I’ve defined a simple Django app that includes the following model: (Technically yes, that could have been an ImageField.) In a template, it’s easy enough to include the MEDIA_URL value (duly coded in settings.py) as a prefix to the thumbnail URL. The following works fine: Using DRF, I’ve defined a HyperlinkedModelSerializer descendant called ProjectSerializer: And I’ve defined a very straightforward
Tag: django
ImproperlyConfigured: There are duplicate field(s) in PackageAdmin.fieldsets
I have a very simple model : here is the admin.py : This implementation give me the following error : Any idea why ? If I let the second ‘fields’ empty, I don’t get the error. But if I let the first ‘fields’ empty, I still have this error. Answer Björn Kristinsson did solve my issue (see in the comment
How to limit choices of ForeignKey choices for Django raw_id_field
How do you limit what choices are shown for ForeignKey fields in Django’s admin when they’re displayed using the raw_id_fields option? When rendered as a select box, it’s simple to define a custom ModelForm to set that field’s queryset value with the choices to want. However, this queryset appears to be completely ignored when rendered using raw_id_fields. It generates a
Django – login required for POST but not GET?
So I’m trying to do a site with user editable tags on items. I want users to be able to edit tags on the page only if they are logged in, but everyone should be able to view the page. The edits are done through a modelform. I can’t use the login_required decorator on the whole view because then only
Best way of linking to a page in Django
I managed to create a URL tag for my index. But right now I’m confused how to add links to other pages. I put this on my urls.py The next thing I put this tag into the href: But what if I wanted to create a new page and would be linking to it. How would I do it the
AttributeError ‘tuple’ object has no attribute ‘get’
I have a Django application. But i can’t an error that i have been struggling with for some time now. And this is the traceback django provides me. I have a hard time figuring out why this error occurs. How can i find out where that tuple is in my code? The view: Answer You are returning a tuple here:
Access-Control-Allow-Origin in Django app
I’m developing a Phonegap app for my Django based app, but when trying to make Ajax calls I get this error: How can I make it so my Django app allows cross origin for some urls? Here’s my Ajax code: Answer Django by default does not provide the headers necessary to provide cross origin. The easiest way would be to
Convert Django Model object to dict with all of the fields intact
How does one convert a django Model object to a dict with all of its fields? All ideally includes foreign keys and fields with editable=False. Let me elaborate. Let’s say I have a django model like the following: In the terminal, I have done the following: I want to convert this to the following dictionary: Questions with unsatisfactory answers: Django:
Django got an unexpected keyword argument
I’m trying to create an archive so I pass to the view the arguments year and month. However, I get an error with the code below and I can’t figure out what it means and how to solve it: What can be wrong? views.py urls.py Update: models.py Template Update 2: error Answer There is something wrong with your parameters names
How to use Query Strings for filtering Querysets in Django?
I created my “API” using REST framework, now I am trying to do filtering for it. This is how my models.py looks like: I would like to see “all passengers in particular workspace” or “all passengers in particular airline” etc in my JSON file. Here is my serializers.py: And views.py: I would like to use Filtering against query parameter but