如何使用 Apache HTTP 客户端将复杂参数传递给 POST 请求？

Question

我尝试用这样的正文发送 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)。所以我需要另一个 class 来代替。

有什么办法可以把methodPropertiesList加到urlParameters上吗？

Answer 1

您的请求看起来像一个 json 结构，所以 post 数据如下：

 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);

参考：

Answer 2

这个问题有一个众所周知的方法。在大多数情况下，您将创建自己的对象来描述您要在 HttpPost 中发送的内容。所以你会得到类似的东西：

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);

    StringEntity strEntity = new StringEntity(gson.toJson(resource));
    post.setEntity(strEntity);

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

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

这通常是您处理具有嵌套结构的更复杂 json 对象的方式。您必须为要发送的资源创建 classes（在您的示例中它可能是一个 class 并在其中使用 map，但通常您会为嵌套的资源创建一个 class如果它具有特定的结构，它也是对象）。为了更好地了解整个画面，请阅读本教程：https://www.baeldung.com/jackson-mapping-dynamic-object

Answer 3

使用 Sushil Mittal 答案是我的解决方案

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

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

    HttpPost post = new HttpPost(HOST);  
    Gson gson = new Gson();

    Params params = new Params(method, methodProperties);
    StringEntity entity = new StringEntity(gson.toJson(params));   
    post.setEntity(entity);

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

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

  class Params {

    String method;   
    Map<String, String> methodProperties;

    public Params(String method, Map<String, String> methodProperties) {
      this.method = method;
      this.methodProperties = methodProperties;
    }

    //getters
  }

如何使用 Apache HTTP 客户端将复杂参数传递给 POST 请求？

How to pass complex parameter to POST request using Apache HTTP client?

java

post

apache-httpcomponents

apache-httpclient-4.x