Mixing
Ordering activerecord relationship from a specific starting value
Let's say I get a photo_id 34
This photo is part of a specific photographers asset current_photographer.photos.all : 1, 23, 24, 34, 78
How is it possible to get the relationship current_photographer.photos.all result starting by the specific photo ID provided => 34, 78, 1, 23, 24 ?
EDIT: I want to keep the general order of photos. Each photo is preceeded by the previously created photo.(this is for a photo carousel)
ruby-on-rails
asked Nov 20 '18 at 13:54
MaxenceMaxence
6451616
add a comment |
Let's say I get a photo_id 34
This photo is part of a specific photographers asset current_photographer.photos.all : 1, 23, 24, 34, 78
How is it possible to get the relationship current_photographer.photos.all result starting by the specific photo ID provided => 34, 78, 1, 23, 24 ?
EDIT: I want to keep the general order of photos. Each photo is preceeded by the previously created photo.(this is for a photo carousel)
ruby-on-rails
asked Nov 20 '18 at 13:54
MaxenceMaxence
6451616
add a comment |
Let's say I get a photo_id 34
This photo is part of a specific photographers asset current_photographer.photos.all : 1, 23, 24, 34, 78
How is it possible to get the relationship current_photographer.photos.all result starting by the specific photo ID provided => 34, 78, 1, 23, 24 ?
EDIT: I want to keep the general order of photos. Each photo is preceeded by the previously created photo.(this is for a photo carousel)
ruby-on-rails
asked Nov 20 '18 at 13:54
MaxenceMaxence
6451616
Let's say I get a photo_id 34
This photo is part of a specific photographers asset current_photographer.photos.all : 1, 23, 24, 34, 78
How is it possible to get the relationship current_photographer.photos.all result starting by the specific photo ID provided => 34, 78, 1, 23, 24 ?
EDIT: I want to keep the general order of photos. Each photo is preceeded by the previously created photo.(this is for a photo carousel)
ruby-on-rails
ruby-on-rails
asked Nov 20 '18 at 13:54
MaxenceMaxence
6451616
asked Nov 20 '18 at 13:54
MaxenceMaxence
6451616
edited Nov 20 '18 at 14:11
Maxence
asked Nov 20 '18 at 13:54
MaxenceMaxence
6451616
asked Nov 20 '18 at 13:54
MaxenceMaxence
6451616
asked Nov 20 '18 at 13:54
MaxenceMaxence
6451616
6451616
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You could create two result sets, one for >= 34 and one for < 34, and then combine them.
current_photographer.photos.where('id >= ?', 34) + current_photographer.photos.where('id < ?', 34)
These could be scopes:
scope :above_id, ->(id) { where('id >= ?', id) }
scope :below_id, ->(id) { where('id < ?', id) }
then: current_photographer.photos.above_id(34) + current_photographer.photos.below_id(34)
or add a class method using the scopes:
def self.above_below_id(id)
above_id(id) + below_id(id)
end
answered Nov 20 '18 at 14:33
abaxabax
66938
I was focused on finding the solution with a single query.. But your solution definitely does the job. thanks
– Maxence
Nov 20 '18 at 14:37
add a comment |
I think you would be better implementing in the application space. I don't really see how it can be done in a meaningful way in the database level.
You can fetch the records in their original order and then manipulate the result.
current_photographer.photos.partition {|photo| photo.id < 34 }.reverse.flatten
answered Nov 20 '18 at 14:38
xlembourasxlembouras
6,78442435
Thanks a lot. Your solution does work in a single query. I have already accepted the other answer but yours is slightly cleaner. Manipulating a single query is the way to go.
– Maxence
Nov 20 '18 at 14:44
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could create two result sets, one for >= 34 and one for < 34, and then combine them.
current_photographer.photos.where('id >= ?', 34) + current_photographer.photos.where('id < ?', 34)
These could be scopes:
scope :above_id, ->(id) { where('id >= ?', id) }
scope :below_id, ->(id) { where('id < ?', id) }
then: current_photographer.photos.above_id(34) + current_photographer.photos.below_id(34)
or add a class method using the scopes:
def self.above_below_id(id)
above_id(id) + below_id(id)
end
answered Nov 20 '18 at 14:33
abaxabax
66938
I was focused on finding the solution with a single query.. But your solution definitely does the job. thanks
– Maxence
Nov 20 '18 at 14:37
add a comment |
You could create two result sets, one for >= 34 and one for < 34, and then combine them.
current_photographer.photos.where('id >= ?', 34) + current_photographer.photos.where('id < ?', 34)
These could be scopes:
scope :above_id, ->(id) { where('id >= ?', id) }
scope :below_id, ->(id) { where('id < ?', id) }
then: current_photographer.photos.above_id(34) + current_photographer.photos.below_id(34)
or add a class method using the scopes:
def self.above_below_id(id)
above_id(id) + below_id(id)
end
answered Nov 20 '18 at 14:33
abaxabax
66938
I was focused on finding the solution with a single query.. But your solution definitely does the job. thanks
– Maxence
Nov 20 '18 at 14:37
add a comment |
You could create two result sets, one for >= 34 and one for < 34, and then combine them.
current_photographer.photos.where('id >= ?', 34) + current_photographer.photos.where('id < ?', 34)
These could be scopes:
scope :above_id, ->(id) { where('id >= ?', id) }
scope :below_id, ->(id) { where('id < ?', id) }
then: current_photographer.photos.above_id(34) + current_photographer.photos.below_id(34)
or add a class method using the scopes:
def self.above_below_id(id)
above_id(id) + below_id(id)
end
answered Nov 20 '18 at 14:33
abaxabax
66938
You could create two result sets, one for >= 34 and one for < 34, and then combine them.
current_photographer.photos.where('id >= ?', 34) + current_photographer.photos.where('id < ?', 34)
These could be scopes:
scope :above_id, ->(id) { where('id >= ?', id) }
scope :below_id, ->(id) { where('id < ?', id) }
then: current_photographer.photos.above_id(34) + current_photographer.photos.below_id(34)
or add a class method using the scopes:
def self.above_below_id(id)
above_id(id) + below_id(id)
end
answered Nov 20 '18 at 14:33
abaxabax
66938
edited Nov 20 '18 at 14:38
answered Nov 20 '18 at 14:33
abaxabax
66938
answered Nov 20 '18 at 14:33
abaxabax
66938
answered Nov 20 '18 at 14:33
abaxabax
66938
66938
I was focused on finding the solution with a single query.. But your solution definitely does the job. thanks
– Maxence
Nov 20 '18 at 14:37
add a comment |
I was focused on finding the solution with a single query.. But your solution definitely does the job. thanks
– Maxence
Nov 20 '18 at 14:37
I was focused on finding the solution with a single query.. But your solution definitely does the job. thanks
– Maxence
Nov 20 '18 at 14:37
I was focused on finding the solution with a single query.. But your solution definitely does the job. thanks
– Maxence
Nov 20 '18 at 14:37
add a comment |
I think you would be better implementing in the application space. I don't really see how it can be done in a meaningful way in the database level.
You can fetch the records in their original order and then manipulate the result.
current_photographer.photos.partition {|photo| photo.id < 34 }.reverse.flatten
answered Nov 20 '18 at 14:38
xlembourasxlembouras
6,78442435
Thanks a lot. Your solution does work in a single query. I have already accepted the other answer but yours is slightly cleaner. Manipulating a single query is the way to go.
– Maxence
Nov 20 '18 at 14:44
add a comment |
I think you would be better implementing in the application space. I don't really see how it can be done in a meaningful way in the database level.
You can fetch the records in their original order and then manipulate the result.
current_photographer.photos.partition {|photo| photo.id < 34 }.reverse.flatten
answered Nov 20 '18 at 14:38
xlembourasxlembouras
6,78442435
Thanks a lot. Your solution does work in a single query. I have already accepted the other answer but yours is slightly cleaner. Manipulating a single query is the way to go.
– Maxence
Nov 20 '18 at 14:44
add a comment |
I think you would be better implementing in the application space. I don't really see how it can be done in a meaningful way in the database level.
You can fetch the records in their original order and then manipulate the result.
current_photographer.photos.partition {|photo| photo.id < 34 }.reverse.flatten
answered Nov 20 '18 at 14:38
xlembourasxlembouras
6,78442435
I think you would be better implementing in the application space. I don't really see how it can be done in a meaningful way in the database level.
You can fetch the records in their original order and then manipulate the result.
current_photographer.photos.partition {|photo| photo.id < 34 }.reverse.flatten
answered Nov 20 '18 at 14:38
xlembourasxlembouras
6,78442435
answered Nov 20 '18 at 14:38
xlembourasxlembouras
6,78442435
answered Nov 20 '18 at 14:38
xlembourasxlembouras
6,78442435
answered Nov 20 '18 at 14:38
xlembourasxlembouras
6,78442435
6,78442435
Thanks a lot. Your solution does work in a single query. I have already accepted the other answer but yours is slightly cleaner. Manipulating a single query is the way to go.
– Maxence
Nov 20 '18 at 14:44
add a comment |
Thanks a lot. Your solution does work in a single query. I have already accepted the other answer but yours is slightly cleaner. Manipulating a single query is the way to go.
– Maxence
Nov 20 '18 at 14:44
Thanks a lot. Your solution does work in a single query. I have already accepted the other answer but yours is slightly cleaner. Manipulating a single query is the way to go.
– Maxence
Nov 20 '18 at 14:44
Thanks a lot. Your solution does work in a single query. I have already accepted the other answer but yours is slightly cleaner. Manipulating a single query is the way to go.
– Maxence
Nov 20 '18 at 14:44
add a comment |
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