[英]How can I get multiple alphabetic aggregates base on values in a single field and each buckets' count
我有一个字段:name.keyword。 我希望聚合(或其中的一组按层次结构细分)结果如下所示:
all_names - a count of all distinct values for the name.keyword field
names_a_through_e - a count of all distinct values that are limited to those values that start with "A" through those that start with "E"
names_a - a count of all distinct values that are limited to those that start with "A"
names - a list of each of these names and their count
names_b - a count of all distinct values that are limited to those that start with "B"
names - a list of each of these names and their count
...
names_f_through_j - a count of all distinct values that are limited to those values that start with "F" through those that start with "J"
names_f - a count of all distinct values that are limited to those that start with "F"
names - a list of each of these names and their count
names_g - a count of all distinct values that are limited to those that start with "G"
names - a list of each of these names and their count
...
...
我当然可以涵盖最内部的列表,如下所示:
"a_names": {
"terms": {
"field": "name.keyword",
"include": "A.*",
"size": 100,
"order": {"_term": "asc"}
}
}
但这并没有给我一个级别的计数 - name.keyword 字段中带有 A* 的所有文档的总和。
或者有没有办法在单个字段上执行这种结构化分桶,这样可以像我描述的那样很好地组织布局?
您可以使用sum_bucket 聚合来计算存储桶中的文档总数。
询问
{
"size": 0,
"aggs": {
"a_names": {
"terms": {
"field": "name.keyword",
"include": "A.*",
"size": 100,
"order": {
"_term": "asc"
}
}
},
"totalcount": {
"sum_bucket": {
"buckets_path": "a_names._count"
}
}
}
}
结果:
"hits" : {
"total" : {
"value" : 3,
"relation" : "eq"
},
"max_score" : null,
"hits" : [ ]
},
"aggregations" : {
"a_names" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "Abc",
"doc_count" : 1,
"Count" : {
"value" : 1
}
},
{
"key" : "Acf",
"doc_count" : 1,
"Count" : {
"value" : 1
}
}
]
},
"totalcount" : {
"value" : 2.0
}
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.