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Deletion populator and multiple new records #128
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I found out that using the default example for deletion Populator does not allow to have multiple new items added to collection. In this case all new fragments do not have `id`. So find matcher will return the wrong item since `song.id.to_s == fragment["id"].to_s` (nil == nil) will always be true from 2nd run. I suggest we additonally check for id presence.
alex-lairan
reviewed
Apr 16, 2018
| @@ -269,7 +269,7 @@ You can implement your own deletion. | |||
| collection :songs, | |||
| populator: ->(fragment:, **) { | |||
| # find out if incoming song is already added. | |||
| item = songs.find { |song| song.id.to_s == fragment["id"].to_s } | |||
| item = songs.find { |song| song.id.present? && song.id.to_s == fragment["id"].to_s } | |||
There was a problem hiding this comment.
The #present? method not exist.
1.present?
NoMethodError: undefined method `present?' for 1:Fixnum
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I found out that using the default example for deletion Populator does not allow to have multiple new items added to collection.
In this case all new fragments do not have
id. So find matcher will return the wrong item sincesong.id.to_s == fragment["id"].to_s(nil == nil) will always be true from 2nd run.I suggest we additonally check for id presence.