Php 仅在SQLite android数据库中保存一行的id

我的android应用程序正在向服务器和SQlite数据库获取和写入数据

我正在使用下面的查询从服务器获取jobaddress.id=1的值。SELECTstatement中的值在android应用程序的UI中显示得非常完美，但是当我按Save时，我需要将服务器的id 1保存在本地数据库中，而不是服务器的值，而不在UI中显示id

仅用于从服务器检索数据的PHP sript查询：

$tsql = "SELECT     tbl_manufacturers.manufacturers_name, tbl_appliances_models.appliances_models_name, tbl_appliances.appliances_serial, tbl_appliances.appliances_id, jobaddress.id
        FROM        jobaddress INNER JOIN
                  tbl_appliances ON jobaddress.id = tbl_appliances.appliances_jobaddress_id LEFT OUTER JOIN
                  tbl_appliances_models INNER JOIN
                  tbl_manufacturers ON tbl_appliances_models.appliances_models_manufacturers_id = tbl_manufacturers.manufacturers_id ON 
                  tbl_appliances.appliances_models_id = tbl_appliances_models.appliances_models_id
        WHERE     (tbl_appliances.appliances_companies_id = 1) AND (jobaddress.id = 1)";

使用JSON解析器显示数据：

public void getJobAddress() 
{
    String result = null;
    InputStream isr = null;
    try
    {
    // Create a new HttpClient and Post Header
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://datanetbeta.multi-trade.co.uk/tablet/getJobAddress.php");
    HttpResponse response = httpclient.execute(httppost);
    HttpEntity entity = response.getEntity();
    isr = entity.getContent();
    }
    catch(Exception e)
    {
        Log.e("Log_tag", "Error in hhtp connection " + e.toString());

    }

    //convert Response to string
    try
    {
         BufferedReader reader = new BufferedReader(new InputStreamReader(isr,"UTF-8"), 8);
         StringBuilder sb = new StringBuilder();
         String line = null;
         while((line = reader.readLine()) != null)
         {

             sb.append(line + "\n");
         }

         isr.close();
         result = sb.toString();
    }
    catch(Exception e)
    {
        Log.e("log_tag", "Error converting result " + e.toString());
    }

try
    {

        JSONArray jArray = new JSONArray(result);

        for(int i=0; i < jArray.length(); i++)
        {
            JSONObject json = jArray.getJSONObject(0);
             s = json.getString("address1");
             t = json.getString("address2");
             u = json.getString("postcode");
        }

        tvJbAddrs1.setText(s);
        if(t == null)
        {
            tvJbAddrs2.setText("");
        }
        else
        {
            tvJbAddrs2.setText(t);
        }

        tvJbPostcode.setText(u);


    }


catch (Exception e)

{
Log.e("log_tag", "Error Parsing Data " + e.toString()); 

}

} //getJobAddress() ends

---

如何显示数据，请提供更多代码。请立即查看。我已经使用JSON解析器更新并显示了代码JSON解析器与此有什么关系？你存钱到底有什么问题？只需将ID插入DB表。