Elastic Search, is it possible to get the index of a specific document in a set of search results?

Question

I know the identifier of the specific document that I want to show in my result set, but I do not know on which page of the results it will be included. Is it possible to use an elastic search to tell him to return the page on which a particular document is located?

My guess is that it's impossible. My current approach is to run a query after loading only the document identifiers and return a very large (all) result set for the query. I found the identifier in this list, then run the query again, loading all the data that I want to show on the page. I would prefer not to run the query twice if I can avoid it.

Answer 1

I use the JAVA API and I get the index, type, identifier and source as follows.

SearchResponse response = client.prepareSearch().execute().actionGet(); SearchHit[] documents = response.getHits().getHits(); for(SearchHit document : documents) { System.out.println("document index :: " + document.getIndex()); System.out.println("document type :: " + document.getType()); System.out.println("document id :: " + document.getId()); System.out.println("document source JSON :: " + document.getSourceAsString()); } 

Answer 2

I faced the same problem. We need to find out what position of the clip is in some list.

It works with the assumption that we sort by score and by id document. Without sorting id or anything else as a second sorting rule, this will not work.

Here is an overview of the PHP code for our solution:

 $clipScore = $this->getScore($clip); $lowestPossibleHigherScore = $this->getLowestPossibleHigherScore($clipScore); $countHigherScore = $this->countClipsWithMinScore($lowestPossibleHigherScore); $countSameScoreAndHigherId = $this->countClipsWithMinScore($clipScore, $clip->getId()); $countHigherScoreAndHigherId = $countHigherScore; if ($countHigherScore > 0) { $countHigherScoreAndHigherId = $this->countClipsWithMinScore($lowestPossibleHigherScore, $clip->getId()); } $position = $countHigherScore + ($countSameScoreAndHigherId - $countHigherScoreAndHigherId); return $position; 

And here are some notes why this works;)

 /** * Considered test cases =D * * | Position | Score | ID | * | 0 | 4.0 | 3 | * | 1 | 3.2 | 1 | * | 2 | 3.1 | 6 | * | 3 | 2.5 | 5 | * | 4 | 2.5 | 4 | * | 5 | 2.5 | 2 | * | 6 | 1.2 | 7 | * * findPosition(ID = 4) * * countAllMinScore(2.501) = 3 (A) // so it best position is 3 (4th place) * countAllMinScoreAndBiggerId(2.5, 4) = 2 (C) // "two" of same score and higher ID * countAllMinScoreAndBiggerId(2.501, 4) = 1 (D) // "one" of higher score and higher ID * * $position (how to get "4"?) = 3 (A) + 1 (C - D) ??? YES !!!! * * // next case * findPosition(ID = 5) * * countAllMinScore(2.501) = 3 (A) * countAllMinScoreAndBiggerId(2.5, 5) = 1 (C) * countAllMinScoreAndBiggerId(2.501, 5) = 1 (D) * * $position (how to get "3"?) = 3 (A) + 0 (C - D) = 3 ??? YES !!!! * * // next case * findPosition(ID = 2) * * countAllMinScore(2.501) = 3 (A) * countAllMinScoreAndBiggerId(2.5, 2) = 5 (C) * countAllMinScoreAndBiggerId(2.501, 2) = 3 (D) * * $position (how to get "5"?) = 3 (A) + 2 (C - D) = 5 ??? YES !!!! * * /// next case * findPosition(ID = 3) * * countAllMinScore(4.001) = 0 (A) * countAllMinScoreAndBiggerId(4.0, 3) = 0 (C) * countAllMinScoreAndBiggerId(4.001, 3) = 0 (D) * * $position (how to get "0"?) = 0 (A) + 0 (C - D) = 0 ??? YES !!!! * * /// next case * findPosition(ID = 7) * * countAllMinScore(1.201) = 6 (A) * countAllMinScoreAndBiggerId(1.2, 7) = 0 (C) * countAllMinScoreAndBiggerId(1.201, 7) = 0 (D) * * $position (how to get "6"?) = 6 (A) + 0 (C - D) = 6 ??? YES !!!! * * * /// next case * * | Position | Score | ID | * | 0 | 4.0 | 3 | * | 1 | 4.0 | 1 | * | 2 | 3.1 | 6 | * | 3 | 2.5 | 5 | * | 4 | 2.5 | 4 | * | 5 | 2.5 | 2 | * | 6 | 1.2 | 7 | * * findPosition(ID = 3) * * countAllMinScore(4.001) = 0 (A) * countAllMinScoreAndBiggerId(4.0, 3) = 0 (C) * countAllMinScoreAndBiggerId(4.001, 3) = 0 (D) * * $position (how to get "0"?) = 0 (A) + 0 (C - D) = 0 ??? YES !!!! * * /// next case * findPosition(ID = 1) * * countAllMinScore(4.001) = 0 (A) * countAllMinScoreAndBiggerId(4.0, 1) = 1 (C) * countAllMinScoreAndBiggerId(4.001, 1) = 0 (D) * * $position (how to get "1"?) = 0 (A) + 1 (C - D) = 1 ??? YES !!!! */