我的django管理员遇到了一些重大的表现问题。基于我有多少内联的重复查询。
models.py
class Setting(models.Model):
name = models.CharField(max_length=50, unique=True)
class Meta:
ordering = ('name',)
def __str__(self):
return self.name
class DisplayedGroup(models.Model):
name = models.CharField(max_length=30, unique=True)
position = models.PositiveSmallIntegerField(default=100)
class Meta:
ordering = ('priority',)
def __str__(self):
return self.name
class Machine(models.Model):
name = models.CharField(max_length=20, unique=True)
settings = models.ManyToManyField(
Setting, through='Arrangement', blank=True
)
class Meta:
ordering = ('name',)
def __str__(self):
return self.name
class Arrangement(models.Model):
machine = models.ForeignKey(Machine, on_delete=models.CASCADE)
setting = models.ForeignKey(Setting, on_delete=models.CASCADE)
displayed_group = models.ForeignKey(
DisplayedGroup, on_delete=models.PROTECT,
default=1)
priority = models.PositiveSmallIntegerField(
default=100,
help_text='Smallest number will be displayed first'
)
class Meta:
ordering = ('priority',)
unique_together = (("machine", "setting"),)
admin.py
class ArrangementInline(admin.TabularInline):
model = Arrangement
extra = 1
class MachineAdmin(admin.ModelAdmin):
inlines = (ArrangementInline,)
如果我在内联表单上添加了3个设置，另外还有1个设置，我有大约10个重复查询
SELECT "corps_setting"."id", "corps_setting"."name", "corps_setting"."user_id", "corps_setting"."tagged", "corps_setting"."created", "corps_setting"."modified" FROM "corps_setting" ORDER BY "corps_setting"."name" ASC
- Duplicated 5 times
SELECT "corps_displayedgroup"."id", "corps_displayedgroup"."name", "corps_displayedgroup"."color", "corps_displayedgroup"."priority", "corps_displayedgroup"."created", "corps_displayedgroup"."modified" FROM "corps_displayedgroup" ORDER BY "corps_displayedgroup"."priority" ASC
- Duplicated 5 times.
有人可以告诉我，我在这里做错了什么吗？我花了3天的时间试图自己解决问题而没有运气。
当我有一台机器的大约50个设置内联时问题会变得更糟，我会有~100个查询。
答案 0 :(得分：5)
我已经根据@ makaveli的答案组装了一个通用的解决方案，评论中似乎没有提到问题：
class CachingModelChoicesFormSet(forms.BaseInlineFormSet):
"""
Used to avoid duplicate DB queries by caching choices and passing them all the forms.
To be used in conjunction with `CachingModelChoicesForm`.
"""
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
sample_form = self._construct_form(0)
self.cached_choices = {}
try:
model_choice_fields = sample_form.model_choice_fields
except AttributeError:
pass
else:
for field_name in model_choice_fields:
if field_name in sample_form.fields and not isinstance(
sample_form.fields[field_name].widget, forms.HiddenInput):
self.cached_choices[field_name] = [c for c in sample_form.fields[field_name].choices]
def get_form_kwargs(self, index):
kwargs = super().get_form_kwargs(index)
kwargs['cached_choices'] = self.cached_choices
return kwargs
class CachingModelChoicesForm(forms.ModelForm):
"""
Gets cached choices from `CachingModelChoicesFormSet` and uses them in model choice fields in order to reduce
number of DB queries when used in admin inlines.
"""
@property
def model_choice_fields(self):
return [fn for fn, f in self.fields.items()
if isinstance(f, (forms.ModelChoiceField, forms.ModelMultipleChoiceField,))]
def __init__(self, *args, **kwargs):
cached_choices = kwargs.pop('cached_choices', {})
super().__init__(*args, **kwargs)
for field_name, choices in cached_choices.items():
if choices is not None and field_name in self.fields:
self.fields[field_name].choices = choices
您需要做的就是从CachingModelChoicesForm继承您的模型，并在您的内联类中使用CachingModelChoicesFormSet：
class ArrangementInlineForm(CachingModelChoicesForm):
class Meta:
model = Arrangement
exclude = ()
class ArrangementInline(admin.TabularInline):
model = Arrangement
extra = 50
form = ArrangementInlineForm
formset = CachingModelChoicesFormSet
答案 1 :(得分：2)
这在Django中是非常正常的行为 - 它不会为你做优化，但它为你自己提供了不错的工具。并且不要冒汗，100个查询并不是一个大问题（我在一个页面上看到了16k查询），需要立即修复。但是如果你的数据量会迅速增加，那么当然应该处理它是明智的。
您将使用的主要武器是查询集方法select_related()和prefetch_related()。真的没有必要深入研究它们，因为它们有很好的记录here，但只是一般指针：
当您查询的对象只有一个相关对象（FK或one2one）时使用select_related()
当您查询的对象有多个相关对象（FK或M2M的另一端）时使用prefetch_related()
如何在Django管理员中使用它们，你问？小学，亲爱的沃森。覆盖管理页面方法get_queryset(self, request)，所以它看起来像这样：
from django.contrib import admin
class SomeRandomAdmin(admin.ModelAdmin):
def get_queryset(self, request):
return super().get_queryset(request).select_related('field1', 'field2').prefetch_related('field3')
编辑：阅读完您的评论后，我意识到我对您的问题的初步解释是绝对错误的。我也有针对您的问题的多种解决方案，这就是：
我大部分时间都在使用的简单建议：只需用raw_id_field小部件替换Django默认选择小部件，不会进行任何查询。只需在内联管理员中设置raw_id_fields = ('setting', 'displayed_group')即可。
但是，如果你不想摆脱选择框，我可以提供一些半hacky代码来完成这个技巧，但是相当冗长且不太漂亮。我们的想法是覆盖创建表单的formset，并为formset中的这些字段指定选项，以便它们只从数据库中查询一次。
这就是：
from django import forms
from django.contrib import admin
from app.models import Arrangement, Machine, Setting, DisplayedGroup
class ChoicesFormSet(forms.BaseInlineFormSet):
setting_choices = list(Setting.objects.values_list('id', 'name'))
displayed_group_choices = list(DisplayedGroup.objects.values_list('id', 'name'))
def _construct_form(self, i, **kwargs):
kwargs['setting_choices'] = self.setting_choices
kwargs['displayed_group_choices'] = self.displayed_group_choices
return super()._construct_form(i, **kwargs)
class ArrangementInlineForm(forms.ModelForm):
class Meta:
model = Arrangement
exclude = ()
def __init__(self, *args, **kwargs):
setting_choices = kwargs.pop('setting_choices', [((), ())])
displayed_group_choices = kwargs.pop('displayed_group_choices', [((), ())])
super().__init__(*args, **kwargs)
# This ensures that you can still save the form without setting all 50 (see extra value) inline values.
# When you save, the field value is checked against the "initial" value
# of a field and you only get a validation error if you've changed any of the initial values.
self.fields['setting'].choices = [('-', '---')] + setting_choices
self.fields['setting'].initial = self.fields['setting'].choices[0][0]
self.fields['setting'].empty_values = (self.fields['setting'].choices[0][0],)
self.fields['displayed_group'].choices = displayed_group_choices
self.fields['displayed_group'].initial = self.fields['displayed_group'].choices[0][0]
class ArrangementInline(admin.TabularInline):
model = Arrangement
extra = 50
form = ArrangementInlineForm
formset = ChoicesFormSet
def get_queryset(self, request):
return super().get_queryset(request).select_related('setting')
class MachineAdmin(admin.ModelAdmin):
inlines = (ArrangementInline,)
admin.site.register(Machine, MachineAdmin)
如果您发现可以改进的内容或有任何疑问，请与我们联系。
答案 2 :(得分：0)
现在，（对that question表示敬意），BaseFormset收到form_kwargs attribute。
可以对接受的答案中的ChoicesFormSet代码进行如下修改：
class ChoicesFormSet(forms.BaseInlineFormSet):
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
setting_choices = list(Setting.objects.values_list('id', 'name'))
displayed_group_choices = list(DisplayedGroup.objects.values_list('id', 'name'))
self.form_kwargs['setting_choices'] = self.setting_choices
self.form_kwargs['displayed_group_choices'] = self.displayed_group_choices
其余代码保持完整，如公认的答案所述：
class ArrangementInlineForm(forms.ModelForm):
class Meta:
model = Arrangement
exclude = ()
def __init__(self, *args, **kwargs):
setting_choices = kwargs.pop('setting_choices', [((), ())])
displayed_group_choices = kwargs.pop('displayed_group_choices', [((), ())])
super().__init__(*args, **kwargs)
# This ensures that you can still save the form without setting all 50 (see extra value) inline values.
# When you save, the field value is checked against the "initial" value
# of a field and you only get a validation error if you've changed any of the initial values.
self.fields['setting'].choices = [('-', '---')] + setting_choices
self.fields['setting'].initial = self.fields['setting'].choices[0][0]
self.fields['setting'].empty_values = (self.fields['setting'].choices[0][0],)
self.fields['displayed_group'].choices = displayed_group_choices
self.fields['displayed_group'].initial = self.fields['displayed_group'].choices[0][0]
class ArrangementInline(admin.TabularInline):
model = Arrangement
extra = 50
form = ArrangementInlineForm
formset = ChoicesFormSet
def get_queryset(self, request):
return super().get_queryset(request).select_related('setting')
class MachineAdmin(admin.ModelAdmin):
inlines = (ArrangementInline,)
admin.site.register(Machine, MachineAdmin)