Django: Order_by multiple fields

I am getting order_by fields in the form of a list. I want to order_by by multiple fields with django orm. List is like below:

orderbyList = ['check-in','check-out','location']

I am writing a query is like:

modelclassinstance.objects.all().order_by(*orderbyList)

Everything i am expecting in a list is dynamic. I don't have predifined set of data.Could some tell me how to write a django ORM with this?

71518 次浏览

Try something like this

modelclassinstance.objects.order_by('check-in', 'check-out', 'location')

You don't need .all() for this

You can also define ordering in your model class

something like

class Meta:
ordering = ['check-in', 'check-out', 'location']

Try this:

listOfInstance = modelclassinstance.objects.all()


for askedOrder in orderbyList:
listOfInstance = listOfInstance.order_by(askedOrder)

Pass orders list in query parameters

eg : yourdomain/?order=location&order=check-out

default_order = ['check-in']  #default order
order = request.GET.getlist('order', default_order)
modelclassinstance.objects.all().order_by(*orderbyList)


What you have to do is chain the querySets, in other words:

classExample.objects.all().order_by(field1).order_by(field2)...

You should pass a list of stringified arguments of the database table columns (as specified by your ORM in the respective models.py file) to the .order_by() method on the return query set like so. shifts = Shift.objects.order_by('start_time', 'employee_first_name'). By notation,.order_by(**args), will help you remember that takes an arbitrary number of arguments.

modelclassinstance.objects.filter(anyfield='value').order_by('check-in', 'check-out', 'location')