "0000000000031415." Can I strip leading zeroes from that string in Routes itself, or do I need to do that substitution in the controller file? If it's possible to do it in routing.py, I'd rather do it there - but I can't quite figure it out from the documentation that I'm reading.

    Read the article

  • Urllib's urlopen breaking on some sites (e.g. StackApps api)

    - by Edan Maor
    I'm using urllib2's urlopen function to try and get a JSON result from the StackOverflow api. The code I'm using: >>> import urllib2 >>> conn = urllib2.urlopen("http://api.stackoverflow.com/0.8/users/") >>> conn.readline() The result I'm getting: '\x1f\x8b\x08\x00\x00\x00\x00\x00\x04\x00\xed\xbd\x07`\x1cI\x96%&/m\xca{\x7fJ\... I'm fairly new to urllib, but this doesn't seem like the result I should be getting. I've tried it in other places and I get what I expect (the same as visiting the address with a browser gives me: a JSON object). Using urlopen on other sites (e.g. "http://google.com") works fine, and gives me actual html. I've also tried using urllib and it gives the same result. I'm pretty stuck, not even knowing where to look to solve this problem. Any ideas?

    Read the article

  • PycURL RESUME_FROM

    - by excid3
    I can't seem to get the RESUME_FROM option to work. Here's some example code that I have been testing with: import os import pycurl import sys def progress(total, existing, upload_t, upload_d): try: frac = float(existing)/float(total) except: frac = 0 sys.stdout.write("\r%s %3i%%" % ("file", frac*100) ) url = "http://launchpad.net/keryx/stable/0.92/+download/keryx_0.92.4.tar.gz" filename = url.split("/")[-1].strip() def test(debug_type, debug_msg): print "debug(%d): %s" % (debug_type, debug_msg) c = pycurl.Curl() c.setopt(pycurl.URL, url) c.setopt(pycurl.FOLLOWLOCATION, 1) c.setopt(pycurl.MAXREDIRS, 5) # Setup writing if os.path.exists(filename): f = open(filename, "ab") c.setopt(pycurl.RESUME_FROM, os.path.getsize(filename)) else: f = open(filename, "wb") c.setopt(pycurl.WRITEDATA, f) #c.setopt(pycurl.VERBOSE, 1) c.setopt(pycurl.DEBUGFUNCTION, test) c.setopt(pycurl.NOPROGRESS, 0) c.setopt(pycurl.PROGRESSFUNCTION, progress) c.perform()

    Read the article

  • Using urllib2 with SOCKS proxy

    - by roddik
    Hello. Is it possible to fetch pages with urllib2 through a SOCKS proxy on a one socks server per opener basic? I've seen the solution using setdefaultproxy method, but I need to have different socks in different openers.

    Read the article

  • (interactive) graph as in graph theory on a web page ?

    - by LB
    Hi, I have to integrate a graph with nodes and edges on a web page. Ideally, i would like to be able to interact with it (like moving the nodes around). Actually, i'm beginning by representing trees, so i would appreciate to be able to collapse subtrees. How can I do that ? I was considering google-visualization api but i wasn't able to find the kind of visualization i'm looking for (org chart doesn't allow to have multiple fathers, if i understood well) I've got no idea of the kind of technology so my tagging may not be really accurate :-). thanks

    Read the article

  • Django and Reportlab Question

    - by Hellnar
    Hello, I have written this small Django view to return pdf. @login_required def code_view(request,myid): try: deal = Deal.objects.get(id=myid) except: raise Http404 header = deal.header code = deal.code response = HttpResponse(mimetype='application/pdf') response['Content-Disposition'] = 'attachment; filename=code.pdf' p = canvas.Canvas(response) p.drawString(10, 800, header) p.drawString(10, 700, code) p.showPage() p.save() return response And my questions: Utf-8 characters are not shown correctly within the pdf. How can I include an image ? How can I include a very basic html such as: . <ul> <li>List One</li> <li>List Two</li> <li>List Three</li> </ul>

    Read the article

  • Django template-printing variables

    - by Hulk
    In django views def add(request): dict{} co_data = optarr dict.update({'co_data' : co_data}) logging.debug(co_data) return render_to_response('scheme/create.html',context_instance=RequestContext(request,{'dict': dict})) And data has the following string 1##2##3##4## And in the template when i say {{co_data}} it doesnt display the values.Please point out whats wrong in the code. Thanks..

    Read the article

  • Py2exe, PyQt4 and Postgre Driver (QPSQL)

    - by Marshall
    Hi, I`m trying to freeze my application using Py2exe. My app uses PyQt4 and it apparently works fine with py2exe. But once I`ve uninstalled PyQt, it shows the following error: QSqlDatabase: QPSQL driver not loaded QSqlDatabase: available driver: QPSQL7 QPSQL Which doesn't make sense at all. If PyQt4 is still installed, it works just fine. This is my py2exe parameters: data_files = [ ('sqldrivers', [ 'C:\Python26\Lib\site-packages\PyQt4\plugins\sqldrivers\qsqlpsql4.dll' ]) ] setup(console=["delivery.py"], options={"py2exe" : {"includes" : ["sip", "PyQt4.QtSql", "PyQt4.QtWebKit", "PyQt4.QtNetwork"]}}, data_files=data_files)

    Read the article

  • Django Admin interface with pickled set

    - by Rosarch
    I have a model that has a pickled set of strings. (It has to be pickled, because Django has no built in set field, right?) class Foo(models.Model): __bar = models.TextField(default=lambda: cPickle.dumps(set()), primary_key=True) def get_bar(self): return cPickle.loads(str(self.__bar)) def set_bar(self, values): self.__bar = cPickle.dumps(values) bar = property(get_bar, set_bar) I would like the set to be editable in the admin interface. Obviously the user won't be working with the pickled string directly. Also, the interface would need a widget for adding/removing strings from a set. What is the best way to go about doing this? I'm not super familiar with Django's admin system. Do I need to build a custom admin widget or something? Update: If I do need a custom widget, this looks helpful: http://www.fictitiousnonsense.com/archives/22

    Read the article

  • django-admin - how to modify ModelAdmin to create multiple objects at once?

    - by skrobul
    let's assume that I have very basic model class Message(models.Model): msg = models.CharField(max_length=30) this model is registered with admin module: class MessageAdmin(admin.ModelAdmin): pass admin.site.register(Message, MessageAdmin) Currently when I go into the admin interface, after clicking "Add message" I have only one form where I can enter the msg. I would like to have multiple forms (formset perhaps) on the "Add page" so I can create multiple messages at once. It's really annoying having to click "Save and add another" every single time. Ideally I would like to achieve something like InlineModelAdmin but it turns out that you can use it only for the models that are related to the object which is edited. What would you recommend to use to resolve this problem?

    Read the article

< Previous Page | 361 362 363 364 365 366 367 368 369 370 371 372  | Next Page >