如何在Elasticsearch中仅返回查询结果的_source字段？

Hey there! Let's solve this—you want to cut through all the default extra fields (like took, timed_out, _shards, max_score, plus metadata like _id/_score in each hit) and only get the _source content from your Elasticsearch queries. Here are the most straightforward ways to do it:

1. Use the filter_path Parameter (Simplest & Most Effective) #

This parameter lets you filter the entire JSON response to keep only the paths you care about. To fetch just the _source entries from your search hits, add this to your request:

Example with URL Parameter: #

GET /your_target_index/_search?filter_path=hits.hits._source

Example with Request Body: #

POST /your_target_index/_search
{
  "filter_path": ["hits.hits._source"],
  "query": {
    // Replace with your actual query (e.g., match, term, etc.)
    "match_all": {}
  }
}

This response will only include the array of _source objects—no extra top-level fields or hit metadata cluttering things up.

2. Combine _source and filter_path (For Explicit Control) #

If you want to explicitly confirm you're fetching the full _source (in case your index has custom default settings), pair _source: true with the filter_path parameter:

POST /your_target_index/_search
{
  "_source": true,
  "filter_path": ["hits.hits._source"],
  "query": {
    "match": { "content": "elasticsearch tips" }
  }
}

Note: _source: true is Elasticsearch's default behavior, so this is optional—but it’s handy if you need to override any configurations that might hide the _source.

3. Skip the fields Parameter for This Use Case #

While you could use fields to fetch all fields, it’s not ideal for getting the full _source. It returns a different structure (field names as keys with array values) instead of the original _source object. Stick to the first two methods for clean, direct results.

That’s all you need! Any of these approaches will give you exactly the _source content you want, without the extra noise.

内容的提问来源于stack exchange，提问作者Tzipporah Raanan