Django: Force select related?

问题

I've created a model, and I'm rendering the default/unmodified model form for it. This alone generates 64 SQL queries because it has quite a few foreign keys, and those in turn have more foreign keys.

Is it possible to force it to always (by default) perform a select_related every time one of these models are returned?

回答1:

You can create a custom manager, and simply override get_queryset for it to apply everywhere. For example:

class MyManager(models.Manager):
    def get_queryset(self):
        return super(MyManager, self).get_queryset().select_related('foo', 'bar')

(Prior to Django 1.6, it was get_query_set).

回答2:

Here's also a fun trick:

class DefaultSelectOrPrefetchManager(models.Manager):
    def __init__(self, *args, **kwargs):
        self._select_related = kwargs.pop('select_related', None)
        self._prefetch_related = kwargs.pop('prefetch_related', None)

        super(DefaultSelectOrPrefetchManager, self).__init__(*args, **kwargs)

    def get_queryset(self, *args, **kwargs):
        qs = super(DefaultSelectOrPrefetchManager, self).get_queryset(*args, **kwargs)

        if self._select_related:
            qs = qs.select_related(*self._select_related)
        if self._prefetch_related:
            qs = qs.prefetch_related(*self._prefetch_related)

        return qs


class Sandwich(models.Model):
    bread = models.ForeignKey(Bread)
    extras = models.ManyToManyField(Extra)

    # ...

    objects = DefaultSelectOrPrefetchManager(select_related=('bread',), prefetch_related=('extras',))

Then you can re-use the manager easily between model classes. As an example use case, this would be appropriate if you had a __unicode__ method on the model which rendered a string that included some information from a related model (or anything else that meant a related model was almost always required).

...and if you really want to get wacky, here's a more generalized version. It allows you to call any sequence of methods on the default queryset with any combination of args or kwargs. There might be some errors in the code, but you get the idea.

from django.db import models


class MethodCalls(object):
    """
    A mock object which logs chained method calls.
    """
    def __init__(self):
        self._calls = []

    def __getattr__(self, name):
        c = Call(self, name)
        self._calls.append(c)
        return c

    def __iter__(self):
        for c in self._calls:
            yield tuple(c)


class Call(object):
    """
    Used by `MethodCalls` objects internally to represent chained method calls.
    """
    def __init__(self, calls_obj, method_name):
        self._calls = calls_obj
        self.method_name = method_name

    def __call__(self, *method_args, **method_kwargs):
        self.method_args = method_args
        self.method_kwargs = method_kwargs

        return self._calls

    def __iter__(self):
        yield self.method_name
        yield self.method_args
        yield self.method_kwargs


class DefaultQuerysetMethodCallsManager(models.Manager):
    """
    A model manager class which allows specification of a sequence of
    method calls to be applied by default to base querysets.
    `DefaultQuerysetMethodCallsManager` instances expose a property
    `default_queryset_method_calls` to which chained method calls can be
    applied to indicate which methods should be called on base querysets.
    """
    def __init__(self, *args, **kwargs):
        self.default_queryset_method_calls = MethodCalls()

        super(DefaultQuerysetMethodCallsManager, self).__init__(*args, **kwargs)

    def get_queryset(self, *args, **kwargs):
        qs = super(DefaultQuerysetMethodCallsManager, self).get_queryset(*args, **kwargs)

        for method_name, method_args, method_kwargs in self.default_queryset_method_calls:
            qs = getattr(qs, method_name)(*method_args, **method_kwargs)

        return qs


class Sandwich(models.Model):
    bread = models.ForeignKey(Bread)
    extras = models.ManyToManyField(Extra)

    # Other field definitions...

    objects = DefaultQuerysetMethodCallsManager()
    objects.default_queryset_method_calls.filter(
        bread__type='wheat',
    ).select_related(
        'bread',
    ).prefetch_related(
        'extras',
    )

The python-mock-inspired MethodCalls object is an attempt at making the API more natural. Some might find that a bit confusing. If so, you could sub out that code for an __init__ arg or kwarg that just accepts a tuple of method call information.

回答3:

Create a custom models.Manager and override all the methods (filter, get etc.) and append select_related onto every query. Then set this manager as the objects attribute on the model.

I would recommend just going through your code and adding the select_related where needed, because doing select_related on everything is going to cause some serious performance issues down the line (and it wouldn't be entirely clear where it's coming from).

来源：https://stackoverflow.com/questions/4892729/django-force-select-related

标签

django

django-orm