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  • Django JSON serializable error

    - by Hulk
    With the following code below, There is an error saying File "/home/user/web_pro/info/views.py", line 184, in headerview, raise TypeError("%r is not JSON serializable" % (o,)) TypeError: <lastname: jerry> is not JSON serializable In the models code header(models.Model): firstname = models.ForeignKey(Firstname) lastname = models.ForeignKey(Lastname) In the views code headerview(request): header = header.objects.filter(created_by=my_id).order_by(order_by)[offset:limit] l_array = [] l_array_obj = [] for obj in header: l_array_obj = [obj.title, obj.lastname ,obj.firstname ] l_array.append(l_array_obj) dictionary_l.update({'Data': l_array}) ; return HttpResponse(simplejson.dumps(dictionary_l), mimetype='application/javascript') what is this error and how to resolve this? thanks..

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  • django-multilingual and switching between languages on template side

    - by israkir
    I am trying to use django-multilingual and setup it properly. But what I found is that everything is clear for django-multilingual except a template usage example. I just started to use django and I don't know, maybe because of this reason, I cannot figure out how to switch between languages on template side. Is there any example that you can give or any 'more' clear source/documentation about this?

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  • Edit/show Primary Key in Django Admin

    - by emcee
    It appears Django hides fields that are flagged Primary Key from being displayed/edited in the Django admin interface. Let's say I'd like to input data in which I may or may not want to specify a primary key. How would I go about displaying primary keys in Django-admin, and how could I make specifying it optional? Many thanks in advance, beloved hive-mind.

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  • Django Admin Actions missing

    - by Andrew C
    One of my Django sites is missing the Django Admin Action bar shown here: http://docs.djangoproject.com/en/dev/ref/contrib/admin/actions/#ref-contrib-admin-actions There is no checkbox next to each row and no Action select box near the top of the page. This is happening on every model. I have several sites running Django 1.1, and they all show the Admin Actions, so it feels like a local configuration issue. Anyone seen this before?

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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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  • Math on Django Templates

    - by Leandro Abilio
    Here's another question about Django. I have this code: views.py cursor = connections['cdr'].cursor() calls = cursor.execute("SELECT * FROM cdr where calldate > '%s'" %(start_date)) result = [SQLRow(cursor, r) for r in cursor.fetchall()] return render_to_response("cdr_user.html", {'calls':result }, context_instance=RequestContext(request)) I use a MySQL query like that because the database is not part of a django project. My cdr table has a field called duration, I need to divide that by 60 and multiply the result by a float number like 0.16. Is there a way to multiply this values using the template tags? If not, is there a good way to do it in my views? My template is like this: {% for call in calls %} <tr class="{% cycle 'odd' 'even' %}"><h3> <td valign="middle" align="center"><h3>{{ call.calldate }}</h3></td> <td valign="middle" align="center"><h3>{{ call.disposition }}</h3></td> <td valign="middle" align="center"><h3>{{ call.dst }}</h3></td> <td valign="middle" align="center"><h3>{{ call.billsec }}</h3></td> <td valign="middle" align="center">{{ (call.billsec/60)*0.16 }}</td></h3> </tr> {% endfor %} The last is where I need to show the value, I know the "(call.billsec/60)*0.16" is impossible to be done there. I wrote it just to represent what I need to show.

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  • Django authentication in django nonrel on GAE

    - by tooba
    I'm using the Django nonrel project on a google app engine project running locally in development. I've created my own models and these are fine when they are saved and retrieved in the datastore. I'm hoping to use django.contrib.auth to provide the user functionality. I can use the shell to create users and these get assigned an ID. When I create one of my own models which references User I have to pass in a user ID as it quite rightly fails otherwise. However, checking via the gae admin interface I can't see the User model in the datastore for the users I've created via the shell. Nor can I retreive the user details from one of my models which references them. Calls to mymodel.user.username return nothing. Nor can I log into admin using the username and password I've set up. I can see saved versions of the models I've made in the gae admin app. I get the impression that users are being created somewhere other than the datastore. Is there something else I need to do to use the standard contrib.auth users with django-nonrel and gae?

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  • Django url parameters

    - by Hulk
    How to pass two paramters in urls in django <script> url=/toolbox/display/" + id + "2"; window.location=url; </script> Also how is this handeled in urls.py (r'^display/(?P<rid>\d+)/(?P<param>\d+)/$', 'table_display'), In views, def table_display(request,rid,param): print param //This should print 2

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  • 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..

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  • How to make custom join query with Django ?

    - by xRobot
    I have these 2 models: genre = ( ('D', 'Dramatic'), ('T', 'Thriller'), ('L', 'Love'), ) class Book(models.Model): title = models.CharField(max_length=100) genre = models.CharField(max_length=1, choices=genre) class Author(models.Model): user = models.ForeignKey(User, unique=True) born = models.DateTimeField('born') book = models.ForeignKey(Book) I need to retrieve first_name and last_name of all authors of dramatic's books. How can I do this in django ?

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  • Can I delete Generic kernel if I use Generic

    - by user206049
    I currently can't update my release as there is not enough space on boot. I just have the one kernel version there, but seem to have both the Generic and Low Latency versions. uname -r just shows 3.8.0-32-lowlatency ls -lah /boot shows -rw-r--r-- 1 root root 899K Oct 2 00:00 abi-3.8.0-32-generic -rw-r--r-- 1 root root 899K Oct 7 09:27 abi-3.8.0-32-lowlatency -rw-r--r-- 1 root root 152K Oct 2 00:00 config-3.8.0-32-generic -rw-r--r-- 1 root root 152K Oct 7 09:27 config-3.8.0-32-lowlatency drwxr-xr-x 3 root root 2.0K Jan 1 1970 efi drwxr-xr-x 5 root root 1.0K Oct 22 10:05 grub -rw-r--r-- 1 root root 32M Oct 22 09:51 initrd.img-3.8.0-32-generic -rw-r--r-- 1 root root 32M Oct 22 10:05 initrd.img-3.8.0-32-lowlatency drwxr-xr-x 2 root root 12K Feb 25 2013 lost+found -rw-r--r-- 1 root root 173K Dec 5 2012 memtest86+.bin -rw-r--r-- 1 root root 175K Dec 5 2012 memtest86+_multiboot.bin -rw------- 1 root root 3.0M Oct 2 00:00 System.map-3.8.0-32-generic -rw------- 1 root root 3.0M Oct 7 09:27 System.map-3.8.0-32-lowlatency -rw------- 1 root root 5.2M Oct 2 00:00 vmlinuz-3.8.0-32-generic -rw------- 1 root root 5.2M Oct 7 09:27 vmlinuz-3.8.0-32-lowlatency So what can I do to allow me to update? Apparently I need 174m on boot and am 40m short.

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  • Accessing django choice field

    - by Hulk
    there is a module as header , from test.models import SEL_VALUES class rubrics_header(models.Model): sel_values = models.IntegerField(choices=SEL_VALUES) So when SEL_VALUES is imported from test.modules.What is the code that has to go in views to get the choices in sel_values . And the test.modules has the following, class SEL_VALUES: vaue = 0 value2 = 1 class Entries(forms.Form) : models.IntegerField(choices=SEL_VALUES) SEL_VALUES = ((ACCESS.value,'NAME'),(ACCESS.value2,'DESIGNATION'))

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  • Django foreign key error

    - by Hulk
    In models the code is as, class header(models.Model): title = models.CharField(max_length = 255) created_by = models.CharField(max_length = 255) def __unicode__(self): return self.id() class criteria(models.Model): details = models.CharField(max_length = 255) headerid = models.ForeignKey(header) def __unicode__(self): return self.id() In views, p_l=header.objects.filter(id=rid) for rows in row_data: row_query =criteria(details=rows,headerid=p_l) row_query.save() In row_query =criteria(details=rows,headerid=p_l) there is an error saying 'long' object is not callable in models.py in __unicode__, What is wrong in the code Thanks..

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  • querying for timestamp field in django

    - by Hulk
    In my views i have the date in the following format s_date=20090106 and e_date=20100106 The model is defined as class Activity(models.Model): timestamp = models.DateTimeField(auto_now_add=True) how to query for the timestamp filed with the above info. Activity.objects.filter(timestamp>=s_date and timestamp<=e_date) Thanks.....

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  • Drupal, Views: display taxonomies as Views titles.

    - by Patrick
    hi, I'm using Views for some nodes, and I want to display a different View title according to which taxonomy tags are selected in my filter. I already have taxonomy field for each node in my view. But this is not what I need. I basically need to display all the currently filtered tags on the top of my view. I was wondering if I can solve adding some line with php, how ? Thanks Update: I'm now using the Views Header field in Views settings, but it only processes html code, not php, so I cannot add taxonomy terms. Is it because of my CCK Editor settings ? thanks

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  • PyDev and Django: PyDev breaking Django shell?

    - by Rosarch
    I've set up a new project, and populated it with simple models. (Essentially I'm following the tut.) When I run python manage.py shell on the command line, it works fine: >python manage.py shell Python 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license" for more information. (InteractiveConsole) >>> from mysite.myapp.models import School >>> School.objects.all() [] Works great. Then, I try to do the same thing in Eclipse (using a Django project that is composed of the same files.) Right click on mysite project Django Shell with Django environment This is the output from the PyDev Console: >>> import sys; print('%s %s' % (sys.executable or sys.platform, sys.version)) C:\Python26\python.exe 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] >>> >>> from django.core import management;import mysite.settings as settings;management.setup_environ(settings) 'path\\to\\mysite' >>> from mysite.myapp.models import School >>> School.objects.all() Traceback (most recent call last): File "<console>", line 1, in <module> File "C:\Python26\lib\site-packages\django\db\models\query.py", line 68, in __repr__ data = list(self[:REPR_OUTPUT_SIZE + 1]) File "C:\Python26\lib\site-packages\django\db\models\query.py", line 83, in __len__ self._result_cache.extend(list(self._iter)) File "C:\Python26\lib\site-packages\django\db\models\query.py", line 238, in iterator for row in self.query.results_iter(): File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 287, in results_iter for rows in self.execute_sql(MULTI): File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 2368, in execute_sql cursor = self.connection.cursor() File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 81, in cursor cursor = self._cursor() File "C:\Python26\lib\site-packages\django\db\backends\sqlite3\base.py", line 170, in _cursor self.connection = Database.connect(**kwargs) OperationalError: unable to open database file What am I doing wrong here?

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  • Django: UserProfile with Unique Foreign Key in Django Admin

    - by lazerscience
    Hi, I have extended Django's User Model using a custom user profile called UserExtension. It is related to User through a unique ForeignKey Relationship, which enables me to edit it in the admin in an inline form! I'm using a signal to create a new profile for every new user: def create_user_profile(sender, instance, created, **kwargs): if created: try: profile, created = UserExtension.objects.get_or_create(user=instance) except: pass post_save.connect(create_user_profile, sender=User) (as described here for example: http://stackoverflow.com/questions/44109/extending-the-user-model-with-custom-fields-in-django) The problem is, that, if I create a new user through the admin, I get an IntegritiyError on saving "column user_id is not unique". It doesnt seem that the signal is called twice, but i guess the admin is trying to save the profile AFTERWARDS? But I need the creation through signal if I create a new user in other parts of the system!

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  • Get the currently saved object in a view in Django

    - by mridang
    Hi, I has a Django view which is accessed through an AJAX call. It's a pretty simple one — all it does is simply pass the request to a form object and save the data. Here's a snippet from my view: form = AddSiteForm(request.user, request.POST) if form.is_valid(): obj = form.save(commit=False) obj.user = request.user obj.save() data['status'] = 'success' data['html'] = render_to_string('site.html', locals(), context_instance=RequestContext(request)) return HttpResponse(simplejson.dumps(data), mimetype='application/json') How do I get the currently saved object (including the internally generated id column) and pass it to the template? Any help guys? Mridang

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  • Sort list of items in Django

    - by mridang
    Hi Guys, I have a Django view called change_priority. I'm posting a request to this view with a commas separated list of values which is basically the order of the items in my model. The data looks like this: 1,4,11,31,2,4,7 I have a model called Items which has two values - id and priority. Upon getting this post request, how can I set the priority of the Item depending upon the list order. So my data in the db would look like. 1,1 4,2 11,3 31,4 2,5 4,6 7,7 Thanks guys.

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  • Django, making a page activate for a fixed time

    - by Hellnar
    Greetings I am hacking Django and trying to test something such as: Like woot.com , I want to sell "an item per day", so only one item will be available for that day (say the default www.mysite.com will be redirected to that item), Assume my urls for calling these items will be such: www.mysite.com/item/<number> my model for item: class Item(models.Model): item_name = models.CharField(max_length=30) price = models.FloatField() content = models.TextField() #keeps all the html content start_time = models.DateTimeField() end_time = models.DateTimeField() And my view for rendering this: def results(request, item_id): item = get_object_or_404(Item, pk=item_id) now = datetime.now() if item.start_time > now: #render and return some "not started yet" error templete elif item.end_time < now: #render and return some "item selling ended" error templete else: # render the real templete for selling this item What would be the efficient and clever model & templete for achieving this ?

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  • django generic view update/create: update works but create raises IntegrityError

    - by smarber
    I'm using CreateView and UpdateView directely into urls.py of my application whose name is dydict. In the file forms.py I'm using ModelForm and I'm exluding a couple of fields from being shown, some of which sould be set when either creating or updating. So, as mentioned in the title, update part works but create part doesn't which is obvious because required fields that I have exluded are sent empty which is not allowed in my case. So the question here is, how should I do to fill exluded fields into the file forms.py so that I don't have to override CreateView? Thanks in advance.

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  • How to open python scripts directly by typing in their name in terminal (Mac OS X)

    - by Haffi112
    I'm working on installing django and running it on my system. I have a problem though, in this tutorial creating a project is explained by running the command django-admin.py startproject mysite My issue is that this doesn't work. I changed to the directory where django-admin.py is located and ran the command chmod +x django-admin.py with no results. I tried adding the directory with the file to my path without results. I ended up fixing my problem with this command python /location/of/django-admin.py startproject mysite which yielded the outcome I expected. My problem is: What do I need to change/configure such that command django-admin.py startproject mysite would be sufficient? Here are some experiments: 21:09~/Desktop/HI/NSN/Polls > django-admin.py startproject mysite -bash: django-admin.py: command not found 21:09~/Desktop/HI/NSN/Polls > ./django-admin.py startproject mysite -bash: ./django-admin.py: No such file or directory 21:09~/Desktop/HI/NSN/Polls > python django-admin.py startproject mysite python: can't open file 'django-admin.py': [Errno 2] No such file or directory 21:09~/Desktop/HI/NSN/Polls > /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1 -bash: /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py: /opt/local/bin: bad interpreter: Permission denied 21:09~/Desktop/HI/NSN/Polls > sudo /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1Password: sudo: unable to execute /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py: Permission denied 21:09~/Desktop/HI/NSN/Polls > sudo /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1sudo: unable to execute /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py: Permission denied 21:09~/Desktop/HI/NSN/Polls > python /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1 21:09~/Desktop/HI/NSN/Polls > ls mysite prufa1 Final edit: The problem is solved, see Ian C's answer for the right solution. Thank you everyone for helping my out, this was very fast!

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  • Django: Validation error in Admin

    - by tomwolber
    NEWBIE ALERT! background: For the first time, I am writing a model that needs to be validated. I cannot have two Items that have overlapping "date ranges". I have everything working, except when I raise forms.ValidationError, I get the yellow screen of death (debug=true) or a 500 page (debug=false). My question: How can I have an error message show up in the Admin (like when you leave a required filed blank)? Sorry for my inexperience, please let me know if I can clarify the question better. Models.py from django.db import models from django import forms from django.forms import ModelForm from django.db.models import Q class Item(models.Model): name = models.CharField(max_length=500) slug = models.SlugField(unique=True) startDate = models.DateField("Start Date", unique="true") endDate = models.DateField("End Date") def save(self, *args, **kwargs): try: Item.objects.get(Q(startDate__range=(self.startDate,self.endDate))|Q(endDate__range=(self.startDate,self.endDate))|Q(startDate__lt=self.startDate,endDate__gt=self.endDate)) #check for validation, which may raise an Item.DoesNotExist error, excepted below #if the validation fails, raise this error: raise forms.ValidationError('Someone has already got that date, or somesuch error message') except Item.DoesNotExist: super(Item,self).save(*args,**kwargs) def __unicode__(self): return self.name def get_absolute_url(self): return "/adtest/%s/" % self.slug

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  • Django - Override admin site's login form

    - by TrojanCentaur
    I'm currently trying to override the default form used in Django 1.4 when logging in to the admin site (my site uses an additional 'token' field required for users who opt in to Two Factor Authentication, and is mandatory for site staff). Django's default form does not support what I need. Currently, I've got a file in my templates/ directory called templates/admin/login.html, which seems to be correctly overriding the template used with the one I use throughout the rest of my site. The contents of the file are simply as below: # admin/login.html: {% extends "login.html" %} The actual login form is as below: # login.html: {% load url from future %}<!DOCTYPE html> <html> <head> <title>Please log in</title> </head> <body> <div id="loginform"> <form method="post" action="{% url 'id.views.auth' %}"> {% csrf_token %} <input type="hidden" name="next" value="{{ next }}" /> {{ form.username.label_tag }}<br/> {{ form.username }}<br/> {{ form.password.label_tag }}<br/> {{ form.password }}<br/> {{ form.token.label_tag }}<br/> {{ form.token }}<br/> <input type="submit" value="Log In" /> </form> </div> </body> </html> My issue is that the form provided works perfectly fine when accessed using my normal login URLs because I supply my own AuthenticationForm as the form to display, but through the Django Admin login route, Django likes to supply it's own form to this template and thus only the username and password fields render. Is there any way I can make this work, or is this something I am just better off 'hard coding' the HTML fields into the form for?

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  • Django tests failing on invalid keyword argument

    - by Darwin Tech
    I have a models.py like so: from django.db import models from django.contrib.auth.models import User from datetime import datetime class UserProfile(models.Model): user = models.OneToOneField(User) def __unicode__(self): return self.user.username class Project(models.Model): user = models.ForeignKey(UserProfile) created = models.DateTimeField(auto_now_add=True) updated = models.DateTimeField(auto_now=True) product = models.ForeignKey('tool.product') module = models.ForeignKey('tool.module') model = models.ForeignKey('tool.model') zipcode = models.IntegerField(max_length=5) def __unicode__(self): return unicode(self.id) And my tests.py: from django.test import TestCase, Client # --- import app models from django.contrib.auth.models import User from tool.models import Module, Model, Product from user_profile.models import Project, UserProfile # --- unit tests --- # class UserProjectTests(TestCase): fixtures = ['admin_user.json'] def setUp(self): self.product1 = Product.objects.create( name='bar', ) self.module1 = Module.objects.create( name='foo', enable=True ) self.model1 = Model.objects.create( module=self.module1, name='baz', enable=True ) self.user1 = User.objects.get(pk=1) ... def test_can_create_project(self): self.project1 = Model.objects.create( user=self.user1, product=self.product1, module=self.module1, model=self.model1, zipcode=90210 ) self.assertEquals(self.project1.zipcode, 90210) But I get a TypeError: 'product' is an invalid keyword argument for this function error. I'm not sure what is failing but I'm guessing something to do with the FK relationships... Any help would be much appreciated.

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