将结果集数据绑定到 Getter Setter 方法以形成 JSON 字符串

Question

我正在尝试使用 GSON 形成一个 JSON。但是我最终 JSON 形成 returns null.My 结果集包含所有可用数据，但是从结果集形成 JSON 时问题仍然存在。

任何帮助，下面是我的代码。

我的 DAO CLASS--ServiceAccess

package com.rest.bean;

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.Statement;
import java.util.ArrayList;

import org.codehaus.jackson.annotate.JsonProperty;

import com.rest.constants.Constants;


public class ServiceAccess 
{
    public ArrayList<ServiceBean> GetServices() throws Exception
    {

        ArrayList<ServiceBean> feedServices = new ArrayList<ServiceBean>();
        try
        {

            Connection con;
            Statement stmt=null;

             Class.forName(Constants.dbClass);
                 con = DriverManager.getConnection("jdbc:oracle:thin:@XX.XX.XX.XXX:152X:orcl10g","XXXX","XXXXX");
             stmt = (Statement) con.createStatement();
             String sqlQuery1="select SSCODE as SERVICECODE,SSDESC as SERVICEDESC from ksubservice order by SSCODE";
             System.out.println("printing sql query1 :"+sqlQuery1);
             ResultSet rst = stmt.executeQuery(sqlQuery1);

             while (rst.next()) 
             {


            ServiceBean ServiceObjects=new ServiceBean();
            ServiceObjects.setServiceCode(rst.getString("SERVICECODE").toString());   
            ServiceObjects.setServiceDesc(rst.getString("SERVICEDESC").toString()); 


        }
        System.out.println(feedServices);
        return feedServices;


        }
        catch(Exception e)
        {
            throw e;
        }



    }


}

我的BeanClass--ServiceBean

package com.rest.bean;

public class ServiceBean 
{

private String ServiceCode;
private String ServiceDesc;

public String getServiceCode()
{
    return ServiceCode;
}

public void setServiceCode(String ServiceCode)
{
    this.ServiceCode=ServiceCode;
}

public String getServiceDesc()
{
    return ServiceDesc; 
}

public void setServiceDesc(String ServiceDesc)
{
    this.ServiceDesc=ServiceDesc;
}

}

我的模型Class--ServiceArray

package com.rest.bean;



import java.util.ArrayList;

import com.rest.database.DBConnection;


public class ServiceArray 
{

    public ArrayList<ServiceBean> GetServices()throws Exception
    {
    ArrayList<ServiceBean> serviceData = null;
    try
    {
     ServiceAccess accessObjects=new ServiceAccess();
     serviceData=accessObjects.GetServices();

    } 
    catch (Exception e) 
    {
     throw e;
    }
    return serviceData;

    }


}

我的 JsonObject 转换器 Class

package com.rest.bean;

import java.util.ArrayList;
import com.google.gson.Gson;
import com.rest.bean.ServiceBean;

public class DataTransformer 
{

    public static String DataFeed(ArrayList<ServiceBean> feedData)
    {
    String feeds = null;
    Gson gson = new Gson();
    feeds = gson.toJson(feedData);
    return feeds;
    }



}

Answer 1

以下是我对 Gson 的最佳使用，也许它可以帮助..

对于此示例，Json 回复采用字符串网络回复的形式：

注意：回复是一个包含 json 数据的字符串

注意：Json类是一个正常的 class 实现，包含与回复中的 Json 元素相匹配的属性

JSONArray array = new HttpManager().getJsonArrayFromReply(reply);


for (int i = 0; i < array.length(); i++) {

            JSONObject jsonObject = null;
            try {
                jsonObject = (JSONObject) array.get(i);
            } catch (JSONException e) {
                return null;
            }

            Gson gson = new Gson();

            JsonClass jsonClass = gson.fromJson(jsonObject.toString(), JsonClass.class);


        }

Answer 2

Json Http 请求

String url = "http://www.myurl.com/data.php";

BufferedReader inStream = null; 
HttpClient httpClient = new DefaultHttpClient();
HttpGet httpRequest = new HttpGet(url);

try {

       HttpResponse response = httpClient.execute(httpRequest);
       inStream = new BufferedReader(
       new InputStreamReader(
       response.getEntity().getContent()));

       StringBuffer buffer = new StringBuffer("");
       String line = "";

       while ((line = inStream.readLine()) != null) {
                buffer.append(line);
            }
            inStream.close();

            reply = buffer.toString();  
        } catch (Exception e) {
                //Handle Exception
        }

如果从 url

中更正后，应该会得到 json 回复

将结果集数据绑定到 Getter Setter 方法以形成 JSON 字符串

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