Android中的jSONArray解析异常问题

时间:2011-07-15 22:09:05

标签: php android mysql json

我在使用JSONArray解析时遇到了一些问题,我正在使用Anddev.org中的示例。这是我的源代码:

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 + "\n");
            }
            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 += "\n\t" + jArreglo.getJSONObject(i);
            }

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

}

这是我的PHP代码

<?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();
?>

例外是:

Java.Lang.String中的值“br”无法转换为JSONArray这是什么意思?

我是Android和PHP的新手。所以,我需要一些帮助。

1 个答案:

答案 0 :(得分:2)

JSON数据必须以[开头,因此从任何字符而不是[开始的任何内容都是错误的。通常<br - bla-bla输出是由PHP代码或零行数据库中的主机或错误输出引起的。

您的JAVA代码没问题。

尝试测试您的PHP代码,并通过更改它来查看输出。 (我假设您已设置SQL数据库并包含表人员)

<?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();
?>

这是您的JSON输出应该如何开始和看起来的示例。这是JSON,在JSON数组中有一个元素

  

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"}]