我希望在嵌套的dict列表中搜索通配符术语,然后获得一个术语列表及其按匹配的通配符分组的uuid。
我的索引中有以下Map:
"mappings": {
"properties": {
"uuid": {
"type": "keyword"
},
"urls": {
"type": "nested",
"properties": {
"url": {
"type": "keyword"
},
"is_visited": {
"type": "boolean"
}
}
}
}
}
和大量的数据,比如:
{
"uuid":"afa9ac03-0723-4d66-ae18-08a51e2973bd"
"urls": [
{
"is_visited": true,
"url": "https://www.google.com"
},
{
"is_visited": false,
"url": "https://www.facebook.com"
},
{
"is_visited": true,
"url": "https://www.twitter.com"
},
]
},
{
"uuid":"4a1c695d-756b-4d9d-b3a0-cf524d955884"
"urls": [
{
"is_visited": true,
"url": "https://www.stackoverflow.com"
},
{
"is_visited": false,
"url": "https://www.facebook.com"
},
{
"is_visited": false,
"url": "https://drive.google.com"
},
{
"is_visited": false,
"url": "https://maps.google.com"
},
]
}
...
我希望通过通配符"*google.com OR *twitter.com"
进行搜索,并获得如下内容:
"hits": [
"*google.com": [
{
"uuid": "4a1c695d-756b-4d9d-b3a0-cf524d955884",
"_source": {
"is_visited": false,
"url": "https://drive.google.com"
}
},
{
"id": "4a1c695d-756b-4d9d-b3a0-cf524d955884",
"_source": {
"is_visited": false,
"url": "https://maps.google.com"
}
},
{
"uuid":"afa9ac03-0723-4d66-ae18-08a51e2973bd",
"_source": {
"is_visited": true,
"url": "https://www.google.com"
}
}
]
"*twitter.com": [
{
"uuid":"afa9ac03-0723-4d66-ae18-08a51e2973bd",
"_source": {
"is_visited": true,
"url": "https://www.twitter.com"
},
},
]
]
这是我的(python)搜索查询:
body = {
#"_source": False,
"size": 100,
"query": {
"nested": {
"path": "urls",
"query":{
"query_string":{
"query": f"urls.url:{urlToSearch}",
}
}
,"inner_hits": {
"size":100 # returns top 100 results
}
}
}
}
但是它返回每个匹配项的命中结果,而不是将它们聚集在一个类似于我想要得到的列表中。
编辑这是我的设置和Map:
{
"settings": {
"analysis": {
"char_filter": {
"my_filter": {
"type": "mapping",
"mappings": [
"- => _",
]
},
},
"analyzer": {
"my_analyzer": {
"tokenizer": "standard",
"char_filter": [
"my_filter"
],
"filter": [
"lowercase",
]
}
}
}
},
"mappings": {
"properties": {
"uuid": {
"type": "keyword"
},
"urls": {
"type": "nested",
"properties": {
"url": {
"type": "keyword"
},
"is_visited": {
"type": "boolean"
}
}
}
}
}
}
1条答案
按热度按时间atmip9wb1#
Elasticsearch不会提供你想要的输出,你设置查询的方式。这个场景是一个聚合。我的建议是应用嵌套查询,并对结果使用聚合。
注意事项wildcard query:
避免以 * 或?开头的模式。这会增加查找匹配项所需的迭代次数,降低搜索性能。
结果: