node.js elasticsearch响应空白_source

Question

我已经使用Elasticsearch和node.js创建了一个示例索引，并使用以下代码设置创建了一个样本索引。

const { Client } = require('@elastic/elasticsearch');
const { ELASTIC_SEARCH } = require('../config');

// Elastic Search Cloud Client Setup
const elasticClient = new Client({
    cloud: { id: ELASTIC_SEARCH.CLOUDID },
    auth: {
        apiKey: ELASTIC_SEARCH.API_KEY
    }
});


async function prepareIndex() {
const merchantIndexExists = await elasticClient.indices.exists({ index: 'index2' }); 
        if (merchantIndexExists) return;
        await elasticClient.indices.create({
            index: 'index2',
            body: {
                mappings: {
                    dynamic: 'strict',
                    properties: {
                        company_name: { type: 'text' },
                        company_email: { type: 'keyword' },
                        name: { type: 'text' },
                        price: { type: 'scaled_float', scaling_factor: 10 },
                        created_date: { type: 'date' },
                        is_delete: { type: 'boolean', doc_values: false },
                        merchant: { type: 'keyword', index: 'true' }
                    }
                }
            }
        });
  }

索引创建后，我添加了以下代码的文档：

const { company_name, company_email, price } = req.body;
        const response = await elasticClient.index({
            index: 'index2',
            document: {
                company_email,
                company_name,
                price
            }
        });

现在，当我从Kibana Cloud Console调用搜索API时，它将返回所有归档的确切搜索结果。喜欢

但是当我通过Postman中的代码击中同一搜索查询时，它将返回空白_source。这是Postman响应

const response = await elasticClient.search({
            index: 'index2',
            query: {
                match_all: {}
            }
        });

​

Answer 1

使用node.js elasticsearch客户端时，您必须将查询包装到身体属性中并将其传递给搜索。

const response = await elasticClient.search({
  index: 'index2',
  body: {
    query: {
      match_all: {}
    }
  }
});