JSON Android PHP如何更正错误或如何显示数据

我是JSON和PHP新手，对Android知之甚少，我测试了以下内容：

我的java代码：

ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            nameValuePairs.add(new BasicNameValuePair("QUERY", "SELECT tblWeather.humidity,tblWeather.rainfall,tblWeather.wDate,tblWeather.wTime,tblStations.IMEI, tblStations.Station, tblWeather.IMEI,tblWeather.msgID,tblWeather.tempture FROM tblStations INNER JOIN tblWeather ON tblStations.IMEI = tblWeather.IMEI WHERE tblStations.IMEI= 013226007289958 ORDER BY msgID DESC LIMIT 0,1")); 
            //Add more parameters as necessary

            //Create the HTTP request
            HttpParams httpParameters = new BasicHttpParams();

            //Setup timeouts
            HttpConnectionParams.setConnectionTimeout(httpParameters, 150000);
            HttpConnectionParams.setSoTimeout(httpParameters, 150000);          

            HttpClient httpclient = new DefaultHttpClient(httpParameters);
            HttpPost httppost = new HttpPost("http://210.14.5.179/aws/getLocal2.php");
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));        
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();

            String result = EntityUtils.toString(entity);

            // Create a JSON object from the request response
            JSONObject jsonObject = new JSONObject(result);

            //Retrieve the data from the JSON object
            Station = jsonObject.getString("Station");
            tempture = jsonObject.getString("tblWeather.tempture");

<?php
if (isset ( $_POST ["QUERY"] ) && $_POST ["QUERY"] != "") {
    /*
     * Following code will list all the products
     */
    $QUERY = $_POST ["QUERY"];

    // array for JSON response
    $response = array ();

    $con = mysqli_connect ( "localhost", "user", "pass", "db" );
    $QUERY = mysqli_real_escape_string ( $con, $QUERY );

    // get all products from products table
    $result = mysqli_query ( $con, $QUERY ) or die ( mysql_error () );

    // check for empty result
    if (mysqli_num_rows ( $result ) > 0) {
        // looping through all results
        // products node
        $response ["tblStations"] = array ();

        while ( $row = mysqli_fetch_array ( $result, MYSQL_ASSOC ) ) {
            // temp user array
            $data = array ();
            // $data ["msgID"] = $row ["msgID"];
            $data ["Station"] = $row ["Station"];
            // $product ["IMEI"] = $row ["IMEI"];
            $data ["Date"] = $row ["wDate"];
            $data ["Time"] = $row ["wTime"];
            $data ["temperature"] = $row ["tempture"];
            $data ["humidity"] = $row ["humidity"];
            $data ["rainfall"] = $row ["rainfall"];

            // push single product into final response array
            array_push ( $response ["tblStations"], $data );
        }
        // success
        $response ["success"] = 1;

        // echoing JSON response
        echo json_encode ( $response );
    } else {
        // no products found
        $response ["success"] = 0;
        $response ["message"] = "No products found";

        // echo no users JSON

        echo json_encode ( $response );
       }
       } else {
    echo "Could not complete query. Missing parameter";
      }

      ?>

我需要帮助。。谢谢

但是，当我使用

POTMAN-REST

客户端时，它将为我提供以下输出：

{"tblStations":[{"Station":"AWS01 - USeP","Date":"2014-03-19","Time":"16:15:01","temperature":"26.9","humidity":"86.4","rainfall":"1.63"}],"success":1}

有人能告诉我出了什么问题吗？

输出

{"tblStations":[{"Station":"AWS01 - USeP","Date":"2014-03-19","Time":"16:15:01","temperature":"26.9","humidity":"86.4","rainfall":"1.63"}],"success":1}

以下是如何初始化JSON解析器：

//Retrieve the data from the JSON object
JSONObject jsonObject = new JSONObject(result);

这将为您提供作为Json对象的整个字符串。从那里，取出一个单独的数组作为JsonArray，如下所示：

JSONArray jsonArray = jsonObject.getJSONArray("tblStations");

要访问每个“站和温度”，您可以使用循环逻辑：

for(int i=0;i<jsonArray.length();i++)
{
JSONObject curr = jsonArray.getJSONObject(i);

String station = curr.getString("Station");
String tempture = curr.getString("temperature");

//Do stuff with the Station and temperature String here
//Add it to a list, print it out, etc.
}

---

for（int i=0；i您的站点值被包装在tblStations数组中。（tblStations.Station）您能给我一些关于这个包装器对象的示例或教程吗？或者您对如何显示数据有什么建议吗？非常感谢..doInBackground（）中有一个错误方法。同时粘贴AsyncTask类公共类DoPOST扩展AsyncTask{Context mContext=null；String strNameToSearch=“”；//结果数据字符串站；String-Tenture；Exception-Exception=null；DoPOST（Context-Context，String-nameToSearch）{mContext=Context；strNameToSearch=nameToSearch；}@Override protected Boolean doInBackground（字符串…arg0）{try{这是异步任务class@user3315040看看我的答案，它对你有用。我有一个问题…这行：String station=jsonObject.getString（“station”）；应该是：String station=curr.getString（“station”）我试过了，先生，最管用的是这个字符串诱惑=curr.getString（“温度”）；
for(int i=0;i<jsonArray.length();i++)
{
JSONObject curr = jsonArray.getJSONObject(i);

String station = curr.getString("Station");
String tempture = curr.getString("temperature");

//Do stuff with the Station and temperature String here
//Add it to a list, print it out, etc.
}