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
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));
其他回答
@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();
}
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();
}
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。
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));
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.