Django ORM Queryset 的缓存机制, 惰性查询简述

在Django的ORM中 必须注意由于QuerySet的 cache导致的数据获取不正确的问题

在哪些情况下不会出发QuerySet缓存?

隐式存储QuerySet（查询语句没有显示赋值给变量而直接进行遍历或截取）

>>> from project.models import ProjectModel
>>>
>>> print([project_instance.name for project_instance in ProjectModel.objects.all()])
['first-test', 'test百度', 'project002']
>>> ProjectModel.objects.filter(name="project002").update(name="project003")
1
>>> print([project_instance.name for project_instance in ProjectModel.objects.all()])
['first-test', 'test百度', 'project003']

而显示的存储QuerSet 并且经过完整遍历才会触发缓存

完整遍历的情况

>>> projects_queryset = ProjectModel.objects.all()
>>> print([project_instance for project_instance in projects_queryset])        # 完整遍历
[<ProjectModel: first-test>, <ProjectModel: test百度>, <ProjectModel: project007>]
>>> ProjectModel.objects.filter(name="project007").update(name="project008")
1
>>> projects_queryset[1:3]
[<ProjectModel: test百度>, <ProjectModel: project007>]  # project007还是缓存的老数据

不遍历的情况

>>> projects_queryset = ProjectModel.objects.all()
>>> projects_queryset
<QuerySet [<ProjectModel: first-test>, <ProjectModel: test百度>, <ProjectModel: project008>]>
>>> ProjectModel.objects.filter(name="project008").update(name="project009")
1
>>> projects_queryset[1:3]
<QuerySet [<ProjectModel: test百度>, <ProjectModel: project009>]>  # 没拿缓存 project009

还有一种场景 也是需要注意的：

获取到单个QuerySet对象后 通过objects update方法修改了部分字段值，此时的QuerySet还是缓存数据

>>> projects_obj = ProjectModel.objects.filter(name="project0011").first()
>>> projects_obj.name
'project0011'
>>> ProjectModel.objects.filter(name="project0011").update(name="project0012")
1
>>> projects_obj.name
'project0011'

有两种方法可以解决这个问题

1. 使用save修改

>>> projects_obj = ProjectModel.objects.filter(name="project0012").first()
>>> projects_obj.name
'project0012'
>>> projects_obj.name = "project0013"
>>> projects_obj.save()
>>> projects_obj.name
'project0013'

1. 使用refresh_from_db()

>>> projects_obj = ProjectModel.objects.filter(name="project0013").first()
>>> projects_obj.name
'project0013'
>>> ProjectModel.objects.filter(name="project0013").update(name="project0014")
1
>>> projects_obj.refresh_from_db()
>>> projects_obj.name
'project0014'