我想在Java中使用JSON制作一个简单的HTTP POST。

假设URL是www.site.com

它接收值{"name":"myname","age":"20"},例如标记为" details "。

我将如何着手创建POST的语法?

我似乎也找不到JSON Javadocs中的POST方法。


当前回答

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。

其他回答

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());

以下是你需要做的:

获取Apache HttpClient,这将使您能够发出所需的请求 用它创建一个HttpPost请求,并添加头应用程序/x-www-form-urlencoded 创建一个StringEntity,并将JSON传递给它 执行调用

代码大致如下(你仍然需要调试它并使它工作):

// @Deprecated HttpClient httpClient = new DefaultHttpClient();
HttpClient httpClient = HttpClientBuilder.create().build();
try {
    HttpPost request = new HttpPost("http://yoururl");
    StringEntity params = new StringEntity("details={\"name\":\"xyz\",\"age\":\"20\"} ");
    request.addHeader("content-type", "application/x-www-form-urlencoded");
    request.setEntity(params);
    HttpResponse response = httpClient.execute(request);
} catch (Exception ex) {
} finally {
    // @Deprecated httpClient.getConnectionManager().shutdown(); 
}

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。

Java 8与apache httpClient 4

CloseableHttpClient client = HttpClientBuilder.create().build();
HttpPost httpPost = new HttpPost("www.site.com");


String json = "details={\"name\":\"myname\",\"age\":\"20\"} ";

        try {
            StringEntity entity = new StringEntity(json);
            httpPost.setEntity(entity);

            // set your POST request headers to accept json contents
            httpPost.setHeader("Accept", "application/json");
            httpPost.setHeader("Content-type", "application/json");

            try {
                // your closeablehttp response
                CloseableHttpResponse response = client.execute(httpPost);

                // print your status code from the response
                System.out.println(response.getStatusLine().getStatusCode());

                // take the response body as a json formatted string 
                String responseJSON = EntityUtils.toString(response.getEntity());

                // convert/parse the json formatted string to a json object
                JSONObject jobj = new JSONObject(responseJSON);

                //print your response body that formatted into json
                System.out.println(jobj);

            } catch (IOException e) {
                e.printStackTrace();
            } catch (JSONException e) {

                e.printStackTrace();
            }

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        }

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));