Is it possible to get 0 count for group by queries when there are no rows for that particular group by












0















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







share|improve this question



























    0















    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







    share|improve this question

























      0












      0








      0








      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







      share|improve this question














      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






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 22 '18 at 7:52









      Manan ShahManan Shah

      63




      63
























          1 Answer
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          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()





          share|improve this answer

























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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

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            0














            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()





            share|improve this answer






























              0














              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()





              share|improve this answer




























                0












                0








                0







                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()





                share|improve this answer















                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()






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 22 '18 at 8:59

























                answered Nov 22 '18 at 7:57









                ruddraruddra

                14.9k32748




                14.9k32748
































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