在Java中使用JSON的HTTP POST

Question

我想在Java中使用JSON進行簡單的HTTP POST。

假設URL是www.site.com

，它需要在標註爲'details'例如價值{"name":"myname","age":"20"}。

我該如何去創建POST的語法？

我也無法在JSON Javadoc中找到POST方法。

Answer 1

這可能是最容易使用的HttpURLConnection。

http://www.xyzws.com/Javafaq/how-to-use-httpurlconnection-post-data-to-web-server/139

您將使用的JSONObject或什麼來構建你的JSON，但不處理網絡;你需要序列化它，然後將它傳遞給HttpURLConnection以POST。

Answer 2

以下是你需要做的：

1. 加深對Apache HttpClient的，這將使你做出了規定要求
2. 創建一個與它的HttpPost請求，並添加標題爲「application/X- WWW的形式，進行了urlencoded」
3. 創建StringEntity，您將通過JSON它
4. 執行調用

釷E碼大致樣子（你仍需要調試它，讓它工作）

//Deprecated 
//HttpClient httpClient = new DefaultHttpClient(); 

HttpClient httpClient = HttpClientBuilder.create().build(); //Use this instead 

try { 

    HttpPost request = new HttpPost("http://yoururl"); 
    StringEntity params =new StringEntity("details={\"name\":\"myname\",\"age\":\"20\"} "); 
    request.addHeader("content-type", "application/x-www-form-urlencoded"); 
    request.setEntity(params); 
    HttpResponse response = httpClient.execute(request); 

    //handle response here... 

}catch (Exception ex) { 

    //handle exception here 

} finally { 
    //Deprecated 
    //httpClient.getConnectionManager().shutdown(); 
} 

Answer 3

@ MOMO的回答爲Apache HttpClient的，版本4.3.1或更高版本。我使用JSON-Java建立我的JSON對象：

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(); 
} 

Answer 4

您可以使用GSON庫到您的Java類轉換成JSON對象。

創建要發送 按照以上示例

{"name":"myname","age":"20"} 

成爲

class pojo1 
{ 
    String name; 
    String age; 
    //generate setter and getters 
} 

一旦你設置pojo1類中的變量，你可以給變量一個POJO類，使用下面的代碼

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); 

這些都是進口

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; 

Answer 5

試試這個代碼：

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(); 
} 

Answer 6

我發現這個問題，尋找有關如何從Java客戶端發送POST請求，谷歌端點解決方案。以上答案很可能是正確的，但在Google Endpoint的情況下不起作用。

Google終端解決方案。

1. 請求主體必須只包含JSON字符串，而不是名稱=值對。

2. 內容類型標頭必須設置爲「application/json」。

   post("http://localhost:8888/_ah/api/langapi/v1/createLanguage", 
           "{\"language\":\"russian\", \"description\":\"dsfsdfsdfsdfsd\"}"); 
   
   
   
   public static void post(String url, String param) 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(param.getBytes(charset)); 
       } 
   
       InputStream response = connection.getInputStream(); 
   } 
   

   它肯定可以使用HttpClient來完成。

Answer 7

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(); 
} 

Answer 8

我建議http-request它基於Apache HTTP API。

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. 
} 

如果你想發送JSON作爲請求體，您可以：

ResponseHandler<String> responseHandler = httpRequest.executeWithBody(yourJsonData); 

我使用前higly建議更換閱讀文檔。