Question

从ElasticSearch获取某个索引的所有_id的最快方法是什么？是否可以使用简单的查询？我的一个索引有大约20,000个文档。

Answer 1

编辑：请阅读@Aleck Landgraf的答案

您只想要elasticsearch-internal _id字段？或者是您文档中的id字段？

对于前者，请尝试

curl http://localhost:9200/index/type/_search?pretty=true -d '
{ 
    "query" : { 
        "match_all" : {} 
    },
    "stored_fields": []
}
'

请注意2017年更新：帖子最初包含"fields": []但从那时起，名称已更改，stored_fields是新值。

结果将仅包含文档的“元数据”

{
  "took" : 7,
  "timed_out" : false,
  "_shards" : {
    "total" : 5,
    "successful" : 5,
    "failed" : 0
  },
  "hits" : {
    "total" : 4,
    "max_score" : 1.0,
    "hits" : [ {
      "_index" : "index",
      "_type" : "type",
      "_id" : "36",
      "_score" : 1.0
    }, {
      "_index" : "index",
      "_type" : "type",
      "_id" : "38",
      "_score" : 1.0
    }, {
      "_index" : "index",
      "_type" : "type",
      "_id" : "39",
      "_score" : 1.0
    }, {
      "_index" : "index",
      "_type" : "type",
      "_id" : "34",
      "_score" : 1.0
    } ]
  }
}

对于后者，如果要在文档中包含字段，只需将其添加到fields数组

curl http://localhost:9200/index/type/_search?pretty=true -d '
{ 
    "query" : { 
        "match_all" : {} 
    },
    "fields": ["document_field_to_be_returned"]
}
'

Answer 2

最好使用scroll and scan来获取结果列表，以便elasticsearch不必对结果进行排名和排序。

使用elasticsearch-dsl python lib，可以通过以下方式完成：

from elasticsearch import Elasticsearch
from elasticsearch_dsl import Search

es = Elasticsearch()
s = Search(using=es, index=ES_INDEX, doc_type=DOC_TYPE)

s = s.fields([])  # only get ids, otherwise `fields` takes a list of field names
ids = [h.meta.id for h in s.scan()]

控制台日志：

GET http://localhost:9200/my_index/my_doc/_search?search_type=scan&scroll=5m [status:200 request:0.003s]
GET http://localhost:9200/_search/scroll?scroll=5m [status:200 request:0.005s]
GET http://localhost:9200/_search/scroll?scroll=5m [status:200 request:0.005s]
GET http://localhost:9200/_search/scroll?scroll=5m [status:200 request:0.003s]
GET http://localhost:9200/_search/scroll?scroll=5m [status:200 request:0.005s]
...

注意：滚动从查询中提取批量结果，并将光标保持打开一段时间（1分钟，2分钟，您可以更新） ; 扫描禁用排序。 scan辅助函数返回一个可以安全迭代的python生成器。

Answer 3

另一个选择

curl 'http://localhost:9200/index/type/_search?pretty=true&fields='

将返回_index，_type，_id和_score。

Answer 4

对于elasticsearch 5.x，您可以使用“_source”字段。

GET /_search
{
    "_source": false,
    "query" : {
        "term" : { "user" : "kimchy" }
    }
}

"fields"已被弃用。（错误：“不再支持字段[字段]，如果字段未存储，请使用[stored_fields]检索存储的字段或_source过滤”）

Answer 5

您也可以在python中执行此操作，它会为您提供正确的列表：

import elasticsearch
es = elasticsearch.Elasticsearch()

res = es.search(
    index=your_index, 
    body={"query": {"match_all": {}}, "size": 30000, "fields": ["_id"]})

ids = [d['_id'] for d in res['hits']['hits']]

Answer 6

受到@ Aleck-Landgraf回答的启发，对我而言，它在标准的elasticsearch python API中直接使用scan函数起作用：

from elasticsearch import Elasticsearch
from elasticsearch.helpers import scan
es = Elasticsearch()
for dobj in scan(es, 
                 query={"query": {"match_all": {}}, "fields" : []},  
                 index="your-index-name", doc_type="your-doc-type"): 
        print dobj["_id"],

Answer 7

详细阐述@ Robert-Lujo和@ Aleck-Landgraf的2个答案（拥有权限的人可以很乐意将其发表评论）：如果您不想打印但是从返回的生成器中获取列表中的所有内容，请使用以下内容：

from elasticsearch import Elasticsearch,helpers
es = Elasticsearch(hosts=[YOUR_ES_HOST])
a=helpers.scan(es,query={"query":{"match_all": {}}},scroll='1m',index=INDEX_NAME)#like others so far

IDs=[aa['_id'] for aa in a]

Answer 8

对于Python用户：python elasticsearch client为滚动API提供了方便的抽象方法：

from elasticsearch import ElasticSearch, helpers
client = ElasticSearch()

query = {
    "query": {
        "match_all": {}
    }
}

scan = helpers.scan(client, index=index, query=query, scroll='1m', size=100)

for doc in scan:
    # do something

Answer 9

我知道这篇文章有很多答案，但是我想结合几个答案来证明我发现最快的答案（无论如何在Python中）。我正在处理亿万个文档，而不是数千个。

helpers类可与sliced scroll一起使用，从而允许执行多线程。就我而言，我还有一个高基数字段可提供（acquired_at）。您会看到我将max_workers设置为14，但是您可能希望根据您的计算机来更改此设置。

此外，我以压缩格式存储文档ID。如果您感到好奇，请you can check how many bytes your doc ids will be并估算最终的转储大小。

# note below I have es, index, and cluster_name variables already set

max_workers = 14
scroll_slice_ids = list(range(0,max_workers))

def get_doc_ids(scroll_slice_id):
    count = 0
    with gzip.open('/tmp/doc_ids_%i.txt.gz' % scroll_slice_id, 'wt') as results_file:
        query = {"sort": ["_doc"], "slice": { "field": "acquired_at", "id": scroll_slice_id, "max": len(scroll_slice_ids)+1}, "_source": False}
        scan = helpers.scan(es, index=index, query=query, scroll='10m', size=10000, request_timeout=600)
        for doc in scan:
            count += 1
            results_file.write((doc['_id'] + '\n'))
            results_file.flush()

    return count 

if __name__ == '__main__':
    print('attempting to dump doc ids from %s in %i slices' % (cluster_name, len(scroll_slice_ids)))
    with futures.ThreadPoolExecutor(max_workers=max_workers) as executor:
        doc_counts = executor.map(get_doc_ids, scroll_slice_ids)

如果要跟踪文件中有多少个ID，可以使用unpigz -c /tmp/doc_ids_4.txt.gz | wc -l。

Answer 10

这正在工作！

def select_ids(self, **kwargs):
    """

    :param kwargs:params from modules
    :return: array of incidents
    """
    index = kwargs.get('index')
    if not index:
        return None

    # print("Params", kwargs)
    query = self._build_query(**kwargs)
    # print("Query", query)

    # get results
    results = self._db_client.search(body=query, index=index, stored_fields=[], filter_path="hits.hits._id")
    print(results)
    ids = [_['_id'] for _ in results['hits']['hits']]
    return ids

Answer 11

Url -> http://localhost:9200/<index>/<type>/_query
http method -> GET
Query -> {"query": {"match_all": {}}, "size": 30000, "fields": ["_id"]}

在ElasticSearch中检索所有_ids的有效方法

11 个答案: