為什么 filterQuery 在用於 JAVA 的 Elastic Search 的高級 REST 客戶端中不起作用？

Question

我正在嘗試創建一個對彈性搜索索引運行模糊搜索的函數。 如果我完全按照索引中的拼寫指定術語，我只會得到匹配項。 如果我故意拼錯了該術語中的單個字母，例如

“博克”

，我想模糊搜索仍應返回相同的匹配項，但它不返回任何匹配項。 同樣，如果我用prefixQuery 或termQuery 替換fuzzyMatch，則搜索僅在給出確切拼寫的情況下返回結果

“鮑勃”

為什么是這樣？ 我該如何解決？ 哪里有解釋這些方法的文檔？

這是我的代碼...

public void searchResults(@PathParam("index_name") String index_name) throws IOException {
    RestHighLevelClient client = createHighLevelRestClient();
    int numberOfSearchHitsToReturn = 100; // defaults to 10
    SearchSourceBuilder sourceBuilder = new SearchSourceBuilder();
    sourceBuilder.query(QueryBuilders.fuzzyQuery("firstname", "Bob"));
    sourceBuilder.from(0);
    sourceBuilder.size(numberOfSearchHitsToReturn);
    sourceBuilder.timeout(new TimeValue(60, TimeUnit.SECONDS));
    SearchRequest searchRequest = new SearchRequest(index_name).source(sourceBuilder);
    SearchResponse searchResponse = client.search(searchRequest, RequestOptions.DEFAULT);
    System.out.print(searchResponse);
    client.close();
}

這是在 Postman 中 Get /index/_search 的結果...

{
    "took": 0,
    "timed_out": false,
    "_shards": {
        "total": 1,
        "successful": 1,
        "skipped": 0,
        "failed": 0
    },
    "hits": {
        "total": {
            "value": 3,
            "relation": "eq"
        },
        "max_score": 1.0,
        "hits": [
            {
                "_index": "contacts",
                "_type": "_doc",
                "_id": "J1NDonABNQ4iHt4UOM4u",
                "_score": 1.0,
                "_source": {}
            },
            {
                "_index": "contacts",
                "_type": "_doc",
                "_id": "153",
                "_score": 1.0,
                "_source": {
                    "firstname": "Bob",
                    "home_city": "San Diego",
                    "home_address": "1029 Loring Street",
                    "home_zip": "92109",
                    "contact_id": "153",
                    "email": "bsmith@gmail.com",
                    "lastname": "Smith",
                    "home_state": "California",
                    "cell_phone": "6192542981"
                }
            },
            {
                "_index": "contacts",
                "_type": "_doc",
                "_id": "154",
                "_score": 1.0,
                "_source": {
                    "firstname": "Alice",
                    "home_city": "Paia",
                    "home_address": "581 Pili Loko Street",
                    "home_zip": "00012",
                    "contact_id": "154",
                    "email": "aHernes@gmail.com",
                    "lastname": "Hernes",
                    "home_state": "Hawaii",
                    "cell_phone": "8083829103"
                }
            }
        ]
    }
}

Answer 1

我相信彈性會讓你有點困惑。

3 個字母術語的模糊度為 1，因此您期望“Bob”返回就足夠了。 但是，我假設您使用默認過濾器“小寫”的標准分析器。

因此計算出的“Boc”和“bob”之間的 Levenshtein 距離是2 ，這就是它沒有返回的原因。

嘗試小寫輸入術語，我打賭會返回“Bob”。

// no results
{
    "query": {
       "fuzzy" : { "firstname" : "Boc" }
    }
}

// "Bob" returned
{
    "query": {
       "fuzzy" : { "firstname" : "boc" }
    }
}

這有意義嗎？

關於您的代碼：

public void searchResults(@PathParam("index_name") String index_name) throws IOException {
    RestHighLevelClient client = createHighLevelRestClient();
    int numberOfSearchHitsToReturn = 100; // defaults to 10
    SearchSourceBuilder sourceBuilder = new SearchSourceBuilder();
    // "Boc".toLowerCase() or simply "boc"
    sourceBuilder.query(QueryBuilders.fuzzyQuery("firstname", "Boc".toLowerCase()));
    sourceBuilder.from(0);
    sourceBuilder.size(numberOfSearchHitsToReturn);
    sourceBuilder.timeout(new TimeValue(60, TimeUnit.SECONDS));
    SearchRequest searchRequest = new SearchRequest(index_name).source(sourceBuilder);
    SearchResponse searchResponse = client.search(searchRequest, RequestOptions.DEFAULT);
    System.out.print(searchResponse);
    client.close();
}