我对聚合有点陌生,我想创建一个等效于以下SQL的
select fullname, natcode, count(1) from table where birthdate = '18-sep-1993' group by fullname, natcode having count(1) > 2 order by count(1) desc
如您所见,结果按全名和natcode分组,计数为2,并按计数排序
我设法形成以下查询:
{
"size": 0,
"aggs": {
"profs": {
"filter": {
"term": {
"birthDate": "18-Sep-1993"
}
},
"aggs": {
"name_count": {
"terms": {
"field": "fullName.raw"
},
"aggs": {
"nat_count": {
"terms": {
"field": "natCode"
},
"aggs": {
"my_filter": {
"bucket_selector": {
"buckets_path": {
"the_doc_count": "_count"
},
"script": {
"source": "params.the_doc_count>2"
}
}
}
}
}
}
}
}
}
}
}
实现的目标: 它按日期进行过滤,对全名(name_count)创建存储桶,对natcode(nat_count)创建子存储桶,并根据文档计数过滤natcode存储桶。
与此有关的问题: 我也可以看到空的name_count存储桶。我只想要具有所需数量的水桶。以下是结果示例
"aggregations": {
"profs": {
"doc_count": 3754,
"name_count": {
"doc_count_error_upper_bound": 4,
"sum_other_doc_count": 3732,
"buckets": [
{
"key": "JOHN SMITH",
"doc_count": 3,
"nat_count": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "111",
"doc_count": 3
}
]
}
},
{
"key": "MIKE CAIN",
"doc_count": 3,
"nat_count": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "205",
"doc_count": 3
}
]
}
},
{
"key": "JULIA ROBERTS",
"doc_count": 2,
"nat_count": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": []
}
},
{
"key": "JAMES STEPHEN COOK",
"doc_count": 2,
"nat_count": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": []
}
}
在结果中,我不希望显示最后两个名字(JULIA ROBERTS和JAMES STEPHEN COOK)
另外缺少什么: 组上的顺序最后计算。我希望显示最多人数的组(全名,natcode)
要求更进一步: 分组需要在另外两个字段上完成,因此就像4个字段一样。
如果我使用了错误的术语,请原谅。希望您对需要什么帮助有所了解。谢谢
答案 0 :(得分:1)
以下是查询的方式。
POST <your_index_name>/_search
{
"size": 0,
"query": {
"bool": {
"filter": {
"term": {
"birthDate": "18-sep-1993"
}
}
}
},
"aggs": {
"groupby_fullname": {
"terms": {
"field": "fullName.raw",
"size": 2000
},
"aggs": {
"natcode_filter": {
"bucket_selector": {
"buckets_path": {
"hits": "groupby_natcode._bucket_count"
},
"script": "params.hits > 0"
}
},
"groupby_natcode": {
"terms": {
"field": "natCode",
"size": 2000,
"min_doc_count": 2
}
}
}
}
}
}
作为最后的选择,我能想到的是做一些基于fullName + "_" + natCode
的select different。因此,基本上,您的密钥将采用JOHN SMITH_111
的形式。确实会给您准确的结果,只是密钥将采用这种形式。
POST <your_index_name>/_search
{
"size":0,
"query":{
"bool":{
"filter":{
"term":{
"birthDate":"18-sep-1993"
}
}
}
},
"aggs":{
"name_count":{
"terms":{
"script":{
"inline":"doc['fullName.raw'].value + params.param + doc['natCode'].value",
"lang":"painless",
"params":{
"param":"_"
}
}
},
"aggs":{
"my_filter":{
"bucket_selector":{
"buckets_path":{
"doc_count":"_count"
},
"script":"params.doc_count > 2"
}
}
}
}
}
}
希望有帮助。