你可以在Apache HTTP中使用以下代码:

String payload = "{\"name\": \"myname\", \"age\": \"20\"}";
post.setEntity(new StringEntity(payload, ContentType.APPLICATION_JSON));

response = client.execute(request);

此外，您还可以创建一个json对象，并像这样在对象中放入字段

HttpPost post = new HttpPost(URL);
JSONObject payload = new JSONObject();
payload.put("name", "myName");
payload.put("age", "20");
post.setEntity(new StringEntity(payload.toString(), ContentType.APPLICATION_JSON));

Java 11标准的HTTP客户端API，实现了HTTP/2和Web Socket，可以在java.net.HTTP找到。*:

String payload = "{\"name\": \"myname\", \"age\": \"20\"}";
HttpClient client = HttpClient.newHttpClient();

HttpRequest request = HttpRequest.newBuilder(URI.create("www.site.com"))
            .header("content-type", "application/json")
            .POST(HttpRequest.BodyPublishers.ofString(payload))
            .build();
    
HttpResponse<String> response = client.send(request, BodyHandlers.ofString());

我推荐在apache http api上构建http-request。

HttpRequest<String> httpRequest = HttpRequestBuilder.createPost(yourUri, String.class)
    .responseDeserializer(ResponseDeserializer.ignorableDeserializer()).build();

public void send(){
   ResponseHandler<String> responseHandler = httpRequest.execute("details", yourJsonData);

   int statusCode = responseHandler.getStatusCode();
   String responseContent = responseHandler.orElse(null); // returns Content from response. If content isn't present returns null. 
}

如果你想发送JSON作为请求体，你可以:

  ResponseHandler<String> responseHandler = httpRequest.executeWithBody(yourJsonData);

我强烈建议在使用前阅读文档。

@momo对Apache HttpClient的回答，4.3.1版本或更高版本。我使用JSON- java构建我的JSON对象:

JSONObject json = new JSONObject();
json.put("someKey", "someValue");    

CloseableHttpClient httpClient = HttpClientBuilder.create().build();

try {
    HttpPost request = new HttpPost("http://yoururl");
    StringEntity params = new StringEntity(json.toString());
    request.addHeader("content-type", "application/json");
    request.setEntity(params);
    httpClient.execute(request);
// handle response here...
} catch (Exception ex) {
    // handle exception here
} finally {
    httpClient.close();
}

使用HttpURLConnection可能是最简单的。

http://www.xyzws.com/Javafaq/how-to-use-httpurlconnection-post-data-to-web-server/139

你将使用JSONObject或其他东西来构造JSON，但不是用来处理网络;你需要序列化它，然后将它传递给一个HttpURLConnection到POST。

你可以在Apache HTTP中使用以下代码:

String payload = "{\"name\": \"myname\", \"age\": \"20\"}";
post.setEntity(new StringEntity(payload, ContentType.APPLICATION_JSON));

response = client.execute(request);

此外，您还可以创建一个json对象，并像这样在对象中放入字段

HttpPost post = new HttpPost(URL);
JSONObject payload = new JSONObject();
payload.put("name", "myName");
payload.put("age", "20");
post.setEntity(new StringEntity(payload.toString(), ContentType.APPLICATION_JSON));