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)

edited Nov 20 '18 at 14:11

asked Nov 20 '18 at 13:54

Maxence

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)

edited Nov 20 '18 at 14:11

asked Nov 20 '18 at 13:54

Maxence

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)

edited Nov 20 '18 at 14:11

asked Nov 20 '18 at 13:54

Maxence

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

edited Nov 20 '18 at 14:11

asked Nov 20 '18 at 13:54

Maxence

6451616

edited Nov 20 '18 at 14:11

asked Nov 20 '18 at 13:54

Maxence

6451616

edited Nov 20 '18 at 14:11

asked Nov 20 '18 at 13:54

Maxence

6451616

asked Nov 20 '18 at 13:54

Maxence

6451616

asked Nov 20 '18 at 13:54

Maxence

6451616

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

edited Nov 20 '18 at 14:38

answered Nov 20 '18 at 14:33

abax

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

xlembouras

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

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

edited Nov 20 '18 at 14:38

answered Nov 20 '18 at 14:33

abax

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

edited Nov 20 '18 at 14:38

answered Nov 20 '18 at 14:33

abax

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

edited Nov 20 '18 at 14:38

answered Nov 20 '18 at 14:33

abax

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

edited Nov 20 '18 at 14:38

answered Nov 20 '18 at 14:33

abax

66938

edited Nov 20 '18 at 14:38

answered Nov 20 '18 at 14:33

abax

66938

answered Nov 20 '18 at 14:33

abax

66938

answered Nov 20 '18 at 14:33

abax

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

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

xlembouras

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

xlembouras

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

xlembouras

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

xlembouras

6,78442435

answered Nov 20 '18 at 14:38

xlembouras

6,78442435

answered Nov 20 '18 at 14:38

xlembouras

6,78442435

answered Nov 20 '18 at 14:38

xlembouras

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

add a comment |

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