Django 1.6 A.objects.all()。order_by(['field'])如果'field'是模型中的默认顺序,django会做什么?

时间:2014-02-10 10:11:47

标签: django

说我有一个问题模型:

class Question(models.Model):
    class Meta:
        ordering = ['title']
    user = models.ForeignKey(User) 
    title = models.CharField()

如果您在Meta中定义排序,它只是一个默认设置,您可以输入较少的单词,因此Question.objects.all().order_by(['title'])Question.objects.all()的效果会相同吗?

1 个答案:

答案 0 :(得分:5)

是的,表现是一样的。指定Meta.ordering的行为与为每个查询附加order_by完全相同。

您可以通过为记录器django.db.backends设置DEBUG级别来观察Django生成的SQL查询。

示例模型:

class ModelA(models.Model):
    dummy = models.TextField()


class ModelB(models.Model):
    dummy = models.TextField()

    class Meta:
        ordering = ['dummy']

SQL查询示例:

>>> import logging
>>> l = logging.getLogger('django.db.backends')
>>> l.setLevel(logging.DEBUG)
>>> l.addHandler(logging.StreamHandler())

>>> from sqlorder.models import ModelA, ModelB
>>> ModelA.objects.all()
(0.111) SELECT "sqlorder_modela"."id", "sqlorder_modela"."dummy" 
FROM "sqlorder_modela" LIMIT 21; args=()
[]

ModelA默认情况下没有排序,因此ModelA.objects.all不会将ORDER BY附加到查询中。您可以手动附加它。

>>> ModelA.objects.order_by('dummy')
(0.001) SELECT "sqlorder_modela"."id", "sqlorder_modela"."dummy" 
FROM "sqlorder_modela" 
ORDER BY "sqlorder_modela"."dummy" ASC LIMIT 21; args=()
[]

ModelB有默认排序。查询与ModelA的查询相同,只需手动添加order_by

>>> ModelB.objects.all()
(0.001) SELECT "sqlorder_modelb"."id", "sqlorder_modelb"."dummy" 
FROM "sqlorder_modelb" 
ORDER BY "sqlorder_modelb"."dummy" ASC LIMIT 21; args=()
[]

<强>更新 默认排序不会向数据库添加任何其他索引:

$ python manage.py sqlall sqlorder
BEGIN;
CREATE TABLE "sqlorder_modela" (
    "id" serial NOT NULL PRIMARY KEY,
    "dummy" text NOT NULL
)
;
CREATE TABLE "sqlorder_modelb" (
    "id" serial NOT NULL PRIMARY KEY,
    "dummy" text NOT NULL
)
;

COMMIT;