How do you create many-to-many class instance relationships in JRuby w/ActiveRecord?
0
I have this basic setup:
class Foo < ActiveRecord::Base
self.primary_key = 'foo_id'
has_and_belongs_to_many :bars
end
class Bar < ActiveRecord::Base
self.primary_key = :bar_id
has_and_belongs_to_many :foos
end
Now I can see all the bars associated with foos using Foo.first.bars or Bar.first.foos and this works as expected.
Where I'm stumped is how to do something like this:
foo_rows = Foo.all
=> (all those rows)
bar_rows = Bar.all
=> (all those rows)
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.some_col
=> "The value from the database"
bar_rows.find { |bar| bar.bar_id == 1 }.some_col = 'a new value'
=> "a new value"
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.some_col
=> "a new value"
But instead that last line says "The value from the database"
How do I achieve the desired behaviour?
ruby activerecord jruby
asked Nov 21 '18 at 23:22
squirrelchewsquirrelchew
186
add a comment |
0
I have this basic setup:
class Foo < ActiveRecord::Base
self.primary_key = 'foo_id'
has_and_belongs_to_many :bars
end
class Bar < ActiveRecord::Base
self.primary_key = :bar_id
has_and_belongs_to_many :foos
end
Now I can see all the bars associated with foos using Foo.first.bars or Bar.first.foos and this works as expected.
Where I'm stumped is how to do something like this:
foo_rows = Foo.all
=> (all those rows)
bar_rows = Bar.all
=> (all those rows)
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.some_col
=> "The value from the database"
bar_rows.find { |bar| bar.bar_id == 1 }.some_col = 'a new value'
=> "a new value"
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.some_col
=> "a new value"
But instead that last line says "The value from the database"
How do I achieve the desired behaviour?
ruby activerecord jruby
asked Nov 21 '18 at 23:22
squirrelchewsquirrelchew
186
add a comment |
0
0
0
I have this basic setup:
class Foo < ActiveRecord::Base
self.primary_key = 'foo_id'
has_and_belongs_to_many :bars
end
class Bar < ActiveRecord::Base
self.primary_key = :bar_id
has_and_belongs_to_many :foos
end
Now I can see all the bars associated with foos using Foo.first.bars or Bar.first.foos and this works as expected.
Where I'm stumped is how to do something like this:
foo_rows = Foo.all
=> (all those rows)
bar_rows = Bar.all
=> (all those rows)
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.some_col
=> "The value from the database"
bar_rows.find { |bar| bar.bar_id == 1 }.some_col = 'a new value'
=> "a new value"
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.some_col
=> "a new value"
But instead that last line says "The value from the database"
How do I achieve the desired behaviour?
ruby activerecord jruby
asked Nov 21 '18 at 23:22
squirrelchewsquirrelchew
186
I have this basic setup:
class Foo < ActiveRecord::Base
self.primary_key = 'foo_id'
has_and_belongs_to_many :bars
end
class Bar < ActiveRecord::Base
self.primary_key = :bar_id
has_and_belongs_to_many :foos
end
Now I can see all the bars associated with foos using Foo.first.bars or Bar.first.foos and this works as expected.
Where I'm stumped is how to do something like this:
foo_rows = Foo.all
=> (all those rows)
bar_rows = Bar.all
=> (all those rows)
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.some_col
=> "The value from the database"
bar_rows.find { |bar| bar.bar_id == 1 }.some_col = 'a new value'
=> "a new value"
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.some_col
=> "a new value"
But instead that last line says "The value from the database"
How do I achieve the desired behaviour?
ruby activerecord jruby
ruby activerecord jruby
asked Nov 21 '18 at 23:22
squirrelchewsquirrelchew
186
asked Nov 21 '18 at 23:22
squirrelchewsquirrelchew
186
asked Nov 21 '18 at 23:22
squirrelchewsquirrelchew
186
asked Nov 21 '18 at 23:22
squirrelchewsquirrelchew
186
asked Nov 21 '18 at 23:22
squirrelchewsquirrelchew
186
186
add a comment |
add a comment |
1 Answer
1
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oldest
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0
Your bar_rows and foo_rows.first.bars are arrays with different objects in memory. Just because the id attribute of one of their elements is equal, it doesn't mean they're the same objects:
bar_rows.find { |bar| bar.bar_id == 1 }.object_id
# => 40057500
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.object_id
# => 40057123
You are changing attribute of one of these objects, there is no reason that the attribute of the second object should be changed.
As for the JRuby part, it does not matter -- MRI would behave the same.
answered Nov 21 '18 at 23:56
Marcin KołodziejMarcin Kołodziej
4,5101315
Yeah, that explains the behaviour. Is there a way to get the desired effect?
– squirrelchew
Nov 22 '18 at 0:04
It's difficult to say what's your desired effect. You can either modify one offoo_rows.first.barsor modify the one frombar_rows, add it tofoo_rows.first.barsand delete the previous one.
– Marcin Kołodziej
Nov 22 '18 at 0:07
I'd expect the bar with the id of 1 to be the same bar that's associated with all the foos, just as it would be in a relational database. This way I can locate a bar through a foo, update it right there, and not have to traverse every foo looking for that bar to update it as well.
– squirrelchew
Nov 22 '18 at 0:18
I don't think there's any way around it. You'll have to find the object directly in the associations of givenfoo. If you change the object somewhere else and want to have the changes reflected infoo, you'll have to reassign the associated item.
– Marcin Kołodziej
Nov 22 '18 at 0:20
That's what I was afraid of, hah! I guess I'll be writing some methods to do that all quick-like. Thanks!
– squirrelchew
Nov 22 '18 at 0:47
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
0
Your bar_rows and foo_rows.first.bars are arrays with different objects in memory. Just because the id attribute of one of their elements is equal, it doesn't mean they're the same objects:
bar_rows.find { |bar| bar.bar_id == 1 }.object_id
# => 40057500
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.object_id
# => 40057123
You are changing attribute of one of these objects, there is no reason that the attribute of the second object should be changed.
As for the JRuby part, it does not matter -- MRI would behave the same.
answered Nov 21 '18 at 23:56
Marcin KołodziejMarcin Kołodziej
4,5101315
Yeah, that explains the behaviour. Is there a way to get the desired effect?
– squirrelchew
Nov 22 '18 at 0:04
It's difficult to say what's your desired effect. You can either modify one offoo_rows.first.barsor modify the one frombar_rows, add it tofoo_rows.first.barsand delete the previous one.
– Marcin Kołodziej
Nov 22 '18 at 0:07
I'd expect the bar with the id of 1 to be the same bar that's associated with all the foos, just as it would be in a relational database. This way I can locate a bar through a foo, update it right there, and not have to traverse every foo looking for that bar to update it as well.
– squirrelchew
Nov 22 '18 at 0:18
I don't think there's any way around it. You'll have to find the object directly in the associations of givenfoo. If you change the object somewhere else and want to have the changes reflected infoo, you'll have to reassign the associated item.
– Marcin Kołodziej
Nov 22 '18 at 0:20
That's what I was afraid of, hah! I guess I'll be writing some methods to do that all quick-like. Thanks!
– squirrelchew
Nov 22 '18 at 0:47
add a comment |
0
Your bar_rows and foo_rows.first.bars are arrays with different objects in memory. Just because the id attribute of one of their elements is equal, it doesn't mean they're the same objects:
bar_rows.find { |bar| bar.bar_id == 1 }.object_id
# => 40057500
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.object_id
# => 40057123
You are changing attribute of one of these objects, there is no reason that the attribute of the second object should be changed.
As for the JRuby part, it does not matter -- MRI would behave the same.
answered Nov 21 '18 at 23:56
Marcin KołodziejMarcin Kołodziej
4,5101315
Yeah, that explains the behaviour. Is there a way to get the desired effect?
– squirrelchew
Nov 22 '18 at 0:04
It's difficult to say what's your desired effect. You can either modify one offoo_rows.first.barsor modify the one frombar_rows, add it tofoo_rows.first.barsand delete the previous one.
– Marcin Kołodziej
Nov 22 '18 at 0:07
I'd expect the bar with the id of 1 to be the same bar that's associated with all the foos, just as it would be in a relational database. This way I can locate a bar through a foo, update it right there, and not have to traverse every foo looking for that bar to update it as well.
– squirrelchew
Nov 22 '18 at 0:18
I don't think there's any way around it. You'll have to find the object directly in the associations of givenfoo. If you change the object somewhere else and want to have the changes reflected infoo, you'll have to reassign the associated item.
– Marcin Kołodziej
Nov 22 '18 at 0:20
That's what I was afraid of, hah! I guess I'll be writing some methods to do that all quick-like. Thanks!
– squirrelchew
Nov 22 '18 at 0:47
add a comment |
0
0
0
Your bar_rows and foo_rows.first.bars are arrays with different objects in memory. Just because the id attribute of one of their elements is equal, it doesn't mean they're the same objects:
bar_rows.find { |bar| bar.bar_id == 1 }.object_id
# => 40057500
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.object_id
# => 40057123
You are changing attribute of one of these objects, there is no reason that the attribute of the second object should be changed.
As for the JRuby part, it does not matter -- MRI would behave the same.
answered Nov 21 '18 at 23:56
Marcin KołodziejMarcin Kołodziej
4,5101315
Your bar_rows and foo_rows.first.bars are arrays with different objects in memory. Just because the id attribute of one of their elements is equal, it doesn't mean they're the same objects:
bar_rows.find { |bar| bar.bar_id == 1 }.object_id
# => 40057500
foo_rows.first.bars.find { |bar| bar.bar_id == 1 }.object_id
# => 40057123
You are changing attribute of one of these objects, there is no reason that the attribute of the second object should be changed.
As for the JRuby part, it does not matter -- MRI would behave the same.
answered Nov 21 '18 at 23:56
Marcin KołodziejMarcin Kołodziej
4,5101315
answered Nov 21 '18 at 23:56
Marcin KołodziejMarcin Kołodziej
4,5101315
answered Nov 21 '18 at 23:56
Marcin KołodziejMarcin Kołodziej
4,5101315
answered Nov 21 '18 at 23:56
Marcin KołodziejMarcin Kołodziej
4,5101315
4,5101315
Yeah, that explains the behaviour. Is there a way to get the desired effect?
– squirrelchew
Nov 22 '18 at 0:04
It's difficult to say what's your desired effect. You can either modify one offoo_rows.first.barsor modify the one frombar_rows, add it tofoo_rows.first.barsand delete the previous one.
– Marcin Kołodziej
Nov 22 '18 at 0:07
I'd expect the bar with the id of 1 to be the same bar that's associated with all the foos, just as it would be in a relational database. This way I can locate a bar through a foo, update it right there, and not have to traverse every foo looking for that bar to update it as well.
– squirrelchew
Nov 22 '18 at 0:18
I don't think there's any way around it. You'll have to find the object directly in the associations of givenfoo. If you change the object somewhere else and want to have the changes reflected infoo, you'll have to reassign the associated item.
– Marcin Kołodziej
Nov 22 '18 at 0:20
That's what I was afraid of, hah! I guess I'll be writing some methods to do that all quick-like. Thanks!
– squirrelchew
Nov 22 '18 at 0:47
add a comment |
Yeah, that explains the behaviour. Is there a way to get the desired effect?
– squirrelchew
Nov 22 '18 at 0:04
It's difficult to say what's your desired effect. You can either modify one offoo_rows.first.barsor modify the one frombar_rows, add it tofoo_rows.first.barsand delete the previous one.
– Marcin Kołodziej
Nov 22 '18 at 0:07
I'd expect the bar with the id of 1 to be the same bar that's associated with all the foos, just as it would be in a relational database. This way I can locate a bar through a foo, update it right there, and not have to traverse every foo looking for that bar to update it as well.
– squirrelchew
Nov 22 '18 at 0:18
I don't think there's any way around it. You'll have to find the object directly in the associations of givenfoo. If you change the object somewhere else and want to have the changes reflected infoo, you'll have to reassign the associated item.
– Marcin Kołodziej
Nov 22 '18 at 0:20
That's what I was afraid of, hah! I guess I'll be writing some methods to do that all quick-like. Thanks!
– squirrelchew
Nov 22 '18 at 0:47
Yeah, that explains the behaviour. Is there a way to get the desired effect?
– squirrelchew
Nov 22 '18 at 0:04
Yeah, that explains the behaviour. Is there a way to get the desired effect?
– squirrelchew
Nov 22 '18 at 0:04
It's difficult to say what's your desired effect. You can either modify one of
foo_rows.first.bars or modify the one from bar_rows, add it to foo_rows.first.bars and delete the previous one.– Marcin Kołodziej
Nov 22 '18 at 0:07
It's difficult to say what's your desired effect. You can either modify one of
foo_rows.first.bars or modify the one from bar_rows, add it to foo_rows.first.bars and delete the previous one.– Marcin Kołodziej
Nov 22 '18 at 0:07
I'd expect the bar with the id of 1 to be the same bar that's associated with all the foos, just as it would be in a relational database. This way I can locate a bar through a foo, update it right there, and not have to traverse every foo looking for that bar to update it as well.
– squirrelchew
Nov 22 '18 at 0:18
I'd expect the bar with the id of 1 to be the same bar that's associated with all the foos, just as it would be in a relational database. This way I can locate a bar through a foo, update it right there, and not have to traverse every foo looking for that bar to update it as well.
– squirrelchew
Nov 22 '18 at 0:18
I don't think there's any way around it. You'll have to find the object directly in the associations of given
foo. If you change the object somewhere else and want to have the changes reflected in foo, you'll have to reassign the associated item.– Marcin Kołodziej
Nov 22 '18 at 0:20
I don't think there's any way around it. You'll have to find the object directly in the associations of given
foo. If you change the object somewhere else and want to have the changes reflected in foo, you'll have to reassign the associated item.– Marcin Kołodziej
Nov 22 '18 at 0:20
That's what I was afraid of, hah! I guess I'll be writing some methods to do that all quick-like. Thanks!
– squirrelchew
Nov 22 '18 at 0:47
That's what I was afraid of, hah! I guess I'll be writing some methods to do that all quick-like. Thanks!
– squirrelchew
Nov 22 '18 at 0:47
add a comment |
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