对于Java 11，你可以使用新的HTTP客户端:

HttpClient client = HttpClient.newHttpClient();
HttpRequest request = HttpRequest.newBuilder()
    .uri(URI.create("http://localhost/api"))
    .header("Content-Type", "application/json")
    .POST(ofInputStream(() -> getClass().getResourceAsStream(
        "/some-data.json")))
    .build();

client.sendAsync(request, BodyHandlers.ofString())
    .thenApply(HttpResponse::body)
    .thenAccept(System.out::println)
    .join();

你可以使用InputStream, String, File中的发布者。将JSON转换为字符串或IS可以用Jackson完成。

您可以使用Gson库将java类转换为JSON对象。

为想要发送的变量创建一个pojo类 如上所述

{"name":"myname","age":"20"}

就变成了

class pojo1
{
   String name;
   String age;
   //generate setter and getters
}

一旦在pojo1类中设置了变量，就可以使用下面的代码发送它

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);

这些是进口

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.HttpClientBuilder;

和GSON

import com.google.gson.Gson;

@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();
}

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());

对于Java 11，你可以使用新的HTTP客户端:

HttpClient client = HttpClient.newHttpClient();
HttpRequest request = HttpRequest.newBuilder()
    .uri(URI.create("http://localhost/api"))
    .header("Content-Type", "application/json")
    .POST(ofInputStream(() -> getClass().getResourceAsStream(
        "/some-data.json")))
    .build();

client.sendAsync(request, BodyHandlers.ofString())
    .thenApply(HttpResponse::body)
    .thenAccept(System.out::println)
    .join();

你可以使用InputStream, String, File中的发布者。将JSON转换为字符串或IS可以用Jackson完成。

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。