弹性搜索从搜索结果中排除嵌套元素，通过 id 获取元素

Question

我有嵌套对象的项目：

          {
                "name": "The Amazon rainforest",
                "id": "610d33da26c25b00191ebcbe",
                "tags": [
                    {
                        "name": "Brazil",
                        "verified": 1
                    },
                    {

                        "name": "new_tag",
                        "verified": 0,
                    }
                ],

            }

在搜索结果中未验证的标签应该省略：

按 id 搜索的输出：610d33da26c25b00191ebcbe

          {
                "name": "The Amazon rainforest",
                "id": "610d33da26c25b00191ebcbe",
                "tags": [
                    {
                        "name": "Brazil",
                        "verified": 1
                    }
                ],

            }

Answer 1

Node.js 版本的答案：

      const { body } = await elasticWrapper.client.search({
        index: ElasticIndexs.Products,
        filter_path:
          'hits.hits._source*, hits.hits.inner_hits.tags.hits.hits._source*',
        body: {
          _source: {
            excludes: ['tags'],
          },
          query: {
            bool: {
              must: [
                {
                  match: {
                    id: req.params.id,
                  },
                },
              ],
              should: [
                {
                  nested: {
                    path: 'tags',
                    query: {
                      term: {
                        'tags.verified': 1,
                      },
                    },
                    inner_hits: {},
                  },
                },
              ],
            },
          },
        },
      });