Mixing
Is it possible to get 0 count for group by queries when there are no rows for that particular group by
For eg:
We have two tables
Note
--------------------
id | text | label_id
1 | abcs | 1
2 | accs | 2
3 | abts | 1
Label
---------
id | name
1 | pro
2 | foo
3 | bar
Now the data I want is
Label Name | Note Count
------------------------
pro | 2
foo | 1
bar | 0
Now if I do this:
Note.objects.all().values('label__name').annotate(count=Count('id', distinct=True))
I wont get the 3rd row, which is (bar, 0). Is there any way to get this information? Condition being it has to be done in a single query and the query should be on Notes model and not start from the Label model
django django-orm
asked Nov 22 '18 at 7:52
Manan ShahManan Shah
63
add a comment |
For eg:
We have two tables
Note
--------------------
id | text | label_id
1 | abcs | 1
2 | accs | 2
3 | abts | 1
Label
---------
id | name
1 | pro
2 | foo
3 | bar
Now the data I want is
Label Name | Note Count
------------------------
pro | 2
foo | 1
bar | 0
Now if I do this:
Note.objects.all().values('label__name').annotate(count=Count('id', distinct=True))
I wont get the 3rd row, which is (bar, 0). Is there any way to get this information? Condition being it has to be done in a single query and the query should be on Notes model and not start from the Label model
django django-orm
asked Nov 22 '18 at 7:52
Manan ShahManan Shah
63
add a comment |
For eg:
We have two tables
Note
--------------------
id | text | label_id
1 | abcs | 1
2 | accs | 2
3 | abts | 1
Label
---------
id | name
1 | pro
2 | foo
3 | bar
Now the data I want is
Label Name | Note Count
------------------------
pro | 2
foo | 1
bar | 0
Now if I do this:
Note.objects.all().values('label__name').annotate(count=Count('id', distinct=True))
I wont get the 3rd row, which is (bar, 0). Is there any way to get this information? Condition being it has to be done in a single query and the query should be on Notes model and not start from the Label model
django django-orm
asked Nov 22 '18 at 7:52
Manan ShahManan Shah
63
For eg:
We have two tables
Note
--------------------
id | text | label_id
1 | abcs | 1
2 | accs | 2
3 | abts | 1
Label
---------
id | name
1 | pro
2 | foo
3 | bar
Now the data I want is
Label Name | Note Count
------------------------
pro | 2
foo | 1
bar | 0
Now if I do this:
Note.objects.all().values('label__name').annotate(count=Count('id', distinct=True))
I wont get the 3rd row, which is (bar, 0). Is there any way to get this information? Condition being it has to be done in a single query and the query should be on Notes model and not start from the Label model
django django-orm
django django-orm
asked Nov 22 '18 at 7:52
Manan ShahManan Shah
63
asked Nov 22 '18 at 7:52
Manan ShahManan Shah
63
asked Nov 22 '18 at 7:52
Manan ShahManan Shah
63
asked Nov 22 '18 at 7:52
Manan ShahManan Shah
63
asked Nov 22 '18 at 7:52
Manan ShahManan Shah
63
63
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Try like this with conditional aggregation:
Note.objects.annotate(
count=Count(Case(
When(label__note_count__gt=0, then=F('id')),
output_field=IntegerField()
)).all()
answered Nov 22 '18 at 7:57
ruddraruddra
14.9k32748
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Try like this with conditional aggregation:
Note.objects.annotate(
count=Count(Case(
When(label__note_count__gt=0, then=F('id')),
output_field=IntegerField()
)).all()
answered Nov 22 '18 at 7:57
ruddraruddra
14.9k32748
add a comment |
Try like this with conditional aggregation:
Note.objects.annotate(
count=Count(Case(
When(label__note_count__gt=0, then=F('id')),
output_field=IntegerField()
)).all()
answered Nov 22 '18 at 7:57
ruddraruddra
14.9k32748
add a comment |
Try like this with conditional aggregation:
Note.objects.annotate(
count=Count(Case(
When(label__note_count__gt=0, then=F('id')),
output_field=IntegerField()
)).all()
answered Nov 22 '18 at 7:57
ruddraruddra
14.9k32748
Try like this with conditional aggregation:
Note.objects.annotate(
count=Count(Case(
When(label__note_count__gt=0, then=F('id')),
output_field=IntegerField()
)).all()
answered Nov 22 '18 at 7:57
ruddraruddra
14.9k32748
edited Nov 22 '18 at 8:59
answered Nov 22 '18 at 7:57
ruddraruddra
14.9k32748
answered Nov 22 '18 at 7:57
ruddraruddra
14.9k32748
answered Nov 22 '18 at 7:57
ruddraruddra
14.9k32748
14.9k32748
add a comment |
add a comment |
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