如何使用 Apache HTTP 客户端将复杂参数传递给 POST 请求？

Question

我尝试发送POST带有这样的正文的请求{  "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 的构造函数BasicNameValuePair适用(String, String)。所以我需要另一堂课。有什么办法可以添加methodPropertiesList吗urlParameters？

Answer 1

您的请求看起来像 json 结构，因此发布如下数据：

 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);

Answer 2

对于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，它可能是一个类并在其中使用映射，但通常您也为嵌套对象创建一个类，如果它具有特定的结构）。

Answer 3

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

  }