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
@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();
}
其他回答
试试下面的代码:
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();
}
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。
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.
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));
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();
}