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  • How to render a POST and make it show up on another page

    - by stack5914
    I'm trying to create a marketplace website similar to craigslist. I created a form according to the Django tutorial "Working with forms", but I don't know how to render information I got from the POST forms. I want to make information(subject,price...etc) that I got from POST show up on another page like this. http://bakersfield.craigslist.org/atq/3375938126.html and, I want the "Subject"(please look at form.py) of this product(eg.1960 French Chair) to show up on another page like this. http://bakersfield.craigslist.org/ata/ } Can I get some advice to handle submitted information? Here's present codes. I'll appreciate all your answers and helps. <-! Here's my codes -- ?forms.py from django import forms class SellForm(forms.Form): subject = forms.CharField(max_length=100) price = forms.CharField(max_length=100) condition = forms.CharField(max_length=100) email = forms.EmailField() body = forms.TextField() ?views.py from django.shortcuts import render, render_to_response from django.http import HttpResponseRedirect from site1.forms import SellForm def sell(request): if request.method =="POST": form =SellForm(request.POST) if form.is_valid(): subject = form.cleaned_data['subject'] price = form.cleaned_data['price'] condition = form.cleaned_data['condition'] email = form.cleaned_data['email'] body = form.cleaned_data['body'] return HttpResponseRedirect('/books/') else: form=SellForm() render(request, 'sell.html',{'form':form,}) ?urls.py from django.conf.urls import patterns, include, url from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', url(r'^sechand/$','site1.views.sell'), url(r'^admin/', include(admin.site.urls)), ) ?sell.html <form action = "/sell/" method = "post">{% csrf_token%} {{ form.as_p }} <input type = "submit" value="Submit" /> </form>

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  • How to stop Django from adding extra html elements to rendered widgets.

    - by stinkypyper
    I have a Django radio button group that renders to HTML as follows: <ul> <li><label for="id_package_id_0"><input type="radio" id="id_package_id_0" value="1" name="package_id" /> Test 256</label></li> <li><label for="id_package_id_1"><input type="radio" id="id_package_id_1" value="2" name="package_id" /> Test 384</label></li> <li><label for="id_package_id_2"><input type="radio" id="id_package_id_2" value="3" name="package_id" /> Test 512</label></li> <li><label for="id_package_id_3"><input type="radio" id="id_package_id_3" value="4" name="package_id" /> Test 768</label></li> <li><label for="id_package_id_4"><input type="radio" id="id_package_id_4" value="5" name="package_id" /> Test 1024</label></li> </ul> I need it to render without being a list. I am a aware of form.as_p, form.as_table, and form.as_ul. They will not help me as they continue to add extra HTML tags. As well, I am not using the form object in it's absolute entirety, just for validation. I am doing a custom template for the form already, but wish to continue to the radio widget.

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  • Django manager for _set in model

    - by Daniel Johansson
    Hello, I'm in the progress of learning Django at the moment but I can't figure out how to solve this problem on my own. I'm reading the book Developers Library - Python Web Development With Django and in one chapter you build a simple CMS system with two models (Story and Category), some generic and custom views together with templates for the views. The book only contains code for listing stories, story details and search. I wanted to expand on that and build a page with nested lists for categories and stories. - Category1 -- Story1 -- Story2 - Category2 - Story3 etc. I managed to figure out how to add my own generic object_list view for the category listing. My problem is that the Story model have STATUS_CHOICES if the Story is public or not and a custom manager that'll only fetch the public Stories per default. I can't figure out how to tell my generic Category list view to also use a custom manager and only fetch the public Stories. Everything works except that small problem. I'm able to create a list for all categories with a sub list for all stories in that category on a single page, the only problem is that the list contains non public Stories. I don't know if I'm on the right track here. My urls.py contains a generic view that fetches all Category objects and in my template I'm using the *category.story_set.all* to get all Story objects for that category, wich I then loop over. I think it would be possible to add a if statement in the template and use the VIEWABLE_STATUS from my model file to check if it should be listed or not. The problem with that solution is that it's not very DRY compatible. Is it possible to add some kind of manager for the Category model too that only will fetch in public Story objects when using the story_set on a category? Or is this the wrong way to attack my problem? Related code urls.py (only category list view): urlpatterns += patterns('django.views.generic.list_detail', url(r'^categories/$', 'object_list', {'queryset': Category.objects.all(), 'template_object_name': 'category' }, name='cms-categories'), models.py: from markdown import markdown import datetime from django.db import models from django.db.models import permalink from django.contrib.auth.models import User VIEWABLE_STATUS = [3, 4] class ViewableManager(models.Manager): def get_query_set(self): default_queryset = super(ViewableManager, self).get_query_set() return default_queryset.filter(status__in=VIEWABLE_STATUS) class Category(models.Model): """A content category""" label = models.CharField(blank=True, max_length=50) slug = models.SlugField() class Meta: verbose_name_plural = "categories" def __unicode__(self): return self.label @permalink def get_absolute_url(self): return ('cms-category', (), {'slug': self.slug}) class Story(models.Model): """A hunk of content for our site, generally corresponding to a page""" STATUS_CHOICES = ( (1, "Needs Edit"), (2, "Needs Approval"), (3, "Published"), (4, "Archived"), ) title = models.CharField(max_length=100) slug = models.SlugField() category = models.ForeignKey(Category) markdown_content = models.TextField() html_content = models.TextField(editable=False) owner = models.ForeignKey(User) status = models.IntegerField(choices=STATUS_CHOICES, default=1) created = models.DateTimeField(default=datetime.datetime.now) modified = models.DateTimeField(default=datetime.datetime.now) class Meta: ordering = ['modified'] verbose_name_plural = "stories" def __unicode__(self): return self.title @permalink def get_absolute_url(self): return ("cms-story", (), {'slug': self.slug}) def save(self): self.html_content = markdown(self.markdown_content) self.modified = datetime.datetime.now() super(Story, self).save() admin_objects = models.Manager() objects = ViewableManager() category_list.html (related template): {% extends "cms/base.html" %} {% block content %} <h1>Categories</h1> {% if category_list %} <ul id="category-list"> {% for category in category_list %} <li><a href="{{ category.get_absolute_url }}">{{ category.label }}</a></li> {% if category.story_set %} <ul> {% for story in category.story_set.all %} <li><a href="{{ story.get_absolute_url }}">{{ story.title }}</a></li> {% endfor %} </ul> {% endif %} {% endfor %} </ul> {% else %} <p> Sorry, no categories at the moment. </p> {% endif %} {% endblock %}

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  • Django ImageField issue with JPEG's

    - by Kieran Lynn
    I am having a major issue with PIL (Python Image Library) in Django and have jumpped through a lot of hoops and have thus far not been able to figure out what the root of the issue is. The problem essentially breaks down to not being able to upload JPEG images through the ImageField in the Django admin. But the issue is not as simple as installing libjpeg. First, I installed PIL (through Buildout) and realized once it was installed that I had not installed libjpeg because JPEG support was not available. Having not setup the server myself, I just assumed that it was not installed and I compiled libjpeg 8 from the source. This ended up in my /usr/local/lib/ directory. I cleared out my Buildout files and rebuilt everything. This time when PIL compiled I had JPEG support. But I went to the Django Admin and tried to upload a JPEG though an ImageField with no luck. I got the "Upload a valid image. The file you uploaded was either not an image or a corrupted image" error. Just as a test I opened up a the Djano shell and ran the following: > import Image > i = Image.open( "/absolute_path/file.jpg" ) > print i <JpegImagePlugin.JpegImageFile image mode=RGB size=940x375 at 0x7F908C529BD8> This runs with no errors and shows that PIL is able to open JPEG's. After doing some reading, I come across this thread: Is it possible to control which libraries apache uses? Looks like PHP also uses libjpeg and is loading before Django, and therefor loading libjpeg 6.2 before. This is show when using lsof: COMMAND PID USER FD TYPE DEVICE SIZE/OFF NODE NAME apache2 2561 www-data mem REG 202,1 146032 639276 /usr/lib/libjpeg.so.62.0.0 So my thought is that I should be using libjpeg 6.2. So I removed libjpeg located in my /usr/local/lib directory. After rereading the PIL installation instructions, I realized that I might not have the dev/header files for libjpeg that PIL needs. So I also uninstalled libjpeg using the aptitude uninstaller (sudo aptitude remove libjpeg62). Then to ensure that I got the header files that PIL needed I installed libjpeg using aptitude: (sudo aptget install libjpeg62-dev). From here I cleaned out my Buildout directory, and reran Buildout, which in turn reinstalled PIL. Once again, I have JPEG support, now using the libjpeg62. So I go to test in the Django Admin. Still no JPEG support. So I wanted to test JPEG support in general and see if the exception was not handled, what kind of error it would throw. So in my homepage view I added the following code to open a JPEG image: import Image i = Image.open( "/absolute_path/file.jpg" ) v = i.verify() Then I pass i to the HTML view just to easily see the output. I deploy these changes to the server and restart. I am surprised not to see an error and get the following output: {{ i }} - <JpegImagePlugin.JpegImageFile image mode=RGB size=940x375 at 0x7F908C529BD8> {{ v }} - None So at this point I am really confused: Why can I successfully open a JPEG while the admin cannot? Am I missing something, is this not an issue with libjpeg? If not an issue with libjpeg, why can I upload a PNG with no issues? Any help would be much appreciated, I have been on this for 2 days debugging with no luck. Setup: 1. Rackspace Cloud Server 2. Ubuntu 10.04 3. Django 1.2.3 (Installed though Buildout) 4. PIL 1.1.7 (Installed though Buildout) 5. libjpeg 6.2 (installed through aptitude (sudo aptget install libjpeg62-dev)

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  • Problem trying to achieve a join using the `comments` contrib in Django

    - by NiKo
    Hi, Django rookie here. I have this model, comments are managed with the django_comments contrib: class Fortune(models.Model): author = models.CharField(max_length=45, blank=False) title = models.CharField(max_length=200, blank=False) slug = models.SlugField(_('slug'), db_index=True, max_length=255, unique_for_date='pub_date') content = models.TextField(blank=False) pub_date = models.DateTimeField(_('published date'), db_index=True, default=datetime.now()) votes = models.IntegerField(default=0) comments = generic.GenericRelation( Comment, content_type_field='content_type', object_id_field='object_pk' ) I want to retrieve Fortune objects with a supplementary nb_comments value for each, counting their respectve number of comments ; I try this query: >>> Fortune.objects.annotate(nb_comments=models.Count('comments')) From the shell: >>> from django_fortunes.models import Fortune >>> from django.db.models import Count >>> Fortune.objects.annotate(nb_comments=Count('comments')) [<Fortune: My first fortune, from NiKo>, <Fortune: Another One, from Dude>, <Fortune: A funny one, from NiKo>] >>> from django.db import connection >>> connection.queries.pop() {'time': '0.000', 'sql': u'SELECT "django_fortunes_fortune"."id", "django_fortunes_fortune"."author", "django_fortunes_fortune"."title", "django_fortunes_fortune"."slug", "django_fortunes_fortune"."content", "django_fortunes_fortune"."pub_date", "django_fortunes_fortune"."votes", COUNT("django_comments"."id") AS "nb_comments" FROM "django_fortunes_fortune" LEFT OUTER JOIN "django_comments" ON ("django_fortunes_fortune"."id" = "django_comments"."object_pk") GROUP BY "django_fortunes_fortune"."id", "django_fortunes_fortune"."author", "django_fortunes_fortune"."title", "django_fortunes_fortune"."slug", "django_fortunes_fortune"."content", "django_fortunes_fortune"."pub_date", "django_fortunes_fortune"."votes" LIMIT 21'} Below is the properly formatted sql query: SELECT "django_fortunes_fortune"."id", "django_fortunes_fortune"."author", "django_fortunes_fortune"."title", "django_fortunes_fortune"."slug", "django_fortunes_fortune"."content", "django_fortunes_fortune"."pub_date", "django_fortunes_fortune"."votes", COUNT("django_comments"."id") AS "nb_comments" FROM "django_fortunes_fortune" LEFT OUTER JOIN "django_comments" ON ("django_fortunes_fortune"."id" = "django_comments"."object_pk") GROUP BY "django_fortunes_fortune"."id", "django_fortunes_fortune"."author", "django_fortunes_fortune"."title", "django_fortunes_fortune"."slug", "django_fortunes_fortune"."content", "django_fortunes_fortune"."pub_date", "django_fortunes_fortune"."votes" LIMIT 21 Can you spot the problem? Django won't LEFT JOIN the django_comments table with the content_type data (which contains a reference to the fortune one). This is the kind of query I'd like to be able to generate using the ORM: SELECT "django_fortunes_fortune"."id", "django_fortunes_fortune"."author", "django_fortunes_fortune"."title", COUNT("django_comments"."id") AS "nb_comments" FROM "django_fortunes_fortune" LEFT OUTER JOIN "django_comments" ON ("django_fortunes_fortune"."id" = "django_comments"."object_pk") LEFT OUTER JOIN "django_content_type" ON ("django_comments"."content_type_id" = "django_content_type"."id") GROUP BY "django_fortunes_fortune"."id", "django_fortunes_fortune"."author", "django_fortunes_fortune"."title", "django_fortunes_fortune"."slug", "django_fortunes_fortune"."content", "django_fortunes_fortune"."pub_date", "django_fortunes_fortune"."votes" LIMIT 21 But I don't manage to do it, so help from Django veterans would be much appreciated :) Hint: I'm using Django 1.2-DEV Thanks in advance for your help.

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  • Permissions restoring from Time Machine - Finder copy vs "cp" copy

    - by Ben Challenor
    Note: this question was starting to sprawl so I rewrote it. I have a folder that I'm trying to restore from a Time Machine backup. Using cp -R works fine, but certain folders cannot be restored with either the Time Machine UI or Finder. Other users have reported similar errors and the cp -R workaround was suggested (e.g. Restoring from Time Machine - Permissions Error). But I wanted to understand: Why cp -R works when the Finder and the Time Machine UI do not. Whether I could prevent the errors by changing file permissions before the backup. There do indeed seem to be some permissions that Finder works with and some that it does not. I've narrowed the errors down to folders with the user ben (that's me) and the group wheel. Here's a simplified reproduction. I have four folders with the owner/group combinations I've seen so far: ben ~/Desktop/test $ ls -lea total 16 drwxr-xr-x 7 ben staff 238 27 Nov 14:31 . drwx------+ 17 ben staff 578 27 Nov 14:29 .. 0: group:everyone deny delete -rw-r--r--@ 1 ben staff 6148 27 Nov 14:31 .DS_Store drwxr-xr-x 3 ben staff 102 27 Nov 14:30 ben-staff drwxr-xr-x 3 ben wheel 102 27 Nov 14:30 ben-wheel drwxr-xr-x 3 root admin 102 27 Nov 14:31 root-admin drwxr-xr-x 3 root wheel 102 27 Nov 14:31 root-wheel Each contains a single file called file with the same owner/group: ben ~/Desktop/test $ cd ben-staff ben ~/Desktop/test/ben-staff $ ls -lea total 0 drwxr-xr-x 3 ben staff 102 27 Nov 14:30 . drwxr-xr-x 7 ben staff 238 27 Nov 14:31 .. -rw-r--r-- 1 ben staff 0 27 Nov 14:30 file In the backup, they look like this: ben /Volumes/Deimos/Backups.backupdb/Ben’s MacBook Air/Latest/Macintosh HD/Users/ben/Desktop/test $ ls -leA total 16 -rw-r--r--@ 1 ben staff 6148 27 Nov 14:34 .DS_Store 0: group:everyone deny write,delete,append,writeattr,writeextattr,chown drwxr-xr-x@ 3 ben staff 102 27 Nov 14:51 ben-staff 0: group:everyone deny add_file,delete,add_subdirectory,delete_child,writeattr,writeextattr,chown drwxr-xr-x@ 3 ben wheel 102 27 Nov 14:51 ben-wheel 0: group:everyone deny add_file,delete,add_subdirectory,delete_child,writeattr,writeextattr,chown drwxr-xr-x@ 3 root admin 102 27 Nov 14:52 root-admin 0: group:everyone deny add_file,delete,add_subdirectory,delete_child,writeattr,writeextattr,chown drwxr-xr-x@ 3 root wheel 102 27 Nov 14:52 root-wheel 0: group:everyone deny add_file,delete,add_subdirectory,delete_child,writeattr,writeextattr,chown Of these, ben-staff can be restored with Finder without errors. root-wheel and root-admin ask for my password and then restore without errors. But ben-wheel does not prompt for my password and gives the error: The operation can’t be completed because you don’t have permission to access “file”. Interestingly, I can restore the file from this folder by dragging it directly to my local drive (instead of dragging its parent folder), but when I do so its permissions are changed to ben/staff. Here are the permissions after the restore for the three folders that worked correctly, and the file from ben-wheel that was changed to ben/staff. ben ~/Desktop/test-restore $ ls -leA total 16 -rw-r--r--@ 1 ben staff 6148 27 Nov 14:46 .DS_Store drwxr-xr-x 3 ben staff 102 27 Nov 14:30 ben-staff -rw-r--r-- 1 ben staff 0 27 Nov 14:30 file drwxr-xr-x 3 root admin 102 27 Nov 14:31 root-admin drwxr-xr-x 3 root wheel 102 27 Nov 14:31 root-wheel Can anyone explain this behaviour? Why do Finder and the Time Machine UI break with the ben / wheel permissions? And why does cp -R work (even without sudo)?

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  • How to setup linux permissions the WWW folder?

    - by Xeoncross
    Updated Summery The /var/www directory is owned by root:root which means that no one can use it and it's entirely useless. Since we all want a web server that actually works (and no-one should be logging in as "root"), then we need to fix this. Only two entities need access. PHP/Perl/Ruby/Python all need access to the folders and files since they create many of them (i.e. /uploads/). These scripting languages should be running under nginx or apache (or even some other thing like FastCGI for PHP). The developers How do they get access? I know that someone, somewhere has done this before. With however-many billions of websites out there you would think that there would be more information on this topic. I know that 777 is full read/write/execute permission for owner/group/other. So this doesn't seem to be needed as it leaves random users full permissions. What permissions are need to be used on /var/www so that... Source control like git or svn Users in a group like "websites" (or even added to "www-data") Servers like apache or lighthttpd And PHP/Perl/Ruby can all read, create, and run files (and directories) there? If I'm correct, Ruby and PHP scripts are not "executed" directly - but passed to an interpreter. So there is no need for execute permission on files in /var/www...? Therefore, it seems like the correct permission would be chmod -R 1660 which would make all files shareable by these four entities all files non-executable by mistake block everyone else from the directory entirely set the permission mode to "sticky" for all future files Is this correct? Update: I just realized that files and directories might need different permissions - I was talking about files above so i'm not sure what the directory permissions would need to be. Update 2: The folder structure of /var/www changes drastically as one of the four entities above are always adding (and sometimes removing) folders and sub folders many levels deep. They also create and remove files that the other 3 entities might need read/write access to. Therefore, the permissions need to do the four things above for both files and directories. Since non of them should need execute permission (see question about ruby/php above) I would assume that rw-rw-r-- permission would be all that is needed and completely safe since these four entities are run by trusted personal (see #2) and all other users on the system only have read access. Update 3: This is for personal development machines and private company servers. No random "web customers" like a shared host. Update 4: This article by slicehost seems to be the best at explaining what is needed to setup permissions for your www folder. However, I'm not sure what user or group apache/nginx with PHP OR svn/git run as and how to change them. Update 5: I have (I think) finally found a way to get this all to work (answer below). However, I don't know if this is the correct and SECURE way to do this. Therefore I have started a bounty. The person that has the best method of securing and managing the www directory wins.

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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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  • Django + FastCGI - randomly raising OperationalError

    - by ibz
    I'm running a Django application. Had it under Apache + mod_python before, and it was all OK. Switched to Lighttpd + FastCGI. Now I randomly get the following exception (neither the place nor the time where it appears seem to be predictable). Since it's random, and it appears only after switching to FastCGI, I assume it has something to do with some settings. Found a few results when googleing, but they seem to be related to setting maxrequests=1. However, I use the default, which is 0. Any ideas where to look for? PS. I'm using PostgreSQL. Might be related to that as well, since the exception appears when making a database query. Thanks. File "/usr/lib/python2.6/site-packages/django/core/handlers/base.py", line 86, in get_response response = callback(request, *callback_args, **callback_kwargs) File "/usr/lib/python2.6/site-packages/django/contrib/admin/sites.py", line 140, in root if not self.has_permission(request): File "/usr/lib/python2.6/site-packages/django/contrib/admin/sites.py", line 99, in has_permission return request.user.is_authenticated() and request.user.is_staff File "/usr/lib/python2.6/site-packages/django/contrib/auth/middleware.py", line 5, in __get__ request._cached_user = get_user(request) File "/usr/lib/python2.6/site-packages/django/contrib/auth/__init__.py", line 83, in get_user user_id = request.session[SESSION_KEY] File "/usr/lib/python2.6/site-packages/django/contrib/sessions/backends/base.py", line 46, in __getitem__ return self._session[key] File "/usr/lib/python2.6/site-packages/django/contrib/sessions/backends/base.py", line 172, in _get_session self._session_cache = self.load() File "/usr/lib/python2.6/site-packages/django/contrib/sessions/backends/db.py", line 16, in load expire_date__gt=datetime.datetime.now() File "/usr/lib/python2.6/site-packages/django/db/models/manager.py", line 93, in get return self.get_query_set().get(*args, **kwargs) File "/usr/lib/python2.6/site-packages/django/db/models/query.py", line 304, in get num = len(clone) File "/usr/lib/python2.6/site-packages/django/db/models/query.py", line 160, in __len__ self._result_cache = list(self.iterator()) File "/usr/lib/python2.6/site-packages/django/db/models/query.py", line 275, in iterator for row in self.query.results_iter(): File "/usr/lib/python2.6/site-packages/django/db/models/sql/query.py", line 206, in results_iter for rows in self.execute_sql(MULTI): File "/usr/lib/python2.6/site-packages/django/db/models/sql/query.py", line 1734, in execute_sql cursor.execute(sql, params) OperationalError: server closed the connection unexpectedly This probably means the server terminated abnormally before or while processing the request.

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  • 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.

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  • Django and App Engine

    - by notnoop
    I wanted to check the status of running Django on the Google App Engine currently and what the benefits of running django on GAE over simply using Webapp. Django main killer feature, IMHO, is the reuseable apps and middleware. Unfortunately, most current Django apps use models or model forms (django-tags, django-reviews, django-profiles, Pinax apps). So what are the remaining features or benefits that django has that can still run in Google App Engine (other than what's disabled: the popular django apps, session and authentication middleware, users and admin, models, etc). Also, is there a list of the Django apps that work in App Engine as well?

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  • review on django book vs django tutorial

    - by momo
    going through both the django book and tutorial, am a bit confused to the differences in approach (aren't they both written by the same people?) can anyone who has experience in both give a short review on them? i have decent python skills (largely untested though), but no experience at all in web apps and am trying to decide which one to stick to. i briefly looked in to practical django projects but that was a bit too complicated for me, my background is primarily bash scripting, the python i know i learned from an instant hacking tutorial and diving into python.

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  • Django: How do I position a page when using Django templates

    - by swisstony
    I have a web page where the user enters some data and then clicks a submit button. I process the data and then use the same Django template to display the original data, the submit button, and the results. When I am using the Django template to display results, I would like the page to be automatically scrolled down to the part of the page where the results begin. This allows the user to scroll back up the page if she wants to change her original data and click submit again. Hopefully, there's some simple way of doing this that I can't see at the moment.

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  • Modify Django Forms

    - by Ninefingers
    Hi All, I've recently been developing on the django platform and have stumbled upon Django Forms (forms.Form/forms.ModelForm) as ways of creating <form> html. Now, this is brilliant for quick stuff but what I'm trying to do is a little bit more complicated. Consider a DateField - my current form has fields for day, month and year and constructs a python date object from that. However, a django form creates a single textbox in which the correct format (say 2010-06-15) must be entered. As another example, for large fields I need to replace <input> with <textarea>. I'd like to take advantage of Django's forms for simple validation but I need something simpler for my users. So my question is: can I intercept the rendering of one of these objects to write out the html as I like? If so, do I have to do all the writing myself or can I only do those objects I wish to re-write? Thanks in advance.

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  • Django Form Preview

    - by Mark Kecko
    I'm trying to use django's FormPreview and I can't get it to work properly. Here's my code: forms.py class MyForm(forms.ModelForm): status = forms.TypedChoiceField( coerce=int, choices=LIST_STATUS, label="type", widget=forms.RadioSelect ) description = forms.CharField(widget = forms.Textarea) stage = forms.CharField() def __init__(self, useradd=None, *args, **kwargs): super(MyForm, self).__init__(*args, **kwargs) self.fields['firm'].label = "Firm" class Meta: model = MyModel fields = ['status', 'description', 'stage'] class MyFormPreview(FormPreview): form_template = 'templates/post.html' preview_template = 'templates/review.html' def process_preview(self, request, cleaned_data): print "processed" def done(self, request, cleaned_data): print "done" # Do something with the cleaned_data, then redirect # to a "success" page. return HttpResponseRedirect('/') urls.py (r'^post/$', MyFormPreview(MyForm)), post.html <form id = "post_ad" action = "" method = "POST" enctype="multipart/form-data"> <table> {{form.as_table}} </table> <input type="submit" name="save" value="Post" /> </form> When I go to /post/ I get the correct form and I fill it out. When I submit the form it goes right back to /post/ but but there are no errors (I've tried displaying {{errors}}) and the form is empty. None of my print statements execute. I'm not sure what I'm missing. Can anyone help me out? I can't find any documentation besides what's on the django site. Also, what's the "preview" variable called that I should use in my preview.html template? {{preview}} or do I just do {{form}} again? -- Answered below. I tried adding 'django.contrib.formtools' to my installed_apps in settings and I tried using the code from the default form templates from django.contrib as suggested below. Still, when I submit the form I go right back to the post template, none of my print statements execute :(

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  • Error in {% markdown %} filter in Django Nonrel

    - by Robert Smith
    I'm having trouble using Markdown in Django Nonrel. I followed this instructions (added 'django.contrib.markup' to INSTALLED_APPS, include {% load markup %} in the template and use |markdown filter after installing python-markdown) but I get the following error: Error in {% markdown %} filter: The Python markdown library isn't installed. In this line: /path/to/project/django/contrib/markup/templatetags/markup.py in markdown they will be silently ignored. """ try: import markdown except ImportError: if settings.DEBUG: raise template.TemplateSyntaxError("Error in {% markdown %} filter: The Python markdown library isn't installed.") ... return force_unicode(value) else: # markdown.version was first added in 1.6b. The only version of markdown # to fully support extensions before 1.6b was the shortlived 1.6a. if hasattr(markdown, 'version'): extensions = [e for e in arg.split(",") if e] It seems obvious that import markdown is causing the problem but when I run: $ python manage.py shell >>> import elementtree >>> import markdown everthing works alright. Running Markdown 2.0.3, Django 1.3.1, Python 2.7. UPDATE: I thought maybe this was an issue related to permissions, so I changed my project via chmod 777 -R, but it didn't work. Ideas? Thanks!

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  • Django Template tag, generating template block tag

    - by Issy
    Hi Guys, Currently a bit stuck, wondering if anyone can assist. I am using django-adminfiles. Which is a near little application. I want to use it to insert images into posts/articles/pages for a site i am building. How django-adminfiles works is it inserts a placeholder i.e <<< ImageFile and this gets rendered using a django template. It also has the feature of inserting custom options i.e (Insert Medium Image) , i figured i would used this to automatically resize images and include it in the post (similar to how WP does it). Django-adminfiles makes use of sorl.thumbnail app to generate thumbnails. So i have tried testing generating thumbnails: The current template that is used to render the inserted image is: {% spaceless %} <img src="{{ upload.upload.url }}" width="{{ upload.width }}" height="{{ upload.height }}" class="{{ options.class }}" class="{{ options.size }}" alt="{% if options.alt %}{{ options.alt }}{% else %}{{ upload.title }}{% endif %}" /> {% endspaceless %} I tried modifying this to: {% load thumbnail %} {% spaceless %} <img src="{% thumbnail upload.upload.url 200x50 %}" width="{{ upload.width }}" height="{{ upload.height }}" class="{{ options.class }}" class="{{ options.size }}" alt="{% if options.alt %}{{ options.alt }}{% else %}{{ upload.title }}{% endif %}" /> {% endspaceless %} I get the error: Exception Value: Caught an exception while rendering: Source file: '/media/uploads/DSC_0014.jpg' does not exist. I figured the thumbnail needs the absolute path so tried putting that in the template, and that works. i.e this works: {% thumbnail '/Users/me/media/uploads/DSC_0014.jpg' 200x50 %} So basically i need to generate the absolute path to the file give the relative path (to web root). You could do this by passing the MEDIA_ROOT setting to the template, but the reason i want to do a template tag is to programmatically set the image size.

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  • Django-South introspection rule doesn't work.

    - by Ory Band
    I'm using Django 1.2.3 and South 0.7.3. I am trying to convert my app (named core) to use Django-South. I have a custom model/field that I'm using, named ImageWithThumbsField. It's basically just the ol' django.db.models.ImageField with some attributes such as height, weight, etc. While trying to ./manage.py convert_to_auth core I receieve South's freezing errors. I have no idea why, I'm Probably missing something... I am using a simple custom Model: from django.db.models import ImageField class ImageWithThumbsField(ImageField): def __init__(self, verbose_name=None, name=None, width_field=None, height_field=None, sizes=None, **kwargs): self.verbose_name=verbose_name self.name=name self.width_field=width_field self.height_field=height_field self.sizes = sizes super(ImageField, self).__init__(**kwargs) And this is my introspection rule, which I add to the top of my models.py: from south.modelsinspector import add_introspection_rules from lib.thumbs import ImageWithThumbsField add_introspection_rules( [ ( (ImageWithThumbsField, ), [], { "verbose_name": ["verbose_name", {"default": None}], "name": ["name", {"default": None}], "width_field": ["width_field", {"default": None}], "height_field": ["height_field", {"default": None}], "sizes": ["sizes", {"default": None}], }, ), ], ["^core/.fields/.ImageWithThumbsField",]) This is the errors I receieve: ! Cannot freeze field 'core.additionalmaterialphoto.photo' ! (this field has class lib.thumbs.ImageWithThumbsField) ! Cannot freeze field 'core.material.photo' ! (this field has class lib.thumbs.ImageWithThumbsField) ! Cannot freeze field 'core.material.formulaimage' ! (this field has class lib.thumbs.ImageWithThumbsField) ! South cannot introspect some fields; this is probably because they are custom ! fields. If they worked in 0.6 or below, this is because we have removed the ! models parser (it often broke things). ! To fix this, read http://south.aeracode.org/wiki/MyFieldsDontWork Does anybody know why? What am I doing wrong?

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  • Complex derived attributes in Django models

    - by rabidpebble
    What I want to do is implement submission scoring for a site with users voting on the content, much like in e.g. reddit (see the 'hot' function in http://code.reddit.com/browser/sql/functions.sql). My submission model currently keeps track of up and down vote totals. Currently, when a user votes I create and save a related Vote object and then use F() expressions to update the Submission object's voting totals. The problem is that I want to update the score for the submission at the same time, but F() expressions are limited to only simple operations (it's missing support for log(), date_part(), sign() etc.) From my limited experience with Django I can see 4 options here: extend F() somehow (haven't looked at the code yet) to support the missing SQL functions; this is my preferred option and seems to fit within the Django framework the best define a scoring function (much like reddit's 'hot' function) in my database, and have Django use the value of that function for the value of the score field; as far as I can tell, #2 is not possible wrap my two step voting process in a suitably isolated transaction so that I can calculate the voting totals in Python and then update the Submission's voting totals without fear that another vote against the submission could be added/changed in the meantime; I'm hesitant to take this route because it seems overly complex - what is a "suitably isolated transaction" in this case anyway? use raw SQL; I would prefer to avoid this entirely -- what's the point of an ORM if I have to revert to SQL for such a common use case as this! (Note that this coming from somebody who loves sprocs, but is using Django for ease of development.) Before I embark on this mission to extend F() (which I'm not sure is even possible), am I about to reinvent the wheel? Is there a more standard way to do this? It seems like such a common use case and yet in an hour of searching I have yet to find a common solution...

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  • Django throws 404 at generic views

    - by x0rg
    I'm trying to get the generic views for a date-based archive working in django. I defined the urls as described in a tutorial, but django returns a 404 error whenever I want to access an url with a variable (such as month or year) in it. It don't even produces a TemplateDoesNotExist-execption. Normal urls without variables work fine. Here's my urlconf: from django.conf.urls.defaults import * from zurichlive.zhl.models import Event info_dict = { 'queryset': Event.objects.all(), 'date_field': 'date', 'allow_future': 'True', } urlpatterns += patterns('django.views.generic.date_based', (r'events/(?P<year>d{4})/(?P<month>[a-z]{3})/(?P<day>w{1,2})/(?P<slug>[-w]+)/$', 'object_detail', dict(info_dict, slug_field='slug',template_name='archive/detail.html')), (r'^events/(?P<year>d{4})/(?P<month>[a-z]{3})/(?P<day>w{1,2})/(?P<slug>[-w]+)/$', 'object_detail', dict(info_dict, template_name='archive/list.html')), (r'^events/(?P<year>d{4})/(?P<month>[a-z]{3})/(?P<day>w{1,2})/$','archive_day',dict(info_dict,template_name='archive/list.html')), (r'^events/(?P<year>d{4})/(?P<month>[a-z]{3})/$','archive_month', dict(info_dict, template_name='archive/list.html')), (r'^events/(?P<year>)/$','archive_year', dict(info_dict, template_name='archive/list.html')), (r'^events/$','archive_index', dict(info_dict, template_name='archive/list.html')), ) When I access /events/2010/may/12/this-is-a-slug I should get to the detail.html template, but instead I get a 404. What am I doing wrong?

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  • WMI permissions: Select CommandLine, ProcessId FROM Win32_Process returns no data for CommandLine

    - by user57935
    Hi all, I am gathering performance data via WMI and would like to avoid having to use an account in the Administrators group for this purpose. The target machine is running Windows Server 2003 with the latest SP/updates. I've done what I believe to be the appropriate configuration to allow our user access to WMI (similar to what is described here: http://msdn.microsoft.com/en-us/library/aa393266.aspx). Here are the specific steps that were followed: Open Administrative Tools - Computer Management: Under Computer Management (Local) Expand Services and Applications, right click WMI Control and select properties. In the Security tab, expand Root, highlight CIMV2, click Security (near bottom of window); add Performance Monitor Users and enable the options : Enable Account and Remote Enable. ­Open Administrative Tools - Component Services: Under Console Root go to Component Services- Computers - Right click My Computer and select properties, select the COM security tab, in “Access Permissions” click "Edit Default" select(or add then select) “Performance Monitor Users” group and allow local access and remote access and click ok. In “Launch and Activation Permissions” click “Edit Default” select(or add then select) “Performance Monitor Users” group and allow Local and Remote Launch and Activation Permissions. ­Open Administrative Tools - Component Services: Under Console Root go to Component Services- Computers - My Computer - DCOM Config - highlight “Windows Management and Instrumentation” right click and select properties, Select the Security tab, Under “Launch and Activation Permissions” select Customize, then click edit, add the “Performance Users Group” and allow local and remote Remote Launch and Remote Activation privileges. I am able to connect remotely via WMI Explorer but when I perform this query: Select CommandLine, ProcessId FROM Win32_Process I get a valid result but every row has an empty CommandLine. If I add the user to the Administrators group and re-run the query, the CommandLine column contains the expected data. It seems there is a permission I am missing somewhere but I am not having much luck tracking it down. Many thanks in advance.

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  • WMI permissions: Select CommandLine, ProcessId FROM Win32_Process returns no data for CommandLine

    - by user57935
    I am gathering performance data via WMI and would like to avoid having to use an account in the Administrators group for this purpose. The target machine is running Windows Server 2003 with the latest SP/updates. I've done what I believe to be the appropriate configuration to allow our user access to WMI (similar to what is described here: http://msdn.microsoft.com/en-us/library/aa393266.aspx). Here are the specific steps that were followed: Open Administrative Tools - Computer Management: Under Computer Management (Local) Expand Services and Applications, right click WMI Control and select properties. In the Security tab, expand Root, highlight CIMV2, click Security (near bottom of window); add Performance Monitor Users and enable the options : Enable Account and Remote Enable. ­Open Administrative Tools - Component Services: Under Console Root go to Component Services- Computers - Right click My Computer and select properties, select the COM security tab, in “Access Permissions” click "Edit Default" select(or add then select) “Performance Monitor Users” group and allow local access and remote access and click ok. In “Launch and Activation Permissions” click “Edit Default” select(or add then select) “Performance Monitor Users” group and allow Local and Remote Launch and Activation Permissions. ­Open Administrative Tools - Component Services: Under Console Root go to Component Services- Computers - My Computer - DCOM Config - highlight “Windows Management and Instrumentation” right click and select properties, Select the Security tab, Under “Launch and Activation Permissions” select Customize, then click edit, add the “Performance Users Group” and allow local and remote Remote Launch and Remote Activation privileges. I am able to connect remotely via WMI Explorer but when I perform this query: Select CommandLine, ProcessId FROM Win32_Process I get a valid result but every row has an empty CommandLine. If I add the user to the Administrators group and re-run the query, the CommandLine column contains the expected data. It seems there is a permission I am missing somewhere but I am not having much luck tracking it down. Many thanks in advance.

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  • Pure-FTPD accounts and permissions for websites

    - by EddyR
    I'm having trouble setting up the appropriate Pure-FTPD accounts and permissions - I have the following sites setup up on my Debian server. /var/www/site1 /var/www/site2 /var/www/wordpress The permissions are 775 for folders and 664 for files. The owner is currently admin:ftpgroup Wordpress also requires special permissions for file uploads in /var/www/wordpress/wp-content/uploads What I need is: a general admin group with access to /var/www a group for each site (site1, site2, wordpress) and a group or user, not www-data (?), with permissions to write files to the wordpress upload folder I ask because restrictions on linux groups (can't have groups in groups) makes it a little bit confusing and also because many of the tutorial sites have conflicting information like, some recommend the use of www-data and some don't. Also, I'm not sure if I understand how Pure-FTP is supposed to work exactly. I create a Pure-FTPD account and assign it a directory (/var/www) and a system user (ftpuser) and group (ftpgroup): Can I assign more than 1 path? For example, if a user requires access to 2 sites. Is it better to assign ftpgroup to all ftp locations and let Pure-FTPD manage account access? Why would anyone have more than 1 ftpuser or ftpgroup? (Doesn't it mean users have access to everyone else's files if they could get there?) Sorry for so many questions at once. I've been reading lots of tutorials but I think they've ended up making me more confused!

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  • copying folder and file permissions from one user to another after switching domains [closed]

    - by emptyspaces
    Please excuse the title, this was the best way I could think to describe this scenario without an entire paragraph. I am using C#. Currently I have a file server running windows server 2003 setup on a domain, we will call this oldDomain, and I have about 500 user accounts with various permissions on this server. Because of restrictions out of my control we are abandoning this domain and using another one that is more dominant within the organization, we will call this newDomain. All of the users that have accounts on oldDomain also have accounts on newDomain, but the usernames are completely different and there is no link between the two. What I am hoping to do is generate a list of all user accounts and this appropriate sid's from AD on the oldDomain, I already have this part done using dsquery and dsget. Then I will have someone go through and match all of the accounts from oldDomain to the correct username on newDomain. Ultimately leaving me with a list of sids from oldDomain and the appropriate username from newDomain. Now I am hoping to copy the file and folder permissions from the old user from oldDomain to the new user on newDomain once I join the server to newDomain. Can anyone tell me what the best way to copy permissions from the sid to the user on newDomain? There are a bunch of articles out there about copying permissions from user a to user b but I wanted to check and see what the recommended practice is here since there are a ton of directories.

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  • chrooted sftp user with write permissions to /var/www

    - by matthew
    I am getting confused about this setup that I am trying to deploy. I hope someone of you folks can lend me a hand: much much appreciated. Background info Server is Debian 6.0, ext3, with Apache2/SSL and Nginx at the front as reverse proxy. I need to provide sftp access to the Apache root directory (/var/www), making sure that the sftp user is chrooted to that path with RWX permissions. All this without modifying any default permission in /var/www. drwxr-xr-x 9 root root 4096 Nov 4 22:46 www Inside /var/www -rw-r----- 1 www-data www-data 177 Mar 11 2012 file1 drwxr-x--- 6 www-data www-data 4096 Sep 10 2012 dir1 drwxr-xr-x 7 www-data www-data 4096 Sep 28 2012 dir2 -rw------- 1 root root 19 Apr 6 2012 file2 -rw------- 1 root root 3548528 Sep 28 2012 file3 drwxr-x--- 6 www-data www-data 4096 Aug 22 00:11 dir3 drwxr-x--- 5 www-data www-data 4096 Jul 15 2012 dir4 drwxr-x--- 2 www-data www-data 536576 Nov 24 2012 dir5 drwxr-x--- 2 www-data www-data 4096 Nov 5 00:00 dir6 drwxr-x--- 2 www-data www-data 4096 Nov 4 13:24 dir7 What I have tried created a new group secureftp created a new sftp user, joined to secureftp and www-data groups also with nologin shell. Homedir is / edited sshd_config with Subsystem sftp internal-sftp AllowTcpForwarding no Match Group <secureftp> ChrootDirectory /var/www ForceCommand internal-sftp I can login with the sftp user, list files but no write action is allowed. Sftp user is in the www-data group but permissions in /var/www are read/read+x for the group bit so... It doesn't work. I've also tried with ACL, but as I apply ACL RWX permissions for the sftp user to /var/www (dirs and files recursively), it will change the unix permissions as well which is what I don't want. What can I do here? I was thinking I could enable the user www-data to login as sftp, so that it'll be able to modify files/dirs that www-data owns in /var/www. But for some reason I think this would be a stupid move securitywise.

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