Question

我希望以Json的格式发送在我的数据库上创建的查询结果。查询返回多个记录，我希望用Json发送所有的记录。我的查询包含字段即。 Restaurant_name，Address，type of food。等等那我该怎么办呢？这是我不完整的代码。

<?php


require_once 'DB_connect.php';
class populatelist{
private $con;
private $conn;

function __construct()
{
$this->con = new DB_connect();
$this->conn = $this->con->connectWithRestaurant();
}
function selectallfields(){
$query = "SELECT * FROM `restaurant_time` LIMIT 50";
$query_exec = $this->conn->query($query);
if($query_exec->num_rows >0){
  $queryresult = $query_exec->fetch_assoc();


}
}
}

?>

我这样做的主要目的是在AndroidListView中检索这些数据。在List中填充它。这是我在android中的代码片段。

class Jsonfetch extends AsyncTask<String,Void,Void>{

    @Override
    protected Void doInBackground(String... params) {
        try {
            URL url = new URL("http://172.16.16.88/orderspot/populatelist.php");
            HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
            httpURLConnection.setRequestMethod("POST");
            httpURLConnection.setAllowUserInteraction(false);
            InputStream inputStream = httpURLConnection.getInputStream();
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
            String temp;
            StringBuilder stringBuilder = new StringBuilder();
            while((temp = bufferedReader.readLine())!=null){
                stringBuilder.append(temp);
            }
            String JsonResponse = stringBuilder.toString();
            try {
               JSONArray new_array =new JSONArray(JsonResponse);
                //int count;
                for(int i = 0, count = new_array.length() ; i < count ; i++){
                    JSONObject jsonObject  = new_array.getJSONObject(i);
                    Resname.add(jsonObject.getString("restname"));
                    Restaddress.add(jsonObject.getString("restadd"));
                    Resttime.add(jsonObject.getString("restime"));
                    images.add(jsonObject.getInt("images"));
                }
            } catch (JSONException e) {
                e.printStackTrace();
            }
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }


        return null;
    }
}

Answer 1

最好像这样使用发送查询结果

 $query['query'] = $yourQueryResult;
 echo json_encode($query);

Answer 2

希望它也能帮到你。

echo json_encode($queryresult);

参考：http://php.net/manual/en/function.json-encode.php

在Json

2 个答案: