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如何将Elasticsarch的结果连接到Spring的REST API响应中

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我已经使用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，我不知道其中的JSON结构 . 我只是想把它还给客户 .

我希望这样的东西能起作用：

@RequestMapping(value = "/search/{index:.*}", method = RequestMethod.GET)
public void 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();
    hits.writeTo(response.getOutputStream());
}

但是最后一行有一个编译错误，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？

java spring elasticsearch

3 回答

  • 1

    您可以更改搜索方法的签名以返回String，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 . 就像是：

    @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();
}
    回复于
  • -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();
}
    回复于
  • 1
    @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 .

    回复于

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