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I have a database of articles with a
submitter = models.ForeignKey(User, editable=False) Where User is imported as follows:
from django.contrib.auth.models import User. I would like to auto insert the current active user to the submitter field when a particular user submits the article.
Anyone have any suggestions?
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First make sure you have SessionMiddleware and AuthenticationMiddleware middlewares added to your MIDDLEWARE_CLASSES setting. request. user will give you a User object representing the currently logged-in user. If a user isn't currently logged in, request.
Django will create or use an autoincrement column named id by default, which is the same as your legacy column.
To automate this process, we can programmatically fetch all the models in the project and register them with the admin interface. Open admin.py file and add this code to it. This will fetch all the models in all apps and registers them with the admin interface.
Just in case anyone is looking for an answer, here is the solution i've found here: http://demongin.org/blog/806/
To summarize: He had an Essay table as follows:
from django.contrib.auth.models import User class Essay(models.Model): title = models.CharField(max_length=666) body = models.TextField() author = models.ForeignKey(User, null=True, blank=True) where multiuser can create essays, so he created a admin.ModelAdmin class as follows:
from myapplication.essay.models import Essay from django.contrib import admin class EssayAdmin(admin.ModelAdmin): list_display = ('title', 'author') fieldsets = [ (None, { 'fields': [('title','body')] } ), ] def save_model(self, request, obj, form, change): if getattr(obj, 'author', None) is None: obj.author = request.user obj.save() If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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