Porting _all field when migrating from ES 1.7.5 to 7.4

Elastic Stack Elasticsearch
1

Hi all

I am in the process of migrating following mapping in ES 1.7.5 to 7.4:

name:
  type: 'string'
  index: 'analyzed'
  include_in_all: true
  fields:
    name_bm25:
      type: 'string'
      index: 'analyzed'
      analyzer: 'english_custom_stems'
      similarity: 'BM25'
      include_in_all: true

as you could see above, I have multi-field name_bm25 of which values is also added to _all field (NOTE: I use include_in_all: true)

So when I migrate to ES 7.x, I was forced to create a new field _all because this field feature is no longer available. Below is my new mapping:

_all:
  type: 'text'
  store: false

name:
  type: 'text'
  copy_to: '_all'
  fields:
    name_bm25:
      type: 'text'
      analyzer: 'english_custom_stems'
      similarity: 'BM25'
      copy_to: '_all' # <= this is not working

I use copy_to to copy field values to _all field which works with name field however the copy_to does not work with term (results after analysis) of multi-field name.name_bm25.

I would like to ask if I want to copy the analysed value of name.name_bm25 to _all, what should I do? OR do I have to do it? (in the case if I misunderstand the copy_all).

So seems to me one solution is to add a new name_bm25 field with analysed field value (either manually or taking advantage of _analyze API), for example:

_all:
  type: 'text'
  store: false

name_bm25: # NOTE: field value must be analysed in advance
  type: 'text'
  copy_to: '_all'

name:
  type: 'text'
  copy_to: '_all'
  fields:
    name_bm25:
      type: 'text'
      analyzer: 'english_custom_stems'
      similarity: 'BM25'

I don't think it is the right approach as it does not feel right to do that manually.

Any help is greatly appreciated. Thanks much in advance

2

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