英文:
django orm multiple points ranking
问题
I've a model name Points which store the user points on the base of it's actions.
class Points(CreateUpdateModelMixin):
class Action(models.TextChoices):
BASE = 'BASE', _('Base')
REVIEW = 'REVIEW', _('Review')
FOLLOW = 'FOLLOW', _('Follow')
VERIFIED_REVIEW = 'VERIFIED_REVIEW', _('Verified Review')
REFERRAL = 'REFERRAL', _('Referral')
ADD = 'ADD', _('Add')
SUBTRACT = 'SUBTRACT', _('Subtract')
user = models.ForeignKey(User, on_delete=models.CASCADE)
points = models.IntegerField()
action = models.CharField(max_length=64, choices=Action.choices, default=Action.BASE)
class Meta:
db_table = "diner_points"
Please note that there are multiple rows for the same user.
For the past few days I'm trying to write a query to get the total_points of the user and also the rank of that user.
Using:
Django 3.2
MySQL 5.7
I want to know input of you guys. Thanks.
I wrote this query and many others like it. But none of them give the results I want.
Let's suppose the data is something like this.
| user | points |
|---|---|
| 771 | 221 |
| 1083 | 160 |
| 1083 | 12 |
| 1083 | 10 |
| 771 | -15 |
| 1083 | 4 |
| 1083 | -10 |
| 124 | 0 |
| 23 | 1771 |
The current query I have written is this...
innerquery = (
DinerPoint.objects
.values("user")
.annotate(total=Sum("points"))
.distinct()
)
query = (
DinerPoint.objects
.annotate(
total = Subquery(
innerquery.filter(user=OuterRef("user")).values("total")
),
rank = Subquery(
DinerPoint.objects
.annotate(
total = Subquery(
innerquery.filter(user=OuterRef("user")).values("total")
),
rank=Func(F("user"), function="Count")
)
.filter(
Q(total__gt=OuterRef("total")) |
Q(total=OuterRef("total"), user__lt=OuterRef("user"))
)
.values("rank")[:1]
)
)
)
query.values('user', 'total', 'rank').distinct().order_by('rank')
But this give the results like this
<QuerySet [
{'user': 23, 'total': 1771, 'rank': 1},
{'user': 1083, 'total': 176, 'rank': 2},
{'user': 771, 'total': 106, 'rank': 8}, <---- Issue beacuse of dups entries
{'user': 124, 'total': 0, 'rank': 9}
]>
I've tried RANK, DENSE RANK and didn't got the results I wanted.
The only way I got the results I wanted I throught the Common Table Expression(CTE). But unfortunately I can't use that because of mysql version 5.7 in produciton.
P.S I'm using the count and greater than beacause of my use case. I have a use case where we have to get rank in the user friends.
The working code using CTE by django_cte (You can ignore this beacuse of mysql 5.7 
)
def get_queryset(user=None, following=False):
if not user:
user = User.objects.get(username="king")
innerquery = (
DinerPoint.objects
.values("user", "user__username", "user__first_name", "user__last_name", "user__profile_fixed_url",
"user__is_influencer", "user__is_verified", "user__instagram_handle")
.annotate(total=Sum("points"))
.distinct()
)
if following:
innerquery = innerquery.filter(Q(user__in=Subquery(user.friends.values('id'))) |
Q(user = user))
basequery = With(innerquery)
subquery = (
basequery.queryset()
.filter(Q(total__gt=OuterRef("total")) |
Q(total=OuterRef("total"), user__lt=OuterRef("user")))
.annotate(rank=Func(F("user"), function="Count"))
.values("rank")
.with_cte(basequery)
)
query = (
basequery.queryset()
.annotate(rank=Subquery(subquery) + 1)
.select_related("user")
.with_cte(basequery)
)
return query
英文:
I've a model name Points which store the user points on the base of it's actions.
class Points(CreateUpdateModelMixin):
class Action(models.TextChoices):
BASE = 'BASE', _('Base')
REVIEW = 'REVIEW', _('Review')
FOLLOW = 'FOLLOW', _('Follow')
VERIFIED_REVIEW = 'VERIFIED_REVIEW', _('Verified Review')
REFERRAL = 'REFERRAL', _('Referral')
ADD = 'ADD', _('Add')
SUBTRACT = 'SUBTRACT', _('Subtract')
user = models.ForeignKey(User, on_delete=models.CASCADE)
points = models.IntegerField()
action = models.CharField(max_length=64, choices=Action.choices, default=Action.BASE)
class Meta:
db_table = "diner_points"
Please note that there are multiple rows for the same user.
For the past few days I'm trying to write a query to get the total_points of the use and also the rank of that user.
Using:
Django 3.2
MySQL 5.7
I want to know input of you guys. Thanks.
I wrote this query and many other like it. But none of them give the results I want.
Let's suppose the data is something like this.
| user | points |
|---|---|
| 771 | 221 |
| 1083 | 160 |
| 1083 | 12 |
| 1083 | 10 |
| 771 | -15 |
| 1083 | 4 |
| 1083 | -10 |
| 124 | 0 |
| 23 | 1771 |
The current query I have written is this...
innerquery = (
DinerPoint.objects
.values("user")
.annotate(total=Sum("points"))
.distinct()
)
query = (
DinerPoint.objects
.annotate(
total = Subquery(
innerquery.filter(user=OuterRef("user")).values("total")
),
rank = Subquery(
DinerPoint.objects
.annotate(
total = Subquery(
innerquery.filter(user=OuterRef("user")).values("total")
),
rank=Func(F("user"), function="Count")
)
.filter(
Q(total__gt=OuterRef("total")) |
Q(total=OuterRef("total"), user__lt=OuterRef("user"))
)
.values("rank")[:1]
)
)
)
query.values('user', 'total', 'rank').distinct().order_by('rank')
But this give the results like this
<QuerySet [
{'user': 23, 'total': 1771, 'rank': 1},
{'user': 1083, 'total': 176, 'rank': 2},
{'user': 771, 'total': 106, 'rank': 8}, <---- Issue beacuse of dups entries
{'user': 124, 'total': 0, 'rank': 9}
]>
I've tried RANK, DENSE RANK and didn't got the results I wanted.
The only way I got the results I wanted I throught the Common Table Expression(CTE). But unfortunately I can't use that because of mysql version 5.7 in produciton.
P.S I'm using the count and greater than beacause of my use case. I have a use case where we have to get rank in the user friends.
The working code using CTE by django_cte (You can ignore this beacuse of mysql 5.7 )
def get_queryset(user=None, following=False):
if not user:
user = User.objects.get(username="king")
innerquery = (
DinerPoint.objects
.values("user", "user__username", "user__first_name", "user__last_name", "user__profile_fixed_url",
"user__is_influencer", "user__is_verified", "user__instagram_handle")
.annotate(total=Sum("points"))
.distinct()
)
if following:
innerquery = innerquery.filter(Q(user__in=Subquery(user.friends.values('id'))) |
Q(user = user))
basequery = With(innerquery)
subquery = (
basequery.queryset()
.filter(Q(total__gt=OuterRef("total")) |
Q(total=OuterRef("total"), user__lt=OuterRef("user")))
.annotate(rank=Func(F("user"), function="Count"))
.values("rank")
.with_cte(basequery)
)
query = (
basequery.queryset()
.annotate(rank=Subquery(subquery) + 1)
.select_related("user")
.with_cte(basequery)
)
return query
答案1
得分: 0
以下是翻译好的部分：
我已经使用 `Func` `expression` 字段完成了这个。对我有效的最终查询如下，以防您正在寻找答案。
Final query
``` py
def get_queryset(self):
following = self.request.query_params.get("following", False)
innerquery = (
DinerPoint.objects.values("user").annotate(total=Sum("points")).distinct()
)
basequery = DinerPoint.objects
if following:
innerquery = innerquery.filter(
Q(user__in=Subquery(self.request.user.friends.values("id")))
| Q(user=self.request.user)
)
basequery = basequery.filter(
Q(user__in=Subquery(self.request.user.friends.values("id")))
| Q(user=self.request.user)
)
query = (
basequery.annotate(
total=Subquery(
innerquery.filter(user=OuterRef("user")).values("total")
),
rank=Subquery(
DinerPoint.objects.annotate(
total=Subquery(
innerquery.filter(user=OuterRef("user")).values("total")
),
rank=Func(
F("user"),
function="Count",
template="%(function)s(DISTINCT %(expressions)s)",
),
)
.filter(
Q(total__gt=OuterRef("total"))
| Q(total=OuterRef("total"), user__lt=OuterRef("user"))
)
.values("rank")
)
+ 1,
)
.values(
"user",
"user__username",
"user__first_name",
"user__last_name",
"user__profile_fixed_url",
"user__is_influencer",
"user__is_verified",
"user__instagram_handle",
"total",
"rank",
)
.distinct()
)
return query
<details>
<summary>英文:</summary>
I have done this using the `Func` `expression` field. The final query which works for me is attached below in case you are looking for an answer.
``` py
rank=Func(
F("user"),
function="Count",
template="%(function)s(DISTINCT %(expressions)s)",
),
Final query
def get_queryset(self):
following = self.request.query_params.get("following", False)
innerquery = (
DinerPoint.objects.values("user").annotate(total=Sum("points")).distinct()
)
basequery = DinerPoint.objects
if following:
innerquery = innerquery.filter(
Q(user__in=Subquery(self.request.user.friends.values("id")))
| Q(user=self.request.user)
)
basequery = basequery.filter(
Q(user__in=Subquery(self.request.user.friends.values("id")))
| Q(user=self.request.user)
)
query = (
basequery.annotate(
total=Subquery(
innerquery.filter(user=OuterRef("user")).values("total")
),
rank=Subquery(
DinerPoint.objects.annotate(
total=Subquery(
innerquery.filter(user=OuterRef("user")).values("total")
),
rank=Func(
F("user"),
function="Count",
template="%(function)s(DISTINCT %(expressions)s)",
),
)
.filter(
Q(total__gt=OuterRef("total"))
| Q(total=OuterRef("total"), user__lt=OuterRef("user"))
)
.values("rank")
)
+ 1,
)
.values(
"user",
"user__username",
"user__first_name",
"user__last_name",
"user__profile_fixed_url",
"user__is_influencer",
"user__is_verified",
"user__instagram_handle",
"total",
"rank",
)
.distinct()
)
return query
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