elasticsearch 中复杂查询的问题 - V2EX
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rqxiao

elasticsearch 中复杂查询的问题

  •  
  •   rqxiao Jun 3, 2020 4760 views
    This topic created in 2156 days ago, the information mentioned may be changed or developed.

    想完成在数据库中这样的操作

    SELECT coupon_type_id, count( * ) FROM coupon_statistics_record WHERE DATE_FORMAT( create_time, "%Y-%m" ) = '2020-05' AND coupon_type_id NOT IN ( '1', '2' ) AND counter_code = '111' GROUP BY coupon_type_id

    现在用 ElasticsearchRepository 去做了聚合,剩下的 DATE_FORMAT 和 not in 不知道该怎么去做

     NativeSearchQueryBuilder queryBuilder = new NativeSearchQueryBuilder(); // 不查询任何结果 queryBuilder.withSourceFilter(new FetchSourceFilter(new String[]{""}, null)); // 1 、添加一个新的聚合,聚合类型为 terms,聚合名称为 brands,聚合字段为 brand queryBuilder.addAggregation( AggregationBuilders.terms("couponTypeIds").field("couponTypeId.keyword")); // 2 、查询,需要把结果强转为 AggregatedPage 类型 AggregatedPage<CouponStatisticsRecordESDto> aggPage = (AggregatedPage<CouponStatisticsRecordESDto>) couponStatisticsRecordESRepository.search(queryBuilder.build()); // 3 、解析 // 3.1 、从结果中取出名为 brands 的那个聚合, // 因为是利用 String 类型字段来进行的 term 聚合,所以结果要强转为 StringTerm 类型 StringTerms agg = (StringTerms) aggPage.getAggregation("couponTypeIds"); // 3.2 、获取桶 List<StringTerms.Bucket> buckets = agg.getBuckets(); // 3.3 、遍历 for (StringTerms.Bucket bucket : buckets) { // 3.4 、获取桶中的 key,即品牌名称 System.out.println(bucket.getKeyAsString()); // 3.5 、获取桶中的文档数量 System.out.println(bucket.getDocCount()); } 
    4 replies    2020-06-03 16:43:22 +08:00
    rqxiao
        1
    rqxiao  
    OP
       Jun 3, 2020
    {
    "mapping": {
    "CouponStatisticsRecord": {
    "properties": {
    "consultantCode": {
    "type": "text",
    "fields": {
    "keyword": {
    "type": "keyword",
    "ignore_above": 256
    }
    }
    },
    "couponId": {
    "type": "text",
    "fields": {
    "keyword": {
    "type": "keyword",
    "ignore_above": 256
    }
    }
    },
    "couponTypeId": {
    "type": "text",
    "fields": {
    "keyword": {
    "type": "keyword",
    "ignore_above": 256
    }
    }
    },
    "createTime": {
    "type": "long"
    },
    "id": {
    "type": "text",
    "fields": {
    "keyword": {
    "type": "keyword",
    "ignore_above": 256
    }
    }
    },
    "memberCode": {
    "type": "text",
    "fields": {
    "keyword": {
    "type": "keyword",
    "ignore_above": 256
    }
    }
    },
    "verifyExtraInfo": {
    "type": "text",
    "fields": {
    "keyword": {
    "type": "keyword",
    "ignore_above": 256
    }
    }
    },
    "verifyTime": {
    "type": "long"
    }
    }
    }
    }
    }
    arloor
        2
    arloor  
       Jun 3, 2020
    看下官方文档中 date 和 date_range 这两个数据类型,相信就能解决
    helloZwq
        3
    helloZwq  
       Jun 3, 2020   1
    GET index/_search
    {
    "query": {
    "bool": {
    "must": [
    {
    "range": {
    "date": {
    "gte": "2020-05-03 11:00:00",
    "lte": "2020-06-03 12:00:00",
    "format":"yyyy-MM-dd HH:mm:ss"
    }
    }
    }
    ],
    "must_not": [
    {
    "terms": {
    "xxx.keyword": [
    "1",
    "2"
    ]
    }
    }
    ]
    }
    }
    }
    ben1024
        4
    ben1024  
       Jun 3, 2020   1
    DATE_FORMAT 可以用 format 来处理
    not in 可以用 query string N OT AND 来处理或 must_not 嵌套处理
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