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问Android中jSONArray解析异常的问题
EN

Stack Overflow用户
提问于 2011-07-15 22:09:04
回答 1查看 2.4K关注 0票数 0

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

代码语言:javascript
复制
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

代码语言:javascript
复制
<?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:

来自Java.Lang.String的值"br“不能转换为JSONArray,这意味着什么?

我是Android和PHP的新手。所以我需要一些帮助。

EN

回答 1

Stack Overflow用户

发布于 2011-07-16 03:16:11

JSON数据必须以[开头,所以从任何字符(而不是[ )开始的任何内容都是错误的。通常,<br-bla输出是由于主机或错误的输出来自您的PHP代码或零行数据库。

您的JAVA代码很好。

尝试测试您的PHP代码,并通过如下更改查看输出。(我假设您已经设置了SQL数据库并包含表人员)

代码语言:javascript
复制
<?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输出应该如何启动和显示的示例。这是JSON数组中有一个元素的JSON

07-16 03:23:41.356: INFO/RESULT (229): {"_id":"1",“ime”:“某事”,"lat_1":"11111111","long_1":"1111111","lat_2":"1111111","long_2":"1111111"}

票数 2
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/6713688

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