如何在 Spring REST API 响应中连接 Elasticsearch 的结果

Question

我使用 Spring 开发了一个 REST API 条目，它在 Elasticsearch 中进行搜索，现在我想返回 ES 找到的任何结果作为响应。我不关心搜索结果，也不知道其中的 JSON 结构。我只是想把它还给客户。

我希望这样的事情会起作用：

@RequestMapping(value = "/search/{index:.*}", method = RequestMethod.GET) 公共无效搜索（@PathVariable 字符串索引，@RequestParam 映射 allRequestParams，HttpServletResponse 响应） 抛出 IOException { BoolQueryBuilder 查询 = QueryBuilders.boolQuery(); for (Map.Entry entry : allRequestParams.entrySet()) { query.should(QueryBuilders.fuzzyQuery(entry.getKey(), entry.getValue())); } SearchResponse 结果 = esClient.prepareSearch("nyc_visionzero") .setTypes("日志") .setQuery(查询) 。执行（） .actionGet(); SearchHits 命中 = results.getHits(); hits.writeTo(response.getOutputStream()); }

但最后一行出现编译错误，因为两个 OutputStream 不兼容。所以我的问题是，将 Elasticsearch 的结果连接到 Spring 的响应中的最简单方法是什么？

Answer 1

我自己设法找到了解决方案：

@RequestMapping(value = "/search/{index:.*}", method = RequestMethod.GET)
public void search(@PathVariable String index, @RequestParam Map<String, String> allRequestParams, HttpServletResponse response)
    throws IOException
{
    BoolQueryBuilder query = QueryBuilders.boolQuery();
    for (Map.Entry<String, String> entry : allRequestParams.entrySet()) {
        query.should(QueryBuilders.fuzzyQuery(entry.getKey(), entry.getValue()));
    }

    SearchResponse results = esClient.prepareSearch("nyc_visionzero")
        .setTypes("logs")
        .setQuery(query)
        .execute()
        .actionGet();

    SearchHits hits = results.getHits();

    ServletOutputStream os = response.getOutputStream();
    XContentBuilder builder = XContentFactory.jsonBuilder(os);
    results.toXContent(builder, ToXContent.EMPTY_PARAMS);
    builder.close();
    os.close();
}

Answer 2

@RequestMapping(value = "/search/{index:.*}", method = RequestMethod.GET)
public ResponseEntity<?> search(@PathVariable String index, @RequestParam Map allRequestParams)
{
BoolQueryBuilder query = QueryBuilders.boolQuery();
for (Map.Entry entry : allRequestParams.entrySet()) {
    query.should(QueryBuilders.fuzzyQuery(entry.getKey(), entry.getValue()));
}

SearchResponse results = esClient.prepareSearch("nyc_visionzero")
    .setTypes("logs")
    .setQuery(query)
    .execute()
    .actionGet();

SearchHits hits = results.getHits();
return new ResponseEntity<>(hits , HttpStatus.OK);
}

使用 ResponseEntity 类，您可以返回任何响应，而无需关心您的结果类型。

Answer 3

您可以更改搜索方法的签名以返回字符串，然后直接将结果作为有效 JSON 返回，而不是尝试写入响应输出流。比如：

@RequestMapping(value = "/search/{index:.*}", method = RequestMethod.GET)
public String search(@PathVariable String index, @RequestParam Map allRequestParams, HttpServletResponse response)
    throws IOException
{
    BoolQueryBuilder query = QueryBuilders.boolQuery();
    for (Map.Entry entry : allRequestParams.entrySet()) {
        query.should(QueryBuilders.fuzzyQuery(entry.getKey(), entry.getValue()));
    }

    SearchResponse results = esClient.prepareSearch("nyc_visionzero")
        .setTypes("logs")
        .setQuery(query)
        .execute()
        .actionGet();

    SearchHits hits = results.getHits();

    // Replacing hits.writeTo(response.getOutputStream()); below
    StringBuilder builder = new StringBuilder();
    SearchHit[] hitsDatas = hits.hits();
    int length = hitsDatas.length;
    builder.append("[");
    for (int i = 0; i < length; i++) {
       if (i == length - 1) {
          builder.append(hitsDatas[i].getSourceAsString());
       } else {
          builder.append(hitsDatas[i].getSourceAsString());
          builder.append(",");
       }
    }
    builder.append("]");
    return builder.toString();
}