如何解决如何以双联接条件联接三个多对多模型
|| 我要对三个模型/表执行此查询:-- get all items that don\'t have a status for this member
SELECT i.*
FROM
mock_item i
LEFT JOIN mock_itemstatusmember ism ON i.id = ism.item_id AND ism.member_id = 2
WHERE
ism.id IS NULL
这是我的模型的外观:
from django.db import models
# think status as a status
class Member(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=255)
class Status(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=255)
no = models.SmallIntegerField()
class Item(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=255)
status = models.ManyToManyField(Status,through=\'ItemStatusMember\')
member = models.ManyToManyField(Member,through=\'ItemStatusMember\')
class ItemStatusMember(models.Model):
\"\"\"
Table that keeps job statuss information for a user.
Tides Items with Members with Statuss
\"\"\"
id = models.AutoField(primary_key=True)
member = models.ForeignKey(Member)
status = models.ForeignKey(Status)
item = models.ForeignKey(Item)
以及他们的数据库样本数据:
mysql> select * from mock_member;
+----+--------+
| id | name |
+----+--------+
| 1 | Stefan |
| 2 | Alex |
| 3 | Diana |
+----+--------+
3 rows in set (0.00 sec)
mysql> select * from mock_status;
+----+---------+----+
| id | name | no |
+----+---------+----+
| 1 | Pending | 1 |
| 2 | Success | 2 |
+----+---------+----+
2 rows in set (0.00 sec)
mysql> select * from mock_item;
+----+----------------+
| id | name |
+----+----------------+
| 1 | My first item |
| 2 | My second item |
| 3 | My third item |
+----+----------------+
3 rows in set (0.00 sec)
mysql> select * from mock_itemstatusmember;
+----+-----------+-----------+---------+
| id | status_id | member_id | item_id |
+----+-----------+-----------+---------+
| 1 | 1 | 1 | 1 |
| 2 | 2 | 1 | 2 |
| 3 | 1 | 2 | 1 |
+----+-----------+-----------+---------+
3 rows in set (0.00 sec)
我需要使用ORM,而不需要ORM的原始SQL查询工具来获取所有没有该成员状态的项目。
问题开头的SQL查询会执行此操作,因此拥有
AND ism.member_id = 2
条件作为左连接部分的第二个条件,而不是移至我知道该怎么做的where
。
解决方法
from django.db.models import Q
Item.objects.filter(~Q(member__id=2))
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