dongqi9125 2011-07-15 22:09
浏览 197

Android中的jSONArray解析异常问题

I have some problems with JSONArray parsing, i'm using the example in Anddev.org. This is my sourcecode:

package net.json.ejemplo;

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import android.app.Activity;
import android.os.Bundle;
import android.util.Log;
import android.widget.LinearLayout;
import android.widget.TextView;


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

   TextView txt;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    // Create a crude view - this should really be set via the layout resources 
    // but since its an example saves declaring them in the XML. 
    LinearLayout rootLayout = new LinearLayout(getApplicationContext()); 
    txt = new TextView(getApplicationContext()); 
    rootLayout.addView(txt); 
    setContentView(rootLayout); 

    // Set the text and call the connect function. 
    txt.setText("Connecting...");
  //call the method to run the data retreival
    txt.setText(getServerData(KEY_121));



}
public static final String KEY_121 = "http://10.0.2.2:8888/conexion2.php"; //i use my real ip here



private String getServerData(String returnString) {
   String result ="";
   InputStream is = null;
    //the year data to send
    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("year","1970"));

    //http post
    try{
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost(KEY_121);
            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());
    }

    //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 + "
");
            }
            is.close();
            result=sb.toString();
            result.trim();
    }catch(Exception e){
            Log.e("log_tag", "Error converting result "+e.toString());
    }
    //parse json data

    try{
            //JSONObject json_data = new JSONObject(result);
            JSONArray jArreglo = new JSONArray(result);
            for(int i=0 ; i<jArreglo.length(); i++)
            {
                JSONObject json_data = jArreglo.getJSONObject(i);
                    Log.i("log_tag","id: "+json_data.getInt("id")+
                            ", name: "+json_data.getString("name")+
                            ", sex: "+json_data.getInt("sex")+
                            ", birthyear: "+json_data.getInt("birthyear")
                    );
                    //Get an output to the screen
                returnString += "
\t" + jArreglo.getJSONObject(i);
            }

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

}

and this is my php code

<?php

      mysql_connect("127.0.0.1","root","123456");

      mysql_select_db("PeopleData");

      $q=mysql_query("SELECT * FROM people WHERE birthyear>'".$_REQUEST['year']."'");

      while($e=mysql_fetch_assoc($q))

              $output[]=$e;

           print(json_encode($output));

    mysql_close();
?>

The exception is:

Value "br" from Java.Lang.String cannot be converted to JSONArray What means this?

I'm new in Android and PHP. So, i need some help.

  • 写回答

1条回答 默认 最新

  • doulei8475 2011-07-16 03:16
    关注

    JSON data must start with [ so anything you get starting with any character rather than [ is erroneous. Often <br-bla-bla output is due the host or bad output from your PHP code or zero-row database.

    Your JAVA code is fine.

    Try testing your PHP code, and see the output by changing it like this. (I assume that you have your SQL database set up and contains table people)

    <?php
    
          mysql_connect("127.0.0.1","root","123456");
    
          mysql_select_db("PeopleData");
    
          $q=mysql_query("SELECT * FROM people WHERE birthyear>'1970'");
    
          while($e=mysql_fetch_assoc($q))
    
                  $output[]=$e;
    
               print(json_encode($output));
    
        mysql_close();
    ?>
    

    This is example how your JSON output should start and look like. This is JSON with one element in the JSON array

    07-16 03:23:41.356: INFO/RESULT HTTP(229):

    [{"_id":"1","ime":"something","lat_1":"11111111","long_1":"1111111","lat_2":"1111111","long_2":"1111111"}]
    
    评论

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