解析JSON数据时出现问题

Question

我已经搜索了一天的解决方案，但找不到适合我的情况的解决方案。 很抱歉，但是我是JSON（自学成才的程序员）的新手，并且我不知道我应该只发布哪些类，因此我将放入所有内容。 我从LogCat收到以下错误：

Error parsing data org.json.JSONException: Value <?xml of type java.lang.String cannot be converted to JSONArray

这是我的课：

package com.example.mytravelbuddy;

import android.os.AsyncTask;
import android.os.Bundle;
import android.app.Activity;
import android.app.ProgressDialog;
import android.util.Log;
import android.view.Menu;
import android.widget.ImageView;
import android.widget.ListAdapter;
import android.widget.ListView;
import android.widget.SimpleAdapter;
import android.widget.Toast;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;

import org.apache.http.NameValuePair;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

public class Itinerary extends Activity {

    // Progress Dialog
    private ProgressDialog pDialog;

    //JSON Parser
    JSONParser jParser = new JSONParser();

    //URL To Get Products
    public static String url = "URL REMOVED"; //Removed my url since i was hosting online

    //JSON Node names
    private static final String TAG_SUCCESS = "success";
    private static final String TAG_ITEMS = "items";
    private static final String TAG_ID = "ID";
    private static final String TAG_LOCATION = "Location";
    private static final String TAG_DESCRIPTION = "Description";
    private static final String TAG_LATITUDE = "Latitude";
    private static final String TAG_LONGITUDE = "Longitude";
    private static final String TAG_TIME = "Time";

    //Array list
    ArrayList<HashMap<String, String>> itemList;

    //Items JSONArray
    JSONArray items = null;


    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_starting_point);

        itemList = new ArrayList<HashMap<String, String>>();

        new LoadAllItems().execute();
    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        // Inflate the menu; this adds items to the action bar if it is present.
        getMenuInflater().inflate(R.menu.starting_point, menu);
        return true;
    }

    class LoadAllItems extends AsyncTask<String, String, String>{

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            pDialog = new ProgressDialog(Itinerary.this);
            pDialog.setMessage("Loading items. Please wait...");
            pDialog.setIndeterminate(false);
            pDialog.setCancelable(false);
            pDialog.show();
        }


        @Override
        protected String doInBackground(String... arg0) {

            //Building Params
            List<NameValuePair> params = new ArrayList<NameValuePair>();        

            //Getting JSON String
            JSONObject json = jParser.makeHttpRequest(url, "GET", params);

            try{
                //Getting array of items
                Log.i("Error","ERROR 1"); //This error message is displayed
                items = json.getJSONArray(TAG_ITEMS); //This is the line that is giving me a problem
                Log.i("Error","ERROR 2"); //This error message is not displayed

                //Looping through
                for(int i = 0; i < items.length();i++){

                    JSONObject c = items.getJSONObject(i);

                    //Storing JSON item in variable
                    String location = c.getString(TAG_LOCATION);
                    String description = c.getString(TAG_DESCRIPTION);
                    String longitude = c.getString(TAG_LONGITUDE);
                    String latitude = c.getString(TAG_LATITUDE);
                    String time = c.getString(TAG_TIME);


                    //Creating HashMap
                    HashMap<String, String> map = new HashMap<String, String>();

                    //Put value in hashmap map
                    map.put(TAG_LOCATION, location);
                    map.put(TAG_DESCRIPTION, description);
                    map.put(TAG_LONGITUDE, longitude);
                    map.put(TAG_LATITUDE, latitude);
                    map.put(TAG_TIME, time);

                    itemList.add(map);
                }
            }catch(JSONException e){
                e.printStackTrace();
            }

            return null;
        }}
}

通过在LogCat中显示错误消息，我发现在哪一行错误。

这是我试图从中获取响应的PHP文件：

 <?php

$user = "root";
$pass = "";
$database = "travel_buddy";
$server = "127.0.0.1";

mysql_connect($server, $user, $pass);
    $db_found = mysql_select_db($database);
    if($db_found){
        echo"DB Found<br>";
    }else{
        echo"DB NOT Found<br>";
    }
echo"Connection Established<br>";
get_details();


function get_details(){
$response = array();
$result = mysql_query("SELECT *FROM adventure");

if(mysql_num_rows($result)>0){
    $response["items"] = array();

    while($row = mysql_fetch_array($result)){
        $info = array();
        $product["ID"] = $row["ID"];
        $product["Location"] = $row["Location"];
        $product["Description"] = $row["Description"];
        $product["Latitude"] = $row["Latitude"];
        $product["Longitude"] = $row["Longitude"];
        $product["Time"] = $row["Time"];

        array_push($response["items"], $product);
    }
        $response["success"] = 1;

        echo json_encode($response);
} else {
    $response["success"] = 0;
    $response["message"] = "No Information Found";

    echo json_encode($response);
}

}

?>

这是我的JSON响应：

{
    "items": [
        {
            "ID": "1",
            "Location": "TEST",
            "Description": "TEST DESC",
            "Latitude": "1",
            "Longitude": "2",
            "Time": "3:00"
        },
        {
            "ID": "2",
            "Location": "TEST2",
            "Description": "TEST2 DESC",
            "Latitude": "1",
            "Longitude": "1",
            "Time": "7:00"
        },
        {
            "ID": "3",
            "Location": "TEST3",
            "Description": "TEST3 DESC",
            "Latitude": "3",
            "Longitude": "4",
            "Time": "12:00"
        }
    ],
    "success": 1
}

这是我的JSON解析器类：

package com.example.mytravelbuddy;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.util.List;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;

import android.util.Log;

public class JSONParser {

    static InputStream is = null;
    static JSONObject jObj = null;
    static String json = "";

    // constructor
    public JSONParser() {

    }

    // function get json from url
    // by making HTTP POST or GET mehtod
    public JSONObject makeHttpRequest(String url, String method,
            List<NameValuePair> params) {

        // Making HTTP request
        try {

            // check for request method
            if(method == "POST"){
                // request method is POST
                // defaultHttpClient
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url);
                httpPost.setEntity(new UrlEncodedFormEntity(params));

                HttpResponse httpResponse = httpClient.execute(httpPost);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();

            }else if(method == "GET"){
                // request method is GET
                DefaultHttpClient httpClient = new DefaultHttpClient();
                String paramString = URLEncodedUtils.format(params, "utf-8");
                url += "?" + paramString;
                HttpGet httpGet = new HttpGet(url);

                HttpResponse httpResponse = httpClient.execute(httpGet);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();
            }           

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            json = sb.toString();
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
        }

        // try parse the string to a JSON object
        try {
            jObj = new JSONObject(json);
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        }

        // return JSON String
        return jObj;

    }
}

任何帮助都会非常有帮助，可以解释为什么会有此错误，因此我不再重复。

Answer 1

这可能会导致问题

mysql_connect($server, $user, $pass);
    $db_found = mysql_select_db($database);
    if($db_found){
        echo"DB Found<br>";
    }else{
        echo"DB NOT Found<br>";
    }
echo"Connection Established<br>";

您的服务正在回响其他事情。 它应该只回显JSON。

---

而是可以记录此类信息

header("Content-Type: application/json");

标头也很重要：

 header("Content-Type: application/json"); 

将其放在脚本的顶部。

编辑：我只是尝试了您的代码，并且我可以解析JSON而不会出现错误。

您的PHP服务器很可能出于某种原因正在发送XML。 您需要再次检查该脚本。 甚至可能报告错误，使编码的json混乱，记录任何错误，但仅回显JSON。

希望这可以帮助。

Answer 2

php页面正在回显其他字符串，而不是json（仅）。

Answer 3

我强烈建议您使用Gson解析您的JSON响应。

对于上面的示例，该类如下所示：

public class ItemList {
    List<Item> items;
}

和Item将是：

public class Item {
    private String ID;
    private String Location;
    private String Description;
    private String Longitude;
    private String Latitude;
    private String Time;
}

然后，解析变得非常琐碎。 将gson-2.2.4.jar放入您的libs文件夹中，然后执行以下操作：

Gson GSON = new Gson();
ItemList itemList = GSON.fromJson(json_string_here, ItemList.class);

而已。 您具有从JSON解析的一流的Java对象。 简单而强大。