Question

I want to do equivalent query as SELECT user_name, item_name FROM users, items WHERE users.favor_item_id = items.item_id, which is to return user_name and item_name pairs. In the database, one user can have multiple favoured items.

我只想问一下一下一下一下一下一下一下一下一下一下,KQQQ是什么?

我的第一项想法是列出所有(用户,赞成项目_id)的电梯,然后看项目_名称和id项目。但看来,它将在使用上述Kingk查询时,复杂性是O(NlogM)(M是ITEM的多个项目),而复杂的是O(N)。

是否在贾戈(或任何办公室管理系统)有更有效的方法这样做?

Answer 1

(或是否有任何管理系统)?

真的? 页: 1

假设合理的地图级(如使用申报):

Base = sqlalchemy.ext.declarative.declarative_base()

class Item(Base):
    __tablename__ =  items 
    id = Column( item_id , Integer, primary_key=True)
    name = Column( item_name , String)

class User(Base):
    __tablename__ =  users 
    name = Column( user_name , String(50), primary_key=True)
    favor_item_id = Column(Integer, ForeignKey(Item.id))

    favor_item = relationship(Item, backref="favored_by")

问题是简单的

>>> print sqlalchemy.orm.Query((User.name, Item.name)).join(Item.favored_by)
SELECT users.user_name AS users_user_name, items.item_name AS items_item_name 
FROM items JOIN users ON items.item_id = users.favor_item_id

Answer 2

假设模型:

class User(models.Model):
    name = models.CharField(max_length=32)
    favorite = models.ForeignKey(Item)

class Item(models.Model):
    name = models.CharField(max_length=32)

相应的Django是:

usersFavorites = User.objects.all().values_list("name", "favorite__name")

这并不使用你提供的田地的确切名称。为此,您仅需要将这些模型添加适当的<代码>db_table和db_column领域。

P.S. 这一特别图谋并不特别好。 IMHO,User and Item should have a many-to-many relationship:

class User(models.Model):
    name = models.CharField(max_length=32)
    favorites = models.ManyToManyField(Item, related_name="users")

class Item(models.Model):
    name = models.CharField(max_length=32)

除非项目只有一个用户,在这种情况下是<代码>。外国Key 项目。

之后,你可以翻一番:

for user in User.objects.all().select_related("favorites"):
    for favorite in user.favorites:
        # use `user` and `favorite` in some way.

Because of the select_related() call, the query is done in one fell swoop in the first for loop.

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