Django：根据最新的子模型字段排序查询集
让我们假设我想显示一个按最新冲刺时间排序的跑步者列表Django：根据最新的子模型字段排序查询集,django,django-models,django-queryset,Django,Django Models,Django Queryset,让我们假设我想显示一个按最新冲刺时间排序的跑步者列表 class Runner(models.Model): name = models.CharField(max_length=255) class Sprint(models.Model): runner = models.ForeignKey(Runner) time = models.PositiveIntegerField() created = models.DateTimeField(auto_no
class Runner(models.Model):
name = models.CharField(max_length=255)
class Sprint(models.Model):
runner = models.ForeignKey(Runner)
time = models.PositiveIntegerField()
created = models.DateTimeField(auto_now_add=True)
SELECT runner.id, runner.name, sprint.time
FROM runner
LEFT JOIN sprint ON (sprint.runner_id = runner.id)
WHERE
sprint.id = (
SELECT sprint_inner.id
FROM sprint as sprint_inner
WHERE sprint_inner.runner_id = runner.id
ORDER BY sprint_inner.created DESC
LIMIT 1
)
OR sprint.id = NULL
ORDER BY sprint.time ASC
我看到的唯一的另一种可能性是自己编写SQL，但我想不惜一切代价避免这种情况。我不认为只有一个查询就可以通过ORM实现这一点，你可以获取一个跑步者列表并使用它添加他们最新的冲刺id，然后过滤和排序这些冲刺
def view_name(request):
spr = Sprint.objects.values('runner', flat=True).order_by(-created).distinct()
runners = []
for s in spr:
latest_sprint = Sprint.objects.filter(runner=s.runner).order_by(-created)[:1]
for latest in latest_sprint:
runners.append({'runner': s.runner, 'time': latest.time})
return render(request, 'page.html', {
'runners': runners,
})
{% for runner in runners %}
{{runner.runner}} - {{runner.time}}
{% endfor %}
>>> from django.db.models import Max
# all runners now have a `last_race` attribute,
# which is the `id` of the last sprint they ran
>>> runners = Runner.objects.annotate(last_race=Max("sprint__id"))
# a list of each runner's last sprint ordered by the the sprint's time,
# we use `select_related` to limit lookup queries later on
>>> results = Sprint.objects.filter(id__in=[runner.last_race for runner in runners])
... .order_by("time")
... .select_related("runner")
# grab the first result
>>> first_result = results[0]
# you can access the runner's details via `.runner`, e.g. `first_result.runner.name`
>>> isinstance(first_result.runner, Runner)
True
# this should only ever execute 2 queries, no matter what you do with the results
>>> from django.db import connection
>>> len(connection.queries)
2
几千条记录并不是那么多，这对于那些类型的数字应该很有效。如果您开始遇到问题，我建议您咬紧牙关，使用原始SQL。问题不是获取最新的sprint，而是按照最新的sprint
时间