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

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。

其他回答

我发现这个问题寻找如何从java客户端发送post请求到谷歌端点的解决方案。以上答案,很可能是正确的,但在谷歌端点的情况下不工作。

谷歌端点的解决方案。

Request body must contains only JSON string, not name=value pair. Content type header must be set to "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(); } It sure can be done using HttpClient as well.

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;

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。

@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();
}

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(); 
}