如何修复“解析数据 org.json.JSONException 时出错：值 <？类型为 java.lang.String 的 xml 无法转换为 JSONArray”在我的程序中

Question

我对android很陌生，我正在尝试制作一个程序来显示我拥有的数据库的结果。因此，当我输入名字时，数据库就会将该人的信息发送给我。然而，当我查看 LogCat 时，它说 “09-09 22：05：39.544：错误/log_tag（8813）：解析数据org.json.JSONException时出错：值

这是我的代码：

public class PS extends Activity {
/** Called when the activity is first created. */

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    //get the two controls we created earlier, also with the resource reference and the id

    final EditText et_Text = (EditText)findViewById(R.id.et_Text);

    //add new KeyListener Callback (to record key input)
    et_Text.setOnKeyListener(new OnKeyListener()
    {
        //function to invoke when a key is pressed
        public boolean onKey(View v, int keyCode, KeyEvent event)
        {
            //check if there is 
            if (event.getAction() == KeyEvent.ACTION_DOWN)
            {
                //check if the right key was pressed
                if (keyCode == KeyEvent.KEYCODE_ENTER)
                {
                    InputStream is = null;
                    String result = "";


                    //the name data to send
                    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
                    nameValuePairs.add(new BasicNameValuePair("name",et_Text.getText().toString()));

                    //http post

                    try{
                            HttpClient httpclient = new DefaultHttpClient();
                            HttpPost httppost = new HttpPost("http://myIPaddress/sampleDB/testSend.php");
                            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                            HttpResponse response = httpclient.execute(httppost);
                            HttpEntity entity = response.getEntity();
                            is = entity.getContent();
                    }catch(Exception e){
                            Log.e("log_tag", "Error in http connection "+e.toString());
                    }

                // At this point is should be set, if it isn't, tell user what went wrong
                    if (is != null) {

                    //convert response to string
                    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();

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

                    //parse json data
                    try{
                            JSONArray jArray = new JSONArray(result);
                            for(int i=0;i<jArray.length();i++){
                                    JSONObject json_data = jArray.getJSONObject(i);
                                    Log.i("log_tag","PersonID: "+json_data.getInt("personID")+
                                            ", FirstName: "+json_data.getString("FirstName")+
                                            ", LastName: "+json_data.getString("LastName")+
                                            ", Age: "+json_data.getInt("Age")
                                    );

                    } 

                    }catch(JSONException e){
                            Log.e("log_tag", "Error parsing data "+e.toString());

                    }} else {Log.i("log_tag", "Something went wrong!"//I don't know what to put here);} ;  




                  et_Text.setText("");                            
                                    //and clear the EditText control
                return true;
                } 


                }

            return false;
        }
    }); 

}
}

这是我的 php 代码：

<?php
$con = mysql_connect("localhost","username","password");
if (!$con)
  {
   die('Could not connect: ' . mysql_error());
   }

 mysql_select_db("_usersDB", $con);

 $q=mysql_query("SELECT * FROM Persons WHERE FirstName='".$_REQUEST['name']."'");
 while($e=mysql_fetch_assoc($q))
    $output[]=$e;

 print(json_encode($output));

 mysql_close($con);
 ?>

它解析的输出是当我输入“Eric”时，它会给我 personID 1、FirstName of Eric、LastName of（我的姓氏）和年龄 15。我不确定你是否有这样的要求......

Answer 1

首先，与世界其他地方共享您的数据库连接详细信息可能并不明智。其次，在 UI 线程上进行网络操作并不是一个好主意。

最后（这是您想知道的），看起来服务器上的输出可能与客户端期望的不同。您可以发布它正在解析的输出吗？一旦您这样做，我就会修改这个答案。

Answer 2

看起来服务器好像在 JSON () 前面添加了 XML 声明。作为第一步，您可能需要检查 Web 服务器输出的 HTTP 流量（通过日志记录或wireshark），以确定问题出在客户端还是服务器上。