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
Java 8与apache httpClient 4
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();
}
其他回答
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。
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。
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());
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();
}
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;