如何使用Apache HTTP客户端将复杂参数传递给POST请求？

Question

我试图用这样的正文发送POST请求

{
  "method": "getAreas",
  "methodProperties": {
      "prop1" : "value1",
      "prop2" : "value2",
   }
}

这是我的代码

static final String HOST = "https://somehost.com";

  public String sendPost(String method,
      Map<String, String> methodProperties) throws ClientProtocolException, IOException {

    HttpPost post = new HttpPost(HOST);

    List<NameValuePair> urlParameters = new ArrayList<>();
    urlParameters.add(new BasicNameValuePair("method", method));

    List<NameValuePair> methodPropertiesList = methodProperties.entrySet().stream()
                .map(entry -> new BasicNameValuePair(entry.getKey(), entry.getValue()))
                .collect(Collectors.toList());

    // ??? urlParameters.add(new BasicNameValuePair("methodProperties", methodPropertiesList));

    post.setEntity(new UrlEncodedFormEntity(urlParameters));

    try (CloseableHttpClient httpClient = HttpClients.createDefault();
        CloseableHttpResponse response = httpClient.execute(post)) {

      return EntityUtils.toString(response.getEntity());
    }
  }

但是BasicNameValuePair的构造函数适用(String, String)。所以我需要另一堂课。

有没有办法将methodPropertiesList添加到urlParameters？

Answer 1

 class pojo1{
   String method;
   Map<String,String> methodProperties;
 }

String postUrl = "www.site.com";// put in your url
Gson gson = new Gson();
HttpClient httpClient = HttpClientBuilder.create().build();
HttpPost post = new HttpPost(postUrl);
StringEntity postingString = new StringEntity(gson.toJson(pojo1));//gson.tojson()    converts your pojo to json
post.setEntity(postingString);
post.setHeader("Content-type", "application/json");
HttpResponse  response = httpClient.execute(post);

ref：HTTP POST using JSON in Java

Answer 2

static final String HOST = "https://somehost.com";

  public String sendPost(String method,
      Map<String, String> methodProperties) throws ClientProtocolException, IOException {

    HttpPost post = new HttpPost(HOST);
    MyResource resource = new MyResource();
    resource.setMethod(method);
    MyNestedResource nestedResource = new MyNestedResource();
    nestedResource.setMethodProperties(methodProperties);
    resource.setNestedResourceMethodProperties(nestedResource);

    post.setEntity(resource);

    try (CloseableHttpClient httpClient = HttpClients.createDefault();
        CloseableHttpResponse response = httpClient.execute(post)) {

      return EntityUtils.toString(response.getEntity());
    }
  }

并且通常这就是您使用嵌套结构处理更复杂的json对象的方式。您必须为要发送的资源创建类（在您的示例中，它可能是一个类，并且在其中使用了map，但是如果嵌套对象具有特定的结构，通常也为嵌套对象创建一个类）。为了更好地了解整个图片，请涵盖本教程：https://www.baeldung.com/jackson-mapping-dynamic-object