在 Java 中使用 JSON 的 HTTP POST

Question

我想在 Java 中使用 JSON 制作一个简单的 HTTP POST。

假设网址是www.site.com

它接受值{"name":"myname","age":"20"}，例如标记为'details'。

我将如何为 POST 创建语法？

我似乎也无法在 JSON Javadocs 中找到 POST 方法。

Answer 1

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

您将使用 JSONObject 或其他任何东西来构建您的 JSON，但不会处理网络；您需要对其进行序列化，然后将其传递给 HttpURLConnection 以进行 POST。

Answer 2

这是你需要做的：

1. 获取 Apache HttpClient，这将使您能够发出所需的请求
2. 用它创建一个HttpPost请求并添加标题application/x-www-form-urlencoded
3. 创建一个StringEntity，您将把 JSON 传递给它
4. 执行调用

代码大致如下（您仍然需要对其进行调试并使其工作）：

// @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(); 
}

Answer 3

@momo 对 Apache HttpClient 版本 4.3.1 或更高版本的回答。我正在使用JSON-Java 来构建我的 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();
}

Answer 4

您可以利用 Gson 库将您的 java 类转换为 JSON 对象。

为要发送的变量创建一个 pojo 类 按照上面的例子

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

变成

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

在 pojo1 类中设置变量后，您可以使用以下代码发送该变量

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;

对于 GSON

import com.google.gson.Gson;

Answer 5

试试这个代码：

HttpClient httpClient = new DefaultHttpClient();

try {
    HttpPost request = new HttpPost("http://yoururl");
    StringEntity params =new StringEntity("details={\"name\":\"myname\",\"age\":\"20\"} ");
    request.addHeader("content-type", "application/json");
    request.addHeader("Accept","application/json");
    request.setEntity(params);
    HttpResponse response = httpClient.execute(request);

    // handle response here...
}catch (Exception ex) {
    // handle exception here
} finally {
    httpClient.getConnectionManager().shutdown();
}

Answer 6

我发现这个问题正在寻找有关如何将发布请求从 Java 客户端发送到 Google Endpoints 的解决方案。以上答案，很可能是正确的，但不适用于 Google Endpoints。

Google 端点解决方案。

1. 请求正文必须只包含 JSON 字符串，而不是名称=值对。

2. 内容类型标头必须设置为“application/json”。

   post("http://localhost:8888/_ah/api/langapi/v1/createLanguage",
                      "{\"language\":\"russian\", \"description\":\"dsfsdfsdfsdfsd\"}");
   
   
   
   public static void post(String url, String json ) throws Exception{
     String charset = "UTF-8"; 
     URLConnection connection = new URL(url).openConnection();
     connection.setDoOutput(true); // Triggers POST.
     connection.setRequestProperty("Accept-Charset", charset);
     connection.setRequestProperty("Content-Type", "application/json;charset=" + charset);
   
     try (OutputStream output = connection.getOutputStream()) {
       output.write(json.getBytes(charset));
     }
   
     InputStream response = connection.getInputStream();
   }
   

   当然也可以使用 HttpClient 来完成。

Answer 7

protected void sendJson(final String play, final String prop) {
     Thread t = new Thread() {
     public void run() {
        Looper.prepare(); //For Preparing Message Pool for the childThread
        HttpClient client = new DefaultHttpClient();
        HttpConnectionParams.setConnectionTimeout(client.getParams(), 1000); //Timeout Limit
        HttpResponse response;
        JSONObject json = new JSONObject();

            try {
                HttpPost post = new HttpPost("http://192.168.0.44:80");
                json.put("play", play);
                json.put("Properties", prop);
                StringEntity se = new StringEntity(json.toString());
                se.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
                post.setEntity(se);
                response = client.execute(post);

                /*Checking response */
                if (response != null) {
                    InputStream in = response.getEntity().getContent(); //Get the data in the entity
                }

            } catch (Exception e) {
                e.printStackTrace();
                showMessage("Error", "Cannot Estabilish Connection");
            }

            Looper.loop(); //Loop in the message queue
        }
    };
    t.start();
}

Answer 8

我推荐 http-request 基于 apache http api。

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. 
}

如果您想发送JSON 作为请求正文，您可以：

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

我强烈建议在使用前阅读文档。

Answer 9

您可以在 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，您可以创建一个 json 对象并将字段放入对象中，如下所示

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));

Answer 10

对于 Java 11，您可以使用新的 HTTP client：

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();

您可以使用来自InputStream、String、File 的发布者。可以使用 Jackson 将 JSON 转换为 String 或 IS。

Answer 11

带有 apache httpClient 4 的 Java 8

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();
        }

Answer 12

实现 HTTP/2 和 Web Socket 的 HTTP 客户端 API 的 Java 11 标准化，可在 java.net.HTTP.* 中找到：

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());