Question

我在Java Swing中创建了一个GUI，想要点击一个API（POST方法）点击一下按钮

http://www.myapi.domain/v1

我必须发送一个JSON对象作为请求参数以及标题，即（Headers [“custom-Header”] ==“AXYZ”）我已将我的字符串转换为JSON对象喜欢

{"machineKey":"","serialNumber":"","name":"","mobile":"","productKey":"","email":""}

我只是做了

public void actionPerformed(ActionEvent e) 
 {
    if (e.getSource() == Submit)
     {

        dispose();
        new Otp().setVisible(true);
     }
      else
   {
        name.setText("");
        email.setText("");
        mobile.setText("");
        machineKey.setText("");
        productKey.setText("");
        serialNumber.setText("");
    }
  String Sname=   name.getText();
  String Semail=  email.getText();
  String Smobile=   mobile.getText();
  String SmachineKey=  machineKey.getText();
  String SproductKey=  productKey.getText();
  String SserialNumber=  serialNumber.getText();    

    JSONObject abc=prepareReqJsonObj(Sname,Semail,Smobile,SmachineKey,SproductKey,SserialNumber);
    System.out.println(abc);

 }

现在，我如何直接从我的GUI按钮发出命令来命中API，我可以通过什么方式实现这一点并获得响应。

提前感谢您的帮助：）

Answer 1

查看this example如何使用HttpClient发布数据。

基本上，您创建一个HttpClient，创建一个POST请求，用您的数据填充该请求，然后提交。

Answer 2

将此添加为所有帖子请求的通用实用程序类

import java.io.BufferedReader;
import java.io.DataOutputStream;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;

public class Utility {

public static String excutePost(String targetURL, String urlParameters)
  {
    URL url;
    HttpURLConnection connection = null;  
    try {
      //Create connection
      url = new URL(targetURL);
      connection = (HttpURLConnection)url.openConnection();
      connection.setRequestMethod("POST");
      connection.setRequestProperty("Content-Type", 
           "application/json");
      connection.setRequestProperty("custom-Header", "XYZ");

      connection.setRequestProperty("Content-Length", "" + 
               Integer.toString(urlParameters.getBytes().length));
      connection.setRequestProperty("Content-Language", "en-US");  

      connection.setUseCaches (false);
      connection.setDoInput(true);
      connection.setDoOutput(true);


      //Send request
      DataOutputStream wr = new DataOutputStream (
                  connection.getOutputStream ());
      wr.writeBytes (urlParameters);
      wr.flush ();
      wr.close ();

      //Get Response    
      InputStream is = connection.getInputStream();
      BufferedReader rd = new BufferedReader(new InputStreamReader(is));
      String line;
      StringBuffer response = new StringBuffer(); 
      while((line = rd.readLine()) != null) {
        response.append(line);
        response.append('\r');
      }
      rd.close();
      return response.toString();

    } catch (Exception e) {

      e.printStackTrace();
      return null;

    } finally {

      if(connection != null) {
        connection.disconnect(); 
      }
    }
  }
 }

和Action Listener中的这个 - ＆gt;

 String Sname=   name.getText();
  String Semail=  emailid.getText();
  String Smobile=   mobile.getText();
  String SmachineKey=  machineKey.getText();
  String SproductKey=  productkey.getText();
  String SserialNumber=  serialNo.getText();    

    JSONObject reqObj=prepareReqJsonObj(Sname,Semail,Smobile,SmachineKey,SproductKey,SserialNumber);

    String reqString= reqObj.toString();
    String APIUrl=  "http://example.com/v1/api";

   String response=  Utility.excutePost(APIUrl, reqString);

   System.out.println(reqObj);
   System.out.println(reqString);
   System.out.println(response);

 }

@SuppressWarnings("unchecked")
public JSONObject prepareReqJsonObj(String s1,String s2,String s3,String s4,String s5,String s6){
 JSONObject jsonobj = new JSONObject();



    jsonobj.put("name", s1);
    jsonobj.put("emailid", s2);
    jsonobj.put("mobile",s3 );
    jsonobj.put("machineKey",s4 );
    jsonobj.put("productkey", s5);
    jsonobj.put("serialNo", s6);

    return jsonobj;

如何通过Java swing GUI发出Post请求

2 个答案: