Re: [Django] #35300: Simple COUNT without OUTER JOIN

[Django]

#35300: Simple COUNT without OUTER JOIN
-+-
 Reporter:  Chrea Chanchhunneng  |Owner:  nobody
 Type:  New feature  |   Status:  closed
Component:  Uncategorized|  Version:  4.2
 Severity:  Normal   |   Resolution:  invalid
 Keywords:  reduce counting  | Triage Stage:
  time,  |  Unreviewed
Has patch:  0|  Needs documentation:  0
  Needs tests:  0|  Patch needs improvement:  0
Easy pickings:  0|UI/UX:  0
-+-
Changes (by Mariusz Felisiak):

 * resolution:   => invalid
 * status:  new => closed

Comment:

 Thanks for the ticket, however I'm not sure what do you want to "count"
 and how do you expect to count related objects without `JOIN`s. You could
 probably use `.aggregate(Count(..))` in a subquery but it's hard to tell
 without a concrete example. Please first ask on the
 [https://code.djangoproject.com/wiki/TicketClosingReasons/UseSupportChannels
 support channels (Trac is not one of them) where folks will be eager to
 help.
-- 
Ticket URL: 
Django 
The Web framework for perfectionists with deadlines.

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[Django] #35300: Simple COUNT without OUTER JOIN

[Django]

#35300: Simple COUNT without OUTER JOIN
-+-
   Reporter:  Chrea  |  Owner:  nobody
  Chanchhunneng  |
   Type:  New| Status:  new
  feature|
  Component: |Version:  4.2
  Uncategorized  |   Keywords:  reduce counting
   Severity:  Normal |  time,
   Triage Stage: |  Has patch:  0
  Unreviewed |
Needs documentation:  0  |Needs tests:  0
Patch needs improvement:  0  |  Easy pickings:  0
  UI/UX:  0  |
-+-
 as in django query, we support `Count() in db.model` and
 `queryset.count()`. But the problem are `Count()` creates the `LEFT OUTER
 JOIN` and `queryset.count()` does not support `OuterRef("pk")`.

 I hope there is a way to just simply count without `LEFT OUTER JOIN` to
 reduce the query time.
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Ticket URL: 
Django 
The Web framework for perfectionists with deadlines.

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