假设我有这个给定的数据
Let's say I have this given data
{
"name" : "ABC",
"favorite_cars" : [ "ferrari","toyota" ]
}, {
"name" : "ABC",
"favorite_cars" : [ "ferrari","toyota" ]
}, {
"name" : "GEORGE",
"favorite_cars" : [ "honda","Hyundae" ]
}
每当我在搜索最喜欢的汽车是丰田的人时查询此数据时,它都会返回此数据
Whenever I query this data when searching for people who's favorite car is toyota, it returns this data
{
"name" : "ABC",
"favorite_cars" : [ "ferrari","toyota" ]
}, {
"name" : "ABC",
"favorite_cars" : [ "ferrari","toyota" ]
}
结果是两条名为 ABC 的记录.如何仅选择不同的文档?我想得到的结果只有这个
the result is Two records of with a name of ABC. How do I select distinct documents only? The result I want to get is only this
{
"name" : "ABC",
"favorite_cars" : [ "ferrari","toyota" ]
}
这是我的查询
{
"fuzzy_like_this_field" : {
"favorite_cars" : {
"like_text" : "toyota",
"max_query_terms" : 12
}
}
}
我正在使用 ElasticSearch 1.0.0.使用 java api 客户端
I am using ElasticSearch 1.0.0. with the java api client
您可以使用 聚合.使用 术语聚合结果将按一个字段分组,例如name
,还提供了该字段每个值的出现次数,并将按此计数对结果进行排序(降序).
You can eliminate duplicates using aggregations. With term aggregation the results will be grouped by one field, e.g. name
, also providing a count of the ocurrences of each value of the field, and will sort the results by this count (descending).
{
"query": {
"fuzzy_like_this_field": {
"favorite_cars": {
"like_text": "toyota",
"max_query_terms": 12
}
}
},
"aggs": {
"grouped_by_name": {
"terms": {
"field": "name",
"size": 0
}
}
}
}
除了 hits
之外,结果还将包含 buckets
,其中 key
中的唯一值和 中的计数>doc_count
:
In addition to the hits
, the result will also contain the buckets
with the unique values in key
and with the count in doc_count
:
{
"took" : 4,
"timed_out" : false,
"_shards" : {
"total" : 5,
"successful" : 5,
"failed" : 0
},
"hits" : {
"total" : 2,
"max_score" : 0.19178301,
"hits" : [ {
"_index" : "pru",
"_type" : "pru",
"_id" : "vGkoVV5cR8SN3lvbWzLaFQ",
"_score" : 0.19178301,
"_source":{"name":"ABC","favorite_cars":["ferrari","toyota"]}
}, {
"_index" : "pru",
"_type" : "pru",
"_id" : "IdEbAcI6TM6oCVxCI_3fug",
"_score" : 0.19178301,
"_source":{"name":"ABC","favorite_cars":["ferrari","toyota"]}
} ]
},
"aggregations" : {
"grouped_by_name" : {
"buckets" : [ {
"key" : "abc",
"doc_count" : 2
} ]
}
}
}
请注意,由于重复消除和结果排序,使用聚合的成本会很高.
Note that using aggregations will be costly because of duplicate elimination and result sorting.
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