我开发了一个连接到远程 mysql 数据库的小应用程序，以便从中返回信息。我基本上使用的是与 mysql 数据库位于同一服务器上的 php 文件；php 文件根据应用程序的用户响应返回数据。这是它的外观（凭据除外）：
<?php
mysql_connect("localhost" ,"my_user","my_password");
mysql_select_db('my_db');
$q=mysql_query("SELECT * FROM people WHERE birthyear>'".$_REQUEST['year']."'");
while($e=mysql_fetch_assoc($q))
$output[]=$e;
print(json_encode($output));
mysql_close();
?>
在 Android 端，代码如下所示：
package com.example.maltinololperson.asynctask;
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.AsyncTask;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.TextView;
public class ReadWebpageAsyncTask extends Activity {
private TextView textView;
InputStream is = null;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
textView = (TextView) findViewById(R.id.TextView01);
}
private class DownloadWebPageTask extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... urls) {
String result = "";
//the year data to send
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("year","1980"));
//http post
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://www.example.org/getAllPeopleBornAfter.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());
}
//convert response to string
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"UTF-8"),8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
// Log.i("log_tag","Line reads: " + line);
}
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","id: "+json_data.getInt("id")+
// ", name: "+json_data.getString("name")+
// ", sex: "+json_data.getInt("sex")+
// ", birthyear: "+json_data.getInt("birthyear")
//);
}
}
catch(JSONException e){
Log.e("log_tag", "Error parsing data "+e.toString());
}
return result;
}
@Override
protected void onPostExecute(String result) {
textView.setText(result);
}
}
public void readWebpage(View view) {
DownloadWebPageTask task = new DownloadWebPageTask();
task.execute();
}
}
一切正常，应用程序似乎完美地连接到服务器，但唯一的问题是返回的数据（在设备模拟上）采用以下形式：[{"id":"1200","name": "John smith", "Sex":"m","birthyear":"1980"},{"id":"7110","name":"Batman":,....}]，我想要看起来像这样：id：1200，姓名：john smith，性别：m，birtheyear：1980等等。由于我是 Android 开发和 JSON 的新手，我无法找到实现这一目标的方法，任何帮助将不胜感激。