在 servlet 中创建 json 对象

Posted 2023-02-26

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【中文标题】在 servlet 中创建 json 对象【英文标题】：Create json object in servlet 【发布时间】：2012-09-04 01:09:51 【问题描述】：

我想将检索到的数据从服务器发送到我的 android 客户端...我使用 json 对象来执行此操作。这是我的 servlet 代码。

import java.io.IOException; 
import java.io.PrintWriter;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import net.sf.json.JSONArray;
import net.sf.json.JSONObject;

    public class AvailabilityResponse extends HttpServlet 

@Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException 

    response.setContentType("application/json");
    PrintWriter out=response.getWriter();

        String br_id;
        br_id=request.getParameter("branchname");

    try
        Class.forName("com.mysql.jdbc.Driver").newInstance();
            Connection con=DriverManager.getConnection("jdbc:mysql://localhost:8888/atmlivedetails","root","root");  
            Statement st=con.createStatement();
            ResultSet rs=st.executeQuery("select atmbrno, atmbrname  from location_stat where act_brname='"+br_id+"'");
            while(rs.next())

        String s = rs.getString("atmbrno");
        String t = rs.getString("atmbrname");

        JSONObject arrayObj = new JSONObject();

        arrayObj.put("atmbrno",s);
        arrayObj.put("atmbrname",t);


        out.print(arrayObj);
        
        rs.close ();
        st.close ();
           
    catch(Exception e)
            out.print(e);

这是我的 android 端代码..

import java.util.ArrayList;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.ParseException;
import org.apache.http.message.BasicNameValuePair;
import org.apache.http.client.HttpClient;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.ClientProtocolException;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import android.app.Activity;
import android.content.Intent;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;
import android.os.Bundle;

public class CheckAvailability extends Activity

Button but1,but2;
EditText brName;
TextView txt1;
String text;


public void onCreate(Bundle savedInstanceState) 
    super.onCreate(savedInstanceState);
    setContentView(R.layout.availability);

    brName =(EditText)findViewById(R.id.editText1);
    but1 = (Button)findViewById(R.id.button5); 
    but2 = (Button)findViewById(R.id.button6);
    txt1 = (TextView)findViewById(R.id.textView3);



    but1.setOnClickListener(new View.OnClickListener()

        public void onClick(View v)

            String result = null;
            InputStream is = null;
            StringBuilder sb=null;

           ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();

           postParameters.add(new BasicNameValuePair("branchname", brName.getText().toString()));

        //http post
                  try
                      HttpClient httpclient = new DefaultHttpClient();
                      HttpPost httppost = new HttpPost("http://10.0.2.2:8080/hello/AvailabilityResponse");
                      httppost.setEntity(new UrlEncodedFormEntity(postParameters));
                      HttpResponse response = httpclient.execute(httppost);
                      HttpEntity entity = response.getEntity();
                      is = entity.getContent();
                  catch(Exception e)
                      Log.e("log_tag", "Error in http connection"+e.toString());
                  

        //convert response to string
                  try
                       BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
                       sb = new StringBuilder();
                       sb.append(reader.readLine() + "\n");
                       String line="0";

                             while ((line = reader.readLine()) != null) 
                                 sb.append(line + "\n");
                             
                             is.close();
                             result=sb.toString();

                  catch(Exception e)
                        Log.e("log_tag", "Error converting result "+e.toString());
                         

              //paring data
                        String atm_id;
                        String atm_name;

                        try
                        JSONArray jArray = new JSONArray(result);
                        JSONObject json_data=null;

                        for(int i=0;i<jArray.length();i++)
                                json_data = jArray.getJSONObject(i);
                                atm_id=json_data.getString("atmbrno");
                                atm_name=json_data.getString("atmbrname");

                              //txt1.setText(atm_name);
                        

                        catch(JSONException e1)
                            Toast.makeText(getBaseContext(), "No DATA Found", Toast.LENGTH_LONG).show();

                        catch (ParseException e1)
                            e1.printStackTrace();
                        





        
);

但是当我运行它时，它总是给我“未找到数据”异常...任何人都可以帮助我吗？？？

【问题讨论】：

在服务器端或安卓端哪里得到“未找到数据”异常？请重新格式化代码.. 删除不必要的导入和过多的代码。使其清晰易读。 @ponraj- 它显示在屏幕上...这里它添加为 Toast.makeText(getBaseContext(), "No DATA Found",Toast.LENGTH_LONG).show(); @Dasaya 在您的 servlet 中尝试该代码并告诉我。谢谢你。我不太了解 *** 的道德规范。 【参考方案1】：

您的 servlet 仅返回 N 个 JSON 对象。但是您对 JSON 数组的响应可能是错误的，请在您的 servlet 中尝试此代码

try
   Class.forName("com.mysql.jdbc.Driver").newInstance();
   Connection con=DriverManager.getConnection("jdbc:mysql://localhost:8888/atmlivedetails","root","root");  
            Statement st=con.createStatement();
            ResultSet rs=st.executeQuery("select atmbrno, atmbrname  from location_stat where act_brname='"+br_id+"'");
            int i=0;
            JSONArray jArray = new JSONArray();
            while(rs.next())

        String s = rs.getString("atmbrno");
        String t = rs.getString("atmbrname");

        JSONObject arrayObj = new JSONObject();

        arrayObj.put("atmbrno",s);
        arrayObj.put("atmbrname",t);

        jArray.add(i,arrayObj);
        i++;
        
        rs.close ();
        st.close ();
        out.print(jArray);

【讨论】：

干得好兄弟...你是最好的..它正在工作:D :D 你的代码中有一个小错误，jArray.put(i,arrayObj);它应该更正为 jArray.add(i,arrayObj);再次感谢 :D :D :D【参考方案2】：

您需要将 json-lib-2.4-jdk15.jar 文件添加到您的项目类路径中，并使用以下代码创建 JSON 对象。

    JSONObject jsonData = new JSONObject();
    jsonData.put("key1", "value1");
    jsonData.put("key2", "value2");
    jsonData.put("key3", "value3");
    System.out.println("JSON data: "+jsonData.toString());

控制台你会得到输出：JSON数据："key1":"value1","key2":"value2","key3":"value3"

【讨论】：

【参考方案3】：

写入outputstream后，需要像这样关闭Printwriter对象。