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  • Nautilus uses different permissions for mounted drives

    - by farhad0011
    I've written two bash scripts to give read-only or read/write access to my NTFS partition: read-only access: sudo umount /media/Data_Drive/ sudo mount -t ntfs-3g -o ro,user,auto,nls=utf8,umask=0000,uid=1000 /dev/sda2 /media/Data_Drive read/write access: sudo umount /media/Data_Drive/ sudo mount -t ntfs-3g -o rw,user,auto,nls=utf8,umask=0000,uid=1000 /dev/sda2 /media/Data_Drive It works perfectly if I only use terminal to work with the files. It also works with Nautilus in read-only mode but not in the read/write mode. In fact, Nautilus gives me an error when I try to copy a file to Data_Drive saying "The destination is read-only". More funny, when I look at the permissions (by right-clicking on Data_Drive and then properties-permissions) I have all the required permissions to write a file in Data_Drive! I am so confused why Nautilus behaves strangely. I appreciate if anybody could solve the mystery!

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  • In Django-pagination Paginate does not working...

    - by mosg
    Hello. Python 2.6.2 django-pagination 1.0.5 Question: How to force pagination work correctly? The problem is that {% paginate %} does not work, but other {% load pagination_tags %} and {% autopaginate object_list 10 %} works! Error message appeared, when I add {% paginate %} into html page: TemplateSyntaxError at /logging Caught an exception while rendering: pagination/pagination.html What I have done: Install django-pagination without any problems. When I do in python import pagination, it's work well. Added pagination to INSTALLED_APP in settings.py: INSTALLED_APPS = ( # ..., 'pagination', ) Added in settings.py: TEMPLATE_CONTEXT_PROCESSORS = ( "django.core.context_processors.auth", "django.core.context_processors.debug", "django.core.context_processors.i18n", "django.core.context_processors.media", "django.core.context_processors.request" ) Also add to settings.py middleware: MIDDLEWARE_CLASSES = ( # ... 'pagination.middleware.PaginationMiddleware', ) Add to top in views.py: from django.template import RequestContext And finally add to my HTML template page lines: {% load pagination_tags %} ... {% autopaginate item_list 50 %} {% for item in item_list %} ... {% endfor %} {% paginate %} Thanks. PS: some edits required, because I can't django code style work well here :)

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  • Login URL using authentication information in Django

    - by fuSi0N
    I'm working on a platform for online labs registration for my university. Login View [project views.py] from django.http import HttpResponse, HttpResponseRedirect, Http404 from django.shortcuts import render_to_response from django.template import RequestContext from django.contrib import auth def index(request): return render_to_response('index.html', {}, context_instance = RequestContext(request)) def login(request): if request.method == "POST": post = request.POST.copy() if post.has_key('username') and post.has_key('password'): usr = post['username'] pwd = post['password'] user = auth.authenticate(username=usr, password=pwd) if user is not None and user.is_active: auth.login(request, user) if user.get_profile().is_teacher: return HttpResponseRedirect('/teachers/'+user.username+'/') else: return HttpResponseRedirect('/students/'+user.username+'/') else: return render_to_response('index.html', {'msg': 'You don\'t belong here.'}, context_instance = RequestContext(request) return render_to_response('login.html', {}, context_instance = RequestContext(request)) def logout(request): auth.logout(request) return render_to_response('index.html', {}, context_instance = RequestContext(request)) URLS #========== PROJECT URLS ==========# urlpatterns = patterns('', (r'^media/(?P<path>.*)$', 'django.views.static.serve', {'document_root': settings.MEDIA_ROOT }), (r'^admin/', include(admin.site.urls)), (r'^teachers/', include('diogenis.teachers.urls')), (r'^students/', include('diogenis.students.urls')), (r'^login/', login), (r'^logout/', logout), (r'^$', index), ) #========== TEACHERS APP URLS ==========# urlpatterns = patterns('', (r'^(?P<username>\w{0,50})/', labs), ) The login view basically checks whether the logged in user is_teacher [UserProfile attribute via get_profile()] and redirects the user to his profile. Labs View [teachers app views.py] from django.http import HttpResponse, HttpResponseRedirect, Http404 from django.shortcuts import render_to_response from django.template import RequestContext from django.contrib.auth.decorators import user_passes_test from django.contrib.auth.models import User from accounts.models import * from labs.models import * def user_is_teacher(user): return user.is_authenticated() and user.get_profile().is_teacher @user_passes_test(user_is_teacher, login_url="/login/") def labs(request, username): q1 = User.objects.get(username=username) q2 = u'%s %s' % (q1.last_name, q1.first_name) q2 = Teacher.objects.get(name=q2) results = TeacherToLab.objects.filter(teacher=q2) return render_to_response('teachers/labs.html', {'results': results}, context_instance = RequestContext(request)) I'm using @user_passes_test decorator for checking whether the authenticated user has the permission to use this view [labs view]. The problem I'm having with the current logic is that once Django authenticates a teacher user he has access to all teachers profiles basically by typing the teachers username in the url. Once a teacher finds a co-worker's username he has direct access to his data. Any suggestions would be much appreciated.

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  • CodeIgniter & Datamapper as frontend, Django Admin as backend, database tables inconsistent

    - by Rasiel
    I created a database for a site i'm doing using Django as the admin backend. However because the server where the site is hosted on, won't be able to support Python, I find myself needing to do the front end in PHP and as such i've decided to use CodeIgniter along with Datamapper to map the models/relationship. However DataMapper requires the tables to be in a specific format for it to work, and Django maps its tables differently, using the App name as the prefix in the table. I've tried using the prefix & join_prefix vars in datamapper but still doesn't map them correctly. Has anyone used a combination of this? and if so how have the fixed the issue of db table names being inconsistent? Is there anything out there that i can use to make them work together?

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  • Django: Summing values of records grouped by foreign key

    - by Dan0
    Hi there In django, given the following models (slightly simplified), I'm struggling to work out how I would get a view including sums of groups class Client(models.Model): api_key = models.CharField(unique=True, max_length=250, primary_key=True) name = models.CharField(unique=True, max_length=250) class Purchase(models.Model): purchase_date = models.DateTimeField() client = models.ForeignKey(SavedClient, to_field='api_key') amount_to_invoice = models.FloatField(null=True) For a given month, I'd like to see e.g. April 2010 For Each Client that purchased this month: * CLient: Name * Total amount of Purchases for this month * Total cost of purchases for this month For each Purchase made by client: * Date * Amount * Etc I've been looking into django annotation, but can't get my head around how to sum values of a field for a particular group over a particular month and send the information to a template as a variable/tag. Any info would be appreciated

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  • Django: Country drop down list?

    - by User
    I have a form for address information. One of the fields is for the address country. Currently this is just a textbox. I would like a drop down list (of ISO 3166 countries) for this. I'm a django newbie so I haven't even used a Django Select widget yet. What is a good way to do this? Hard-code the choices in a file somewhere? Put them in the database? In the template?

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  • 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?

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  • Project name inserted automatically in url when using django template url tag

    - by thebossman
    I am applying the 'url' template tag to all links in my current Django project. I have my urls named like so... url(r'^login/$', 'login', name='site_login'), This allows me to access /login at my site's root. I have my template tag defined like so... <a href="{% url site_login %}"> It works fine, except that Django automatically resolves that url as /myprojectname/login, not /login. Both urls are accessible. Why? Is there an option to remove the projectname? This occurs for all url tags, not just this one.

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  • How to customize a many-to-many inline model in django admin

    - by Jonathan
    I'm using the admin interface to view invoices and products. To make things easy, I've set the products as inline to invoices, so I will see the related products in the invoice's form. As you can see I'm using a many-to-many relationship. In models.py: class Product(models.Model): name = models.TextField() price = models.DecimalField(max_digits=10,decimal_places=2) class Invoice(models.Model): company = models.ForeignKey(Company) customer = models.ForeignKey(Customer) products = models.ManyToManyField(Product) In admin.py: class ProductInline(admin.StackedInline): model = Invoice.products.through class InvoiceAdmin(admin.ModelAdmin): inlines = [FilteredApartmentInline,] admin.site.register(Product, ProductAdmin) The problem is that django presents the products as a table of drop down menus (one per associated product). Each drop down contains all the products listed. So if I have 5000 products and 300 are associated with a certain invoice, django actually loads 300x5000 product names. Also the table is not aesthetic. How can I change it so that it'll just display the product's name in the inline table? Which form should I override, and how?

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  • django cross-site reverse a url

    - by tutuca
    I have a similar question than django cross-site reverse. But i think I can't apply the same solution. I'm creating an app that lets the users create their own site. After completing the signup form the user should be redirected to his site's new post form. Something along this lines: new_post_url = 'http://%s.domain:9292/manage/new_post %site.domain' logged_user = authenticate(username=user.username, password=user.password) if logged_user is not None: login(request, logged_user) return redirect(new_product_url) Now, I know that "new_post_url" is awful and makes babies cry so I need to reverse it in some way. I thought in using django.core.urlresolvers.reverse to solve this but that only returns urls on my domain, and not in the user's newly created site, so it doesn't works for me. So, do you know a better/smarter way to solve this?

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  • Load django template from the database

    - by Björn Lindqvist
    Hello, Im trying to render a django template from a database outside of djangos normal request-response structure. But it appears to be non-trivial due to the way django templates are compiled. I want to do something like this: >>> s = Template.objects.get(pk = 123).content >>> some_method_to_render(s, {'a' : 123, 'b' : 456}) >>> ... the rendered output here ... How do you do this?

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  • How to use multiple flatpages models in a django app?

    - by the_drow
    I have multiple models that can be converted to flatpages but have to have some extra information (For example I have an about us page but I also have a blog). However I understand that there must be only one flatpages model since the middleware only returns the flatpages instance and does not resolve the child models. What do I have to do? EDIT: It seems I need to change the views. Here's the current code: from django.contrib.flatpages.models import FlatPage from django.template import loader, RequestContext from django.shortcuts import get_object_or_404 from django.http import HttpResponse, HttpResponseRedirect from django.conf import settings from django.core.xheaders import populate_xheaders from django.utils.safestring import mark_safe from django.views.decorators.csrf import csrf_protect DEFAULT_TEMPLATE = 'flatpages/default.html' # This view is called from FlatpageFallbackMiddleware.process_response # when a 404 is raised, which often means CsrfViewMiddleware.process_view # has not been called even if CsrfViewMiddleware is installed. So we need # to use @csrf_protect, in case the template needs {% csrf_token %}. # However, we can't just wrap this view; if no matching flatpage exists, # or a redirect is required for authentication, the 404 needs to be returned # without any CSRF checks. Therefore, we only # CSRF protect the internal implementation. def flatpage(request, url): """ Public interface to the flat page view. Models: `flatpages.flatpages` Templates: Uses the template defined by the ``template_name`` field, or `flatpages/default.html` if template_name is not defined. Context: flatpage `flatpages.flatpages` object """ if not url.endswith('/') and settings.APPEND_SLASH: return HttpResponseRedirect("%s/" % request.path) if not url.startswith('/'): url = "/" + url # Here instead of getting the flat page it needs to find if it has a page with a child model. f = get_object_or_404(FlatPage, url__exact=url, sites__id__exact=settings.SITE_ID) return render_flatpage(request, f) @csrf_protect def render_flatpage(request, f): """ Internal interface to the flat page view. """ # If registration is required for accessing this page, and the user isn't # logged in, redirect to the login page. if f.registration_required and not request.user.is_authenticated(): from django.contrib.auth.views import redirect_to_login return redirect_to_login(request.path) if f.template_name: t = loader.select_template((f.template_name, DEFAULT_TEMPLATE)) else: t = loader.get_template(DEFAULT_TEMPLATE) # To avoid having to always use the "|safe" filter in flatpage templates, # mark the title and content as already safe (since they are raw HTML # content in the first place). f.title = mark_safe(f.title) f.content = mark_safe(f.content) # Here I need to be able to configure what I am passing in the context c = RequestContext(request, { 'flatpage': f, }) response = HttpResponse(t.render(c)) populate_xheaders(request, response, FlatPage, f.id) return response

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  • Create Django formset wihtout multiple queries

    - by Martin
    I need to display multiple forms (up to 10) of a model on a page. This is the code I use for to accomplish this. TheFormSet = formset_factory(SomeForm, extra=10) ... formset = TheFormSet(prefix='party') return render_to_response('template.html', { 'formset' : formset, }) The problem is, that it seems to me that Django queries the database for each of the forms in the formset, even though the data displayed in them is the same. Is this the way Formsets work or am I doing something wrong? Is there a way around it inside django or would I have to use JavaScript for a workaround?

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  • django auth_views.login and redirects

    - by Zayatzz
    Hello I could not understand why after logging in from address: http://localhost/en/accounts/login/?next=/en/test/ I get refirected to http://localhost/accounts/profile/ So i ran search in django files and found that this address is the default LOGIN_REDIRECT_URL for django. What i did not understand is why it gets redirected to there. I guessed, that my login form's post address should be : /accounts/login/?next=/en/test/ instead of /accounts/login/ I wrote it into template and it worked. But since the redirect url changes dynamically, how can i make this login post forms address change dynamically too? is there a templatetag for that or something? Alan

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  • Calling a method from within a django model save() override

    - by Jonathan
    I'm overriding a django model save() method. Within the override I'm calling another method of the same class and instance which calculates one of the instance's fields based on other fields of the same instance. class MyClass(models.Model): field1 = models.FloatField() field2 = models.FloatField() field3 = models.FloatField() def calculateField1(self) self.field1 = self.field2 + self.field3 def save(self, *args, **kwargs): self.calculateField1() super(MyClass, self).save(*args, **kwargs) The override method is called when I change the model in admin. Alas I've discovered that within calculateField1() field2 and field3 have the values of the instance from before I edited them in admin. If I enter the instance again in admin and save again, only then field1 receives the correct value as field2 and field3 are already updated. Is this the correct behavior on django's side? If yes, then how can I use the new values within calculateField1? I cannot implement the calculation within the save() as calculateField1() actually quite long and I need it to be called from elsewhere.

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  • Django QuerySet filter + order_by + limit

    - by handsofaten
    So I have a Django app that processes test results, and I'm trying to find the median score for a certain assessment. I would think that this would work: e = Exam.objects.all() total = e.count() median = int(round(total / 2)) median_exam = Exam.objects.filter(assessment=assessment.id).order_by('score')[median:1] median_score = median_exam.score But it always returns an empty list. I can get the result I want with this: e = Exam.objects.all() total = e.count() median = int(round(total / 2)) exams = Exam.objects.filter(assessment=assessment.id).order_by('score') median_score = median_exam[median].score I would just prefer not to have to query the entire set of exams. I thought about just writing a raw MySQL query that looks something like: SELECT score FROM assess_exam WHERE assessment_id = 5 ORDER BY score LIMIT 690,1 But if possible, I'd like to stay within Django's ORM. Mostly, it's just bothering me that I can't seem to use order_by with a filter and a limit. Any ideas?

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  • IE 8 Errors for Group Permissions for access

    - by Jeff P.
    Hello, The company I work for is about to begin testing for a transition from IE6 to IE8. One of our main concerns is about using permissions to control intranet access to internal sites,etc. This was a problem we encountered during testing for IE7 and scrapped it all together. However, IE6 has more than its foot in the grave, its head is peeking out. Microsoft is only supporting IE6 until 2014 when they also abandon support for Windows XP. IE7 just came up with a box asking for you to enter your credentials. It didn't matter if you had permissions, it always said access denied. We have yet to begin testing but I was hoping someone has encountered this issue as well and may be able to shed some light on the subject.

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  • Umbraco Permissions Script - Secure Version

    - by Vizioz Limited
    Back in May I blogged about how to set Permissions for Umbraco using SetACL to set the appropriate directory permissions based on the installation recommendations.Recently I have been working on a site for a client who wanted every security item to be locked down as tightly as possible. And so I modified the script based on the Umbraco security best practices, I thought I'd share it with everyone, if I have missed anything, or if anyone has any suggestions on how to improve this, please let me know :)Please refer to my previous post regarding the SetAcl command line application that you will need.I suggest you save the following into a batch file called: umbPermSecure.batecho offREM Script to setup the Security Permissions for an Umbraco siteREM This script will give your machine Network Service the minimum rights requiredREM for Umbraco to workREM I suggest you update this script to also remove any users who do not need REM access to the web foldersREM **** Pre-requisites ****REM You will need to download - http://setacl.sourceforge.net/REM It is assumed that you have stored SetACL in a directory called, C:\SetACL ifREM not, you will need to modify the script.REM **** Usage ****REM You need to pass in the path for the root of your Umbraco directoryREM E.g. umbPermSecure.bat C:\inetpub\umbracoroot@echo umbPermSecure.bat - Script to set Umbraco File and Directory Permissions@echo based on the Umbraco Security Best Practices Document (13th March 2009)@echo Published by Chris Houston - 19th October 2009@echo http://blog.vizioz.com@echo Adding READ only access SetACL.exe -on "%1" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\web.config" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\bin" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\umbraco" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"@echo Adding READ and EXECUTE access SetACL.exe -on "%1\app_code" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read_ex" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\usercontrols" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read_ex" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"@echo Adding READ, WRITE and MODIFY access SetACL.exe -on "%1\config" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -ace "n:%computername%\NETWORK SERVICE;p:change" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\css" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -ace "n:%computername%\NETWORK SERVICE;p:change" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\data" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -ace "n:%computername%\NETWORK SERVICE;p:change" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\masterpages" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -ace "n:%computername%\NETWORK SERVICE;p:change" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\media" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -ace "n:%computername%\NETWORK SERVICE;p:change" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\python" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -ace "n:%computername%\NETWORK SERVICE;p:change" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\scripts" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -ace "n:%computername%\NETWORK SERVICE;p:change" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"SetACL.exe -on "%1\xslt" -ot file -actn ace -ace "n:%computername%\NETWORK SERVICE;p:read" -ace "n:%computername%\NETWORK SERVICE;p:change" -actn clear -clr "dacl,sacl" -log "c:\setacl\log.txt"

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  • no sudo permissions to /etc/hostname

    - by Pagis
    When I try to edit my /etc/hostname file, using sudo nano /etc/hostname (or any other editor) I get the following: Error writing /etc/hostname: Permission denied I then tried: sudo chmod u+w ./hostname And got the error: chmod: changing permissions of `./hostname': Operation not permitted I get the same error trying to add write permissions to any use or group. This is what ls shows: -rw-r--r-- 1 root root 22 2012-06-18 12:25 /etc/hostname Any idea why and what can I do about it? I'm running Ubuntu 11.10.

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  • Django | django-socialregistration error

    - by MMRUser
    I'm trying to add the facebook connect feature to my site, I decided to use django socialregistration.All are setup including pyfacebook, here is my source code. settings.py MIDDLEWARE_CLASSES = ( 'django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'facebook.djangofb.FacebookMiddleware', 'socialregistration.middleware.FacebookMiddleware', ) urls.py (r'^callback/$', 'fbproject.fbapp.views.callback'), views.py def callback(request): return render_to_response('canvas.fbml') Template <html> <body> {% load facebook_tags %} {% facebook_button %} {% facebook_js %} </body> </html> but when I point to the URL, I'm getting this error Traceback (most recent call last): File "C:\Python26\lib\site-packages\django\core\servers\basehttp.py", line 279, in run self.result = application(self.environ, self.start_response) File "C:\Python26\lib\site-packages\django\core\servers\basehttp.py", line 651, in __call__ return self.application(environ, start_response) File "C:\Python26\lib\site-packages\django\core\handlers\wsgi.py", line 241, in __call__ response = self.get_response(request) File "C:\Python26\lib\site-packages\django\core\handlers\base.py", line 73, in get_response response = middleware_method(request) File "build\bdist.win32\egg\socialregistration\middleware.py", line 13, in process_request request.facebook.check_session(request) File "C:\Python26\lib\site-packages\facebook\__init__.py", line 1293, in check_session self.session_key_expires = int(params['expires']) ValueError: invalid literal for int() with base 10: 'None' Django 1.1.1 *Python 2.6.2*

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  • Manditory read-only fields in django

    - by jamida
    I'm writing a test "grade book" application. The models.py file is shown below. class Student(models.Model): name = models.CharField(max_length=50) parent = models.CharField(max_length=50) def __unicode__(self): return self.name class Grade(models.Model): studentId = models.ForeignKey(Student) finalGrade = models.CharField(max_length=3) I'd like to be able to change the final grade for several students in a modelformset but for now I'm just trying one student at a time. I'm also trying to create a form for it that shows the student name as a field that can not be changed, the only thing that can be changed here is the finalGrade. So I used this trick to make the studentId read-only. class GradeROForm(ModelForm): studentId = forms.ModelChoiceField(queryset=Student.objects.all()) def __init__(self, *args, **kwargs): super(GradeROForm,self).__init__(*args, **kwargs) instance = getattr(self, 'instance', None) if instance and instance.id: self.fields['studentId'].widget.attrs['disabled']='disabled' def clean_studentId(self): instance = getattr(self,'instance',None) if instance: return instance.studentId else: return self.cleaned_data.get('studentId',None) class Meta: model=Grade And here is my view: def modifyGrade(request,student): student = Student.objects.get(name=student) mygrade = Grade.objects.get(studentId=student) if request.method == "POST": myform = GradeROForm(data=request.POST, instance=mygrade) if myform.is_valid(): grade = myform.save() info = "successfully updated %s" % grade.studentId else: myform=GradeROForm(instance=mygrade) return render_to_response('grades/modifyGrade.html',locals()) This displays the form like I expect, but when I hit "submit" I get a form validation error for the student field telling me this field is required. I'm guessing that, since the field is "disabled", the value is not being reported in the POST and for reasons unknown to me the instance isn't being used in its place. I'm a new Django/Python programmer, but quite experienced in other languages. I can't believe I've stumbled upon such a difficult to solve problem in my first significant django app. I figure I must be missing something. Any ideas?

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  • Django Upload form to S3 img and form validation

    - by citadelgrad
    I'm fairly new to both Django and Python. This is my first time using forms and upload files with django. I can get the uploads and saves to the database to work fine but it fails to valid email or check if the users selected a file to upload. I've spent a lot of time reading documentation trying to figure this out. Thanks! views.py def submit_photo(request): if request.method == 'POST': def store_in_s3(filename, content): conn = S3Connection(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY) bucket = conn.create_bucket(AWS_STORAGE_BUCKET_NAME) mime = mimetypes.guess_type(filename)[0] k = Key(bucket) k.key = filename k.set_metadata("Content-Type", mime) k.set_contents_from_file(content) k.set_acl('public-read') if imghdr.what(request.FILES['image_url']): qw = request.FILES['image_url'] filename = qw.name image = filename content = qw.file url = "http://bpd-public.s3.amazonaws.com/" + image data = {image_url : url, user_email : request.POST['user_email'], user_twittername : request.POST['user_twittername'], user_website : request.POST['user_website'], user_desc : request.POST['user_desc']} s = BeerPhotos(data) if s.is_valid(): #import pdb; pdb.set_trace() s.save() store_in_s3(filename, content) return HttpResponseRedirect(reverse('photos.views.thanks')) return s.errors else: return errors else: form = BeerPhotoForm() return render_to_response('photos/submit_photos.html', locals(),context_instance=RequestContext(request) forms.py class BeerPhotoForm(forms.Form): image_url = forms.ImageField(widget=forms.FileInput, required=True,label='Beer',help_text='Select a image of no more than 2MB.') user_email = forms.EmailField(required=True,help_text='Please type a valid e-mail address.') user_twittername = forms.CharField() user_website = forms.URLField(max_length=128,) user_desc = forms.CharField(required=True,widget=forms.Textarea,label='Description',) template.html <div id="stylized" class="myform"> <form action="." method="post" enctype="multipart/form-data" width="450px"> <h1>Photo Submission</h1> {% for field in form %} {{ field.errors }} {{ field.label_tag }} {{ field }} {% endfor %} <label><span>Click here</span></label> <input type="submit" class="greenbutton" value="Submit your Photo" /> </form> </div>

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  • Invalidating Memcached Keys on save() in Django

    - by Zack
    I've got a view in Django that uses memcached to cache data for the more highly trafficked views that rely on a relatively static set of data. The key word is relatively: I need invalidate the memcached key for that particular URL's data when it's changed in the database. To be as clear as possible, here's the meat an' potatoes of the view (Person is a model, cache is django.core.cache.cache): def person_detail(request, slug): if request.is_ajax(): cache_key = "%s_ABOUT_%s" % settings.SITE_PREFIX, slug # Check the cache to see if we've already got this result made. json_dict = cache.get(cache_key) # Was it a cache hit? if json_dict is None: # That's a negative Ghost Rider person = get_object_or_404(Person, display = True, slug = slug) json_dict = { 'name' : person.name, 'bio' : person.bio_html, 'image' : person.image.extra_thumbnails['large'].absolute_url, } cache.set(cache_key) # json_dict will now exist, whether it's from the cache or not response = HttpResponse() response['Content-Type'] = 'text/javascript' response.write(simpljson.dumps(json_dict)) # Make sure it's all properly formatted for JS by using simplejson return response else: # This is where the fully templated response is generated What I want to do is get at that cache_key variable in it's "unformatted" form, but I'm not sure how to do this--if it can be done at all. Just in case there's already something to do this, here's what I want to do with it (this is from the Person model's hypothetical save method) def save(self): # If this is an update, the key will be cached, otherwise it won't, let's see if we can't find me try: old_self = Person.objects.get(pk=self.id) cache_key = # Voodoo magic to get that variable old_key = cache_key.format(settings.SITE_PREFIX, old_self.slug) # Generate the key currently cached cache.delete(old_key) # Hit it with both barrels of rock salt # Turns out this doesn't already exist, let's make that first request even faster by making this cache right now except DoesNotExist: # I haven't gotten to this yet. super(Person, self).save() I'm thinking about making a view class for this sorta stuff, and having functions in it like remove_cache or generate_cache since I do this sorta stuff a lot. Would that be a better idea? If so, how would I call the views in the URLconf if they're in a class?

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