Leave last/newest records in table by a combination of fields












0















How can I leave one record (last/newest) for each resource_owner_id and scopes combination and remove other records?



For example, for resource_owner_id=3 and scopes=driver it should be only record with id=1357.resource_owner_id=5 and scopes=driver - with id=1034



I know that I can use .ids, after that get all records by id and scope (scopes are enum - 'driver' and 'passenger'), after that use .pop and remove all remaining records.



Maybe there is a more elegant solution?



enter image description here










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  • What DB are you using?

    – Marcin Kołodziej
    Nov 27 '18 at 16:12











  • @MarcinKołodziej PostgreSQL

    – ViT-Vetal-
    Nov 27 '18 at 16:40
















0















How can I leave one record (last/newest) for each resource_owner_id and scopes combination and remove other records?



For example, for resource_owner_id=3 and scopes=driver it should be only record with id=1357.resource_owner_id=5 and scopes=driver - with id=1034



I know that I can use .ids, after that get all records by id and scope (scopes are enum - 'driver' and 'passenger'), after that use .pop and remove all remaining records.



Maybe there is a more elegant solution?



enter image description here










share|improve this question























  • What DB are you using?

    – Marcin Kołodziej
    Nov 27 '18 at 16:12











  • @MarcinKołodziej PostgreSQL

    – ViT-Vetal-
    Nov 27 '18 at 16:40














0












0








0








How can I leave one record (last/newest) for each resource_owner_id and scopes combination and remove other records?



For example, for resource_owner_id=3 and scopes=driver it should be only record with id=1357.resource_owner_id=5 and scopes=driver - with id=1034



I know that I can use .ids, after that get all records by id and scope (scopes are enum - 'driver' and 'passenger'), after that use .pop and remove all remaining records.



Maybe there is a more elegant solution?



enter image description here










share|improve this question














How can I leave one record (last/newest) for each resource_owner_id and scopes combination and remove other records?



For example, for resource_owner_id=3 and scopes=driver it should be only record with id=1357.resource_owner_id=5 and scopes=driver - with id=1034



I know that I can use .ids, after that get all records by id and scope (scopes are enum - 'driver' and 'passenger'), after that use .pop and remove all remaining records.



Maybe there is a more elegant solution?



enter image description here







ruby-on-rails activerecord






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asked Nov 27 '18 at 14:16









ViT-Vetal-ViT-Vetal-

1,13911025




1,13911025













  • What DB are you using?

    – Marcin Kołodziej
    Nov 27 '18 at 16:12











  • @MarcinKołodziej PostgreSQL

    – ViT-Vetal-
    Nov 27 '18 at 16:40



















  • What DB are you using?

    – Marcin Kołodziej
    Nov 27 '18 at 16:12











  • @MarcinKołodziej PostgreSQL

    – ViT-Vetal-
    Nov 27 '18 at 16:40

















What DB are you using?

– Marcin Kołodziej
Nov 27 '18 at 16:12





What DB are you using?

– Marcin Kołodziej
Nov 27 '18 at 16:12













@MarcinKołodziej PostgreSQL

– ViT-Vetal-
Nov 27 '18 at 16:40





@MarcinKołodziej PostgreSQL

– ViT-Vetal-
Nov 27 '18 at 16:40












1 Answer
1






active

oldest

votes


















0














SQL Fiddle



ActiveRecord version:



Model.select("DISTINCT ON(resource_owner_id, scopes) *")
.order("resource_owner_id, scopes, id")


Read more about DISTINCT ON here.






share|improve this answer
























  • Thanks! But I need leave in table this records and remove all other records. Should I just add this lineModel.where.not(id:@models.map(&:id)).destroy_all?

    – ViT-Vetal-
    Nov 28 '18 at 10:06











  • Ah, I thought that by remove you mean remove them from the returned records. Yes, that would work, depending on what callbacks you have, you might consider delete_all as it would be much faster.

    – Marcin Kołodziej
    Nov 28 '18 at 10:18











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














SQL Fiddle



ActiveRecord version:



Model.select("DISTINCT ON(resource_owner_id, scopes) *")
.order("resource_owner_id, scopes, id")


Read more about DISTINCT ON here.






share|improve this answer
























  • Thanks! But I need leave in table this records and remove all other records. Should I just add this lineModel.where.not(id:@models.map(&:id)).destroy_all?

    – ViT-Vetal-
    Nov 28 '18 at 10:06











  • Ah, I thought that by remove you mean remove them from the returned records. Yes, that would work, depending on what callbacks you have, you might consider delete_all as it would be much faster.

    – Marcin Kołodziej
    Nov 28 '18 at 10:18
















0














SQL Fiddle



ActiveRecord version:



Model.select("DISTINCT ON(resource_owner_id, scopes) *")
.order("resource_owner_id, scopes, id")


Read more about DISTINCT ON here.






share|improve this answer
























  • Thanks! But I need leave in table this records and remove all other records. Should I just add this lineModel.where.not(id:@models.map(&:id)).destroy_all?

    – ViT-Vetal-
    Nov 28 '18 at 10:06











  • Ah, I thought that by remove you mean remove them from the returned records. Yes, that would work, depending on what callbacks you have, you might consider delete_all as it would be much faster.

    – Marcin Kołodziej
    Nov 28 '18 at 10:18














0












0








0







SQL Fiddle



ActiveRecord version:



Model.select("DISTINCT ON(resource_owner_id, scopes) *")
.order("resource_owner_id, scopes, id")


Read more about DISTINCT ON here.






share|improve this answer













SQL Fiddle



ActiveRecord version:



Model.select("DISTINCT ON(resource_owner_id, scopes) *")
.order("resource_owner_id, scopes, id")


Read more about DISTINCT ON here.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 27 '18 at 16:55









Marcin KołodziejMarcin Kołodziej

4,4901315




4,4901315













  • Thanks! But I need leave in table this records and remove all other records. Should I just add this lineModel.where.not(id:@models.map(&:id)).destroy_all?

    – ViT-Vetal-
    Nov 28 '18 at 10:06











  • Ah, I thought that by remove you mean remove them from the returned records. Yes, that would work, depending on what callbacks you have, you might consider delete_all as it would be much faster.

    – Marcin Kołodziej
    Nov 28 '18 at 10:18



















  • Thanks! But I need leave in table this records and remove all other records. Should I just add this lineModel.where.not(id:@models.map(&:id)).destroy_all?

    – ViT-Vetal-
    Nov 28 '18 at 10:06











  • Ah, I thought that by remove you mean remove them from the returned records. Yes, that would work, depending on what callbacks you have, you might consider delete_all as it would be much faster.

    – Marcin Kołodziej
    Nov 28 '18 at 10:18

















Thanks! But I need leave in table this records and remove all other records. Should I just add this lineModel.where.not(id:@models.map(&:id)).destroy_all?

– ViT-Vetal-
Nov 28 '18 at 10:06





Thanks! But I need leave in table this records and remove all other records. Should I just add this lineModel.where.not(id:@models.map(&:id)).destroy_all?

– ViT-Vetal-
Nov 28 '18 at 10:06













Ah, I thought that by remove you mean remove them from the returned records. Yes, that would work, depending on what callbacks you have, you might consider delete_all as it would be much faster.

– Marcin Kołodziej
Nov 28 '18 at 10:18





Ah, I thought that by remove you mean remove them from the returned records. Yes, that would work, depending on what callbacks you have, you might consider delete_all as it would be much faster.

– Marcin Kołodziej
Nov 28 '18 at 10:18




















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