Django如何为PostgreSQL数据库中现有的许多表定义模型 [英] django how to define models for existing many to many tables in postgresql database

问题描述

我有一个具有多对多关系的现有PostgreSQL数据库，它通过如下的联接表进行处理.

I have an existing PostgreSQL database with a many to many relationship, it is handled through a join table as below.

我要应用ManyToMany字段类型，这会绕过此联接/中间表吗?如何在我的models.py中正确配置它?我宁愿保留中间表.

I'm looking to apply the ManyToMany Field type, does this bypass this join/intermediate table? How do I correctly configure it in my models.py? I'd rather keep the intermediate table.

# models.py
class Sample(models.Model):
    sample_id = models.AutoField(primary_key=True)
...

class JoinSampleContainer(models.Model):
    id = models.AutoField(primary_key=True)
    container_id = models.ForeignKey(Container, db_column='container_id', on_delete = models.PROTECT)
    sample_id = models.ForeignKey(Sample, db_column='sample_id', on_delete = models.PROTECT)
...

class Container(models.Model):
    container_id = models.AutoField(primary_key=True)

推荐答案

在您的其中一个模型(例如 Sample )上定义 ManyToManyField ，并指定 through 选项作为此处记录:

Define the ManyToManyField on one of your models (e.g. Sample) specifying a through option as documented here:

class Sample(models.Model):
    id = ...
    containers = models.ManyToManyField(Container, through='JoinSampleContainer', through_fields=('sample_id', 'container_id'),
        related_name='samples')

注意:您应该为模型中的字段命名，以提高可读性(并使用 db_column 指定所使用的DB列).使用 id 代替 sample_id ，使用 sample.id 代替 sample.sample_id 更具可读性.并使用 sample 代替 sample_id ，在通过模型上分别使用 container 代替 container_id .

Note: You should name the fields in your models for readability (and use db_column to specify the DB column that is used). Use id instead of sample_id, it's much more readable to use sample.id instead of sample.sample_id. And use sample instead of sample_id, resp container instead of container_id on the through model.

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