java - 如何将 json 放入 HTTPPost 请求正文中，然后在响应中读取正文

Question

我正在尝试使用正文发出 HTTP Post 请求。我想知道如何将 json 对象放入请求正文中，然后读取响应正文。

HttpClient httpClient = HttpClientBuilder.create().build();
    String responseString = null;
    try {
        HttpPost request = new HttpPost(link);
        log.info("Executing request " + request.getRequestLine());
        StringEntity params = new StringEntity(data);
        request.addHeader("content-type", "application/json");
        request.setEntity(params);
        HttpResponse response = httpClient.execute(request);
        HttpEntity entity = response.getEntity();
        responseString = EntityUtils.toString(entity, "UTF-8");
    } catch (Exception e) {
        log.info("Exception thrown while executing HTTPPostRequest");
    }
    return responseString;

这是我当前用来发出请求的代码片段。如果使用“StringEntity”是将数据放入正文中的正确选择。 我得到空正文，它应该返回包含 json 数据的正文

Answer 1

只要你的“数据”是 json 就应该没问题。

HttpClient httpClient = HttpClientBuilder.create().build();
    String responseString = null;
    try {
        HttpPost request = new HttpPost(link);
        log.info("Executing request " + request.getRequestLine());
        StringEntity content = new StringEntity(data);
        content.setContentType("application/json"); 
        //or request.setHeader("Content-type", "application/json");
        request.setEntity(content);
        HttpResponse response = httpClient.execute(request);
        HttpEntity entity = response.getEntity();
        responseString = EntityUtils.toString(entity, "UTF-8");
    } catch (Exception e) {
        log.info("Exception thrown while executing HTTPPostRequest");
    }
    return responseString;