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• Django, wsgi, py. what's the difference?

  - by Kenny

  I'm trying to get a django application up and running on my cpanel system. I've installed mod_wsgi, and am following the guide here: http://www.nerdydork.com/setting-up-django-on-a-whm-cpanel-vps-liquidweb.html However, I'm now confused as I don't know what to do next. The application has .py files, and I am able to run it via this: python manage.py runserver 211.144.131.148:8000 However, that's via command line and binds to port 8000. I want to use Apache instead. The question is, that tutorial doesn't go further into how to get apache to recognize .py files and run the application as I want it. What do I do next?

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• Google App Engine (python): TemplateSyntaxError: 'for' statements with five words should end in 'rev

  - by Phil

  This is using the web app framework, not Django. The following template code is giving me an TemplateSyntaxError: 'for' statements with five words should end in 'reversed' error when I try to render a dictionary. I don't understand what's causing this error. Could somebody shed some light on it for me? {% for code, name in charts.items %} <option value="{{code}}">{{name}}</option> {% endfor %} I'm rendering it using the following: class GenerateChart(basewebview): def get(self): values = {"datepicker":True} values["charts"] = {"p3": "3D Pie Chart", "p": "Segmented Pied Chart"} self.render_page("generatechart.html", values) class basewebview(webapp.RequestHandler): ''' Base class for all webapp.RequestHandler type classes ''' def render_page(self, filename, template_values=dict()): filename = "%s/%s" % (_template_dir, filename) path = os.path.join(os.path.dirname(__file__), filename) self.response.out.write(template.render(path, template_values))

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• Hooking up Sproutcore frontend and custom Python backend

  - by Suvir

  Hello everyone, I am building a web-based application. The frontend has been designed in Sproutcore. For the backend, we have our own python API which handles all transactions with multiple databases. What is the best way to hook up the front-end with the back-end. AFAIK django is pretty monolithic (correct me if i am wrong) and it would be cumbersome if I dont use its native ORM...I would prefer a python-based solution..any ideas? thanks! Suvir

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• Why don't these class attributes register?

  - by slypete

  I have a factory method that generates django form classes like so: def get_indicator_form(indicator, patient): class IndicatorForm(forms.Form): #These don't work! indicator_id = forms.IntegerField(initial=indicator.id, widget=forms.HiddenInput()) patient_id = forms.IntegerField(initial=patient.id, widget=forms.HiddenInput()) def __init__(self, *args, **kwargs): forms.Form.__init__(self, *args, **kwargs) self.indicator = indicator self.patient = patient #These do! setattr(IndicatorForm, 'indicator_id', forms.IntegerField(initial=indicator.id, widget=forms.HiddenInput())) setattr(IndicatorForm, 'patient_id', forms.IntegerField(initial=patient.id, widget=forms.HiddenInput())) for field in indicator.indicatorfield_set.all(): setattr(IndicatorForm, field.name, copy(field.get_field_type())) return type('IndicatorForm', (forms.Form,), dict(IndicatorForm.__dict__)) I'm trying to understand why the top form field declarations don't work, but the setattr method below does work. I'm fairly new to python, so I suspect it's some language feature that I'm misunderstanding. Can you help me understand why the field declarations at the top of the class don't add the fields to the class? In a possibly related note, when these classes are instantiated, instance.media returns nothing even though some fields have widgets with associated media. Thanks, Pete

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• Search over multiple fields

  - by schneck

  Hi there, I think I don't unterstand django-haystack properly: I have a data model containing several fields, and I would to have two of them searched: class UserProfile(models.Model): user = models.ForeignKey(User, unique=True, default=None) twitter_account = models.CharField(max_length=50, blank=False) My search index settings: class UserProfileIndex(SearchIndex): text = CharField(document=True, model_attr='user') twitter_account = CharField(model_attr='twitter_account') def get_queryset(self): """Used when the entire index for model is updated.""" return UserProfile.objects.all() But when I perform a search, only the field "username" is searched; "twitter_account" is ignored. When I select the Searchresults via dbshell, the objects contain the correct values for "user" and "twitter_account", but the result page shows a "no results": {% if query %} <h3>Results</h3> {% for result in page.object_list %} <p> <a href="{{ result.object.get_absolute_url }}">{{ result.object.id }}</a> </p> {% empty %} <p>No results</p> {% endfor %} {% endif %} Any ideas?

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• Cache for everybody except staff members.

  - by Oli

  I have a django site where I want to stick an "admin bar" along the top of every non-admin page for staff members. It would contain useful things like page editing tools, etc. The problem comes from me using the @cache_page decorator on lots of pages. If a normal user hits a page, the cached version comes up without the admin bar (even for admin users) and if an admin hits the page first, normal users see the admin bar. I could tediously step through the templates, adding regional cache blocks but there are a lot of templates, and life is altogether too short. Ideally, there would be a way of telling the caching to ignore cache get/set requests from admin users... But I don't know how to best implement that. How would you tackle this problem?

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• How do I prevent a ManyToManyField('self') from linked an object to itself?

  - by dyve

  Consider this model (simplified for this question): class SecretAgentName(models.Model): name = models.CharField(max_length=100) aliases = ManyToManyField('self') I have three names, "James Bond", "007" and "Jason Bourne". "James Bond" and "007" are aliases of each other. This works exactly like I want it to, except for the fact that every instance can also be an alias of itself. This I want to prevent. So, there can be many SecretAgentNames, all can be aliases of each other as long as "James Bond" does not show up as an alias for "James Bond". Can I prevent this in the model definition? If not, can I prevent it anywhere else, preferably so that the Django Admin understands it?

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• Removing a result from Queryset

  - by Enrico

  Is there a simple way to discard/remove the last result in a queryset without affecting the db? I am trying to paginate results in Django, but don't know the total number of objects for a given query. I was planning on using next/previous or older/newer links, so I only need to know if this is the first and/or last page. First is easy to check. To check for the last page I can compare the number of results with the pagesize or make a second query. The first method fails to detect the last page when the number of results in the last set equals the pagesize (ie 100 records get broken into 10 pages with the last page containing exactly 10 results) and I would like to avoid making a second query. My current thought is that I should fetch pagesize + 1 results from the db. If the queryset length equals 11, I know this is not the last page and I want to discard the last result in the queryset before passing the queryset to the template.

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• Quering distinct values throught related model

  - by matheus.emm

  Hi! I have a simple one-to-many (models.ForeignKey) relationship between two of my model classes: class TeacherAssignment(models.Model): # ... some fields year = models.CharField(max_length=4) class LessonPlan(models.Model): teacher_assignment = models.ForeignKey(TeacherAssignment) # ... other fields I'd like to query my database to get the set of distinct years of TeacherAssignments related to at least one LessonPlan. I'm able to get this set using Django query API if I ignore the relation to LessonPlan: class TeacherAssignment(models.Model): # ... model's fields def get_years(self): year_values = self.objects.all().values_list('year').distinct().order_by('-year') return [yv[0] for yv in year_values if len(yv[0]) == 4] Unfortunately I don't know how to express the condition that the TeacherAssignment must be related to at least one LessonPlan. Any ideas how I'd be able to write the query? Thanks in advance.

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• Passing session data to ModelForm inside of ModelAdmin

  - by theactiveactor

  I'm trying to initialize the form attribute for MyModelAdmin class inside an instance method, as follows: class MyModelAdmin(admin.ModelAdmin): def queryset(self, request): MyModelAdmin.form = MyModelForm(request.user) My goal is to customize the editing form of MyModelForm based on the current session. When I try this however, I keep getting an error (shown below). Is this the proper place to pass session data to ModelForm? If so, then what may be causing this error? TypeError at ... Exception Type: TypeError Exception Value: issubclass() arg 1 must be a class Exception Location: /usr/lib/pymodules/python2.6/django/forms/models.py in new, line 185

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• Is it ok to hardcode dynamic links in a permanent view?

  - by meder

  Let's say I wanted to showcase 2-3 clickable buttons on my homepage which will be there permanently. These are links to the css, html, and javascript tag listing pages. Is it fine to just hardcode href=/tags/css and href=/tags/html right in my django templates/view? I won't change them for at least a year or so, meaning I don't think I need to add a column to the tags table to distinguish them - is this common or should I try to make it somewhat dynamic? These tags are in a table but so are 1000 other tags.

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• Saving related model objects

  - by iHeartDucks

  I have two related models (one to many) in my django app and When I do something like this ObjBlog = Blog() objBlog.name = 'test blog' objEntry1 = Entry() objEntry1.title = 'Entry one' objEntry2 = Entry() objEntry2.title = 'Entry Two' objBlog.entry_set.add(objEntry1) objBlog.entry_set.add(objEntry2) I get an error which says "null value in column and it violates the foreign key not null constraint". None of my model objects have been saved. Do I have to save the "objBlog" before I could set the entries? I was hoping I could call the save method on objBlog to save it all. NOTE: I am not creating a blog engine and this is just an example.

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• Select distinct users with referrals

  - by Mark

  I have a bunch of Users. Since Django doesn't really let me extend the default User model, they each have Profiles. The Profiles have a referred_by field (a FK to User). I'm trying to get a list of Users with = 1 referral. Here's what I've got so far Profile.objects.filter(referred_by__isnull=False).values_list('referred_by', flat=True) Which gives me a list of IDs of the users who have referrals... but I need it to be distinct, and I want the User object, not their ID. Or better yet, it would be nice if it could return the number of referrals a user has. Any ideas?

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• Overwrite queryset which builds filter sidebar

  - by cw

  Hi, I'm writing a hockey database/manager. So I have the following models: class Team(models.Model): name = models.CharField(max_length=60) class Game(models.Model): home_team = models.ForeignKey(Team,related_name='home_team') away_team = models.ForeignKey(Team,related_name='away_team') class SeasonStats(models.Model): team = models.ForeignKey(Team) Ok, so my problem is the following. There are a lot of teams, but Stats are just managed for my Club. So if I use "list_display" in the admin backend, I'd like to modify/overwrite the queryset which builds the sidebar for filtering, to just display our home teams as a filter option. Is this somehow possible in Django? I already made a custom form like this class SeasonPlayerStatsAdminForm(forms.ModelForm): team = forms.ModelChoiceField(Team.objects.filter(club__home=True)) So now just the filtering is missing. Any ideas?

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• Enable export to XML via HTTP on a large number of models with child relations

  - by Vasil

  I've a large number of models (120+) and I would like to let users of my application export all of the data from them in XML format. I looked at django-piston, but I would like to do this with minimum code. Basically I'd like to have something like this: GET /export/applabel/ModelName/ Would stream all instances of ModelName in applabel together with it's tree of related objects . I'd like to do this without writing code for each model. What would be the best way to do this?

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• Errors in Decimal Calcs within def clean method?

  - by allanhenderson

  I'm attempting a few simple calculations in a def clean method following validation (basically spitting out a euro conversion of retrieved uk product price on the fly). I keep getting a TypeError. Full error reads: Cannot convert {'product': , 'invoice': , 'order_discount': Decimal("0.00"), 'order_price': {...}, 'order_adjust': None, 'order_value': None, 'DELETE': False, 'id': 92, 'quantity': 8} to Decimal so I guess django is passing through the entire cleaned_data form to Decimal method. Not sure where I'm going wrong - the code I'm working with is: def clean_order_price(self): cleaned_data = self.cleaned_data data = self.data order_price = cleaned_data.get("order_price") if not order_price: try: existing_price = ProductCostPrice.objects.get(supplier=data['supplier'], product_id=cleaned_data['product'], is_latest=True) except ProductCostPrice.DoesNotExist: existing_price = None if not existing_price: raise forms.ValidationError('No match found, please enter new price') else: if data['invoice_type'] == 1: return existing_price.cost_price_gross elif data['invoice_type'] == 2: exchange = EuroExchangeRate.objects.latest('exchange_date') calc = exchange.exchange_rate * float(existing_price.cost_price_gross) calc = Decimal(str(calc)) return calc return cleaned_data If the invoice is of type 2 (a euro invoice) then the system should grab the latest exchange rate and apply that to the matching UK pound price pulled through to get euro result. Should performing a decimal conversion be a problem within def clean method? Thanks

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• Elegant solution for multiple forms on single page

  - by NFicano

  I'm building a web application (in Django) that will accept a search criteria and display a report - once the user is satisfied with the results, save both the criteria and a reference to these objects back to the database. The problem I'm having is finding an elegant solution for having 2 forms: Display (GET) the results of their criteria. Enter in some descriptions, and save (POST) everything back to the database. I'm leaning towards AJAX for the GET stuff and a POST for the save, but I wanted to make sure there wasn't a more elegant solution first.

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• Indexing a method return (depending on Internationalization)

  - by Hedde

  Consider a django model with an IntegerField with some choices, e.g. COLORS = ( (0, _(u"Blue"), (1, _(u"Red"), (2, _(u"Yellow"), ) class Foo(models.Model): # ...other fields... color = models.PositiveIntegerField(choices=COLOR, verbose_name=_(u"color")) My current (haystack) index: class FooIndex(SearchIndex): text = CharField(document=True, use_template=True) color = CharField(model_attr='color') def prepare_color(self, obj): return obj.get_color_display() site.register(Product, ProductIndex) This obviously only works for keyword "yellow", but not for any (available) translations. Question: What's would be a good way to solve this problem? (indexing method returns based on the active language) What I have tried: I created a function that runs a loop over every available language (from settings) appending any translation to a list, evaluating this against the query, pre search. If any colors are matched it converts them backwards into their numeric representation to evaluate against obj.color, but this feels wrong.

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• Not quite nested inlines?

  - by Lynden Shields

  Not quite sure what to call this, it's not quite nested inlines, but is probably related. I have a 3 level hierarchy of objects, A one-to-many B one-to-many C. Therefore, every C implicitly also belongs to an A. class A(models.Model): stuff = models.CharField("Stuff", max_length=50) class B(models.Model): a = models.ForeignKey(A) class C(models.Model): b = models.ForeignKey(B) I would like all C's that belong to an A to be listed on the admin page for A in an in-line. They do not have to show which B they belong to on the same page. Is this possible or is it the same problem as nested inlines anyway? If it's possible, how do I do it? I'm using django 1.3

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• Celery tasks not works with gevent

  - by Novarg

  When i use celery + gevent for tasks that uses subprocess module i'm getting following stacktrace: Traceback (most recent call last): File "/home/venv/admin/lib/python2.7/site-packages/celery/task/trace.py", line 228, in trace_task R = retval = fun(*args, **kwargs) File "/home/venv/admin/lib/python2.7/site-packages/celery/task/trace.py", line 415, in __protected_call__ return self.run(*args, **kwargs) File "/home/webapp/admin/webadmin/apps/loggingquarantine/tasks.py", line 107, in release_mail_task res = call_external_script(popen_obj.communicate) File "/home/webapp/admin/webadmin/apps/core/helpers.py", line 42, in call_external_script return func_to_call(*args, **kwargs) File "/usr/lib64/python2.7/subprocess.py", line 740, in communicate return self._communicate(input) File "/usr/lib64/python2.7/subprocess.py", line 1257, in _communicate stdout, stderr = self._communicate_with_poll(input) File "/usr/lib64/python2.7/subprocess.py", line 1287, in _communicate_with_poll poller = select.poll() AttributeError: 'module' object has no attribute 'poll' My manage.py looks following (doing monkeypatch there): #!/usr/bin/env python from gevent import monkey import sys import os if __name__ == "__main__": if not 'celery' in sys.argv: monkey.patch_all() os.environ.setdefault("DJANGO_SETTINGS_MODULE", "webadmin.settings") from django.core.management import execute_from_command_line sys.path.append(".") execute_from_command_line(sys.argv) Is there a reason why celery tasks act like it wasn't patched properly? p.s. strange thing that my local setup on Macos works fine while i getting such exceptions under Centos (all package versions are the same, init and config scripts too)

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• Using map() on a _set in a template?

  - by Stuart Grimshaw

  I have two models like this: class KPI(models.Model): """KPI model to hold the basic info on a Key Performance Indicator""" title = models.CharField(blank=False, max_length=100) description = models.TextField(blank=True) target = models.FloatField(blank=False, null=False) group = models.ForeignKey(KpiGroup) subGroup = models.ForeignKey(KpiSubGroup, null=True) unit = models.TextField(blank=True) owner = models.ForeignKey(User) bt_measure = models.BooleanField(default=False) class KpiHistory(models.Model): """A historical log of previous KPI values.""" kpi = models.ForeignKey(KPI) measure = models.FloatField(blank=False, null=False) kpi_date = models.DateField() and I'm using RGraph to display the stats on internal wallboards, the handy thing is Python lists get output in a format that Javascript sees as an array, so by mapping all the values into a list like this: def f(x): return float(x.measure) stats = map(f, KpiHistory.objects.filter(kpi=1) then in the template I can simply use {{ stats }} and the RGraph code sees it as an array which is exactly what I want. [87.0, 87.5, 88.5, 90] So my question is this, is there any way I can achieve the same effect using Django's _set functionality to keep the amount of data I'm passing into the template, up until now I've been passing in a single KPI object to be graphed but now I want to pass in a whole bunch so is there anything I can do with _set {{ kpi.kpihistory_set }} dumps the whole model out, but I just want the measure field. I can't see any of the built in template methods that will let me pull out just the single field I want. How have other people handled this situation?

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• What's the best way to run Drupal and Django sites behind the same Varnish server?

  - by Alexis Bellido

  I have a high traffic website running with Drupal and Apache, five web servers behind a Varnish server load balancing. Let's say this site is example.com. I'm using five backends and a director like this in my default.vcl: director balancer round-robin { { .backend = web1; } { .backend = web2; } { .backend = web3; } { .backend = web4; } { .backend = web5; } } Now I'm working on a new Django project that will be a new section of this site running on example.com/new-section. After checking the documentation I found I can do something like this: sub vcl_recv { if (req.url ~ "^/new-section/") { set req.backend = newbackend; } else { set req.backend = default; } } That is, using a different backend for a subdirectory /new-section under the same domain. My question is, how do I make something like this work with my director and load balancing setup? I'm probably going to run two or more web servers (backends) with my new Django project, each one with a mix of Gunicorn, Nginx, and a few Python packages, and would like to put all of those in their own Varnish director to load balance. Is it possible to do use the above approach to decide which director to use?, like this: sub vcl_recv { if (req.url ~ "^/new-section/") { set req.director = newdirector; } else { set req.director = balancer; } } All suggestions welcome. Thanks!

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• Go for Zend framework or Django for a modular web application?

  - by dr. squid

  I am using both Zend framework and Django, and they both have they strengths and weakness, but they are both good framworks in their own way. I do want to create a highly modular web application, like this example: modules: Admin cms articles sections ... ... ... I also want all modules to be self contained with all confid and template files. I have been looking into a way to solve this is zend the last days, but adding one omer level to the module setup doesn't feel right. I am sure this could be done, but should I? I have also included Doctrine to my zend application that could give me even more problems in my module setup! When we are talking about Django this is easy to implement (Easy as in concept, not in implementation time or whatever) and a great way to create web apps. But one of the downsides of Django is the web hosing part. There are some web hosts offering Django support, but not that many.. So then I guess the question is what have the most value; rapid modular development versus hosting options! Well, comments are welcome! Thanks

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• Django: How can I identify the calling view from a template?

  - by bryan

  Short version: Is there a simple, built-in way to identify the calling view in a Django template, without passing extra context variables? Long (original) version: One of my Django apps has several different views, each with its own named URL pattern, that all render the same template. There's a very small amount of template code that needs to change depending on the called view, too small to be worth the overhead of setting up separate templates for each view, so ideally I need to find a way to identify the calling view in the template. I've tried setting up the views to pass in extra context variables (e.g. "view_name") to identify the calling view, and I've also tried using {% ifequal request.path "/some/path/" %} comparisons, but neither of these solutions seems particularly elegant. Is there a better way to identify the calling view from the template? Is there a way to access to the view's name, or the name of the URL pattern? Update 1: Regarding the comment that this is simply a case of me misunderstanding MVC, I understand MVC, but Django's not really an MVC framework. I believe the way my app is set up is consistent with Django's take on MVC: the views describe which data is presented, and the templates describe how the data is presented. It just happens that I have a number of views that prepare different data, but that all use the same template because the data is presented the same way for all the views. I'm just looking for a simple way to identify the calling view from the template, if this exists. Update 2: Thanks for all the answers. I think the question is being overthought -- as mentioned in my original question, I've already considered and tried all of the suggested solutions -- so I've distilled it down to a "short version" now at the top of the question. And right now it seems that if someone were to simply post "No", it'd be the most correct answer :) Update 3: Carl Meyer posted "No" :) Thanks again, everyone.

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• What's the next steps for moving from appengine to full django?

  - by tomcritchlow

  Hey guys, I'm super new to programming and I've been using appengine to help me learn python and general coding. I'm getting better quickly and I'm loving it all the way :) Appengine was awesome for allowing me to just dive into writing my app and getting something live that works (see http://www.7bks.com/). But I'm realising that the longer I continue to learn on appengine the more I'm constraining myself and locking myself into a single system. I'd like to move to developing on full django (since django looks super cool!). What are my next steps? To give you a feel for my level of knowledge: I'm not a unix user I'm not familiar with command line controls (I still use appengine/python completely via the appengine SDK) I've never programmed in anything other than python, anywhere other than appengine I know the word SQL, but don't know what MySQL is really or how to use it. So, specifically: What are the skills I need to learn to get up and running with full django/python? If I'm going to host somewhere else I suppose I'll need to learn some sysadmin type skills (maybe even unix?). Is there anywhere that offers easy hosting (like appengine) but that supports django? I hear such great things about heroku I'm considering switching to RoR and going there I appreciate that I'm likely not quite ready to move away from appengine just yet but I'm a fiercely passionate learner (http://www.7bks.com/blog/179001) and would love it if I knew all the steps I needed to learn so I could set about learning them. At the moment, I don't even know what the steps are I need to learn! Thank you very much. Sorry this isn't a specific programming question but I've looked around and haven't found a good how-to for someone of my level of experience and I think others would appreciate a good roadmap for the things we need to learn to get up and running. Thanks, Tom PS - if anyone is in London and fancies showing me the ropes in person that would be super awesome :)

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