Json 解析错误：Java.lang.String 类型的值 <?xml 无法转换为 JSONArray [重复]

Posted 2023-04-19

【中文标题】Json 解析错误：Java.lang.String 类型的值 <?xml 无法转换为 JSONArray [重复]

【英文标题】：Json parsing error: Value <?xml of type java.lang.String cannot be converted to JSONArray [duplicate]

【发布时间】：2018-01-08 15:58:35

【问题描述】：

我是编程新手，我有一个小问题，我正在尝试制作一个使用 json 来读取 flickr api 的应用程序，但我在环顾四周的描述中收到错误提及，但没有一个解决方案有效，所以我过来问

public class JSONPareser 
    final String TAG = "JSONParser";

    public String makeServiceCall(String reqUrl) 
        String response = null;
        try 
            URL url = new URL(reqUrl);
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setRequestMethod("GET");
            // read the response
            InputStream in = new BufferedInputStream(conn.getInputStream());
            response = convertStreamToString(in);
         catch (MalformedURLException e) 
            Log.e(TAG, "MalformedURLException: " + e.getMessage());
         catch (ProtocolException e) 
            Log.e(TAG, "ProtocolException: " + e.getMessage());
         catch (IOException e) 
            Log.e(TAG, "IOException: " + e.getMessage());
         catch (Exception e) 
            Log.e(TAG, "Exception: " + e.getMessage());
        
        return response;
    

    private String convertStreamToString(InputStream is) throws UnsupportedEncodingException 
        BufferedReader reader = new BufferedReader(new InputStreamReader(is,"UTF-8"),8);
        StringBuilder sb = new StringBuilder();

        String line;
        try 
            while ((line = reader.readLine()) != null) 
                sb.append(line).append('\n');
            
         catch (IOException e) 
            e.printStackTrace();
         finally 
            try 
                is.close();
             catch (IOException e) 
                e.printStackTrace();
            
        
        return sb.toString();
    

这是我的解析器类，这是我调用 json 对象的时候

public class getData extends AsyncTask<Void, Void, Void> 
    @Override
    protected void onPreExecute() 
        super.onPreExecute();
        dialog = new ProgressDialog(MainActivity.this);
        dialog.setMessage("Please wait");
        dialog.setCancelable(false);
        dialog.show();

    

    @Override
    protected Void doInBackground(Void... params) 
        String url = "https://api.flickr.com/services/feeds/photos_public.gne";
        JSONPareser pareser = new JSONPareser();
        String jsonStr = pareser.makeServiceCall(url);
        if (jsonStr != null) 
            try 

                JSONArray jsonArray = new JSONArray(jsonStr);
                for (int i = 0; i < jsonArray.length(); i++) 
                    JSONObject obj = jsonArray.getJSONObject(i);

                    FlickrModel model = new FlickrModel();
                    model.setLink(obj.getString("link"));
                    model.setDescription(obj.getString("description"));
                    model.setTitle(obj.getString("title"));
                    model.setAuthor(obj.getString("author"));
                    model.setTags(obj.getString("tags"));
                    flickrList.add(model);

                
             catch (final JSONException e) 
                Log.e(TAG, "Json parsing error: " + e.getMessage());
                runOnUiThread(new Runnable() 
                    @Override
                    public void run() 
                        Toast.makeText(getApplicationContext(),
                                "Json parsing error: " + e.getMessage(),
                                Toast.LENGTH_LONG)
                                .show();
                    
                );
            
         else 
            Log.e(TAG, "Couldn't get json from server.");
            runOnUiThread(new Runnable() 
                @Override
                public void run() 
                    Toast.makeText(getApplicationContext(),
                            "Couldn't get json from server. Check LogCat for possible errors!",
                            Toast.LENGTH_LONG)
                            .show();
                
            );

        

        return null;
    

    @Override
    protected void onPostExecute(Void result) 
        super.onPostExecute(result);
        if (dialog.isShowing()) 
            dialog.dismiss();
        

非常感谢你

【问题讨论】：

给你一个提示，你不需要来自 onPostExecute 方法的 runOnUiThread；

您能否粘贴响应，因为它显示 xml 响应即将到来。我认为 url 架构不正确或服务器未启动，因此发送 xml 而不是 json 作为响应... JsonParser 的最低要求要解析的是响应字符串必须是 json 格式。

【参考方案1】：

您的服务似乎返回的是 XML 而不是 JSON，请参阅标题