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。
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
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。
其他回答
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();
}
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。
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。
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();
}