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我将我的 android 连接到 MySQL 数据库，它工作正常。我使用 php 文件来连接数据库并进行查询。现在我想显示来自同一个数据库的不同表的数据。我想我必须调用函数 read 但我不知道如何删除数据。也许使用 Web 服务更容易或更好？这是我在另一篇文章中使用的代码

package com.example.onlinetestactivity;

import java.io.IOException;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.util.EntityUtils;
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.widget.TextView;


public class MainActivity extends Activity {
    /** Called when the activity is first created. */
    TextView resultView;
    HttpClient client;
    JSONObject json;
    String Dat;
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        resultView = (TextView) findViewById(R.id.tvjson);
        client = new DefaultHttpClient();
        try {
            json = RedData();
        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

      Dat = json.toString();

        new Read().onPostExecute(Dat);
    }
    public JSONObject RedData() throws ClientProtocolException, IOException, JSONException {

        HttpPost httppost = new HttpPost("http://file.php");     
        HttpResponse r = client.execute(httppost);

        HttpEntity e = r.getEntity();
        String data = EntityUtils.toString(e);
        JSONArray jArray = new JSONArray(data);
        JSONObject last = jArray.getJSONObject(0);
        return last;


    }


    public class Read extends AsyncTask<String, Integer, String> {

        @Override
        protected String doInBackground(String... arg0) {
            // TODO Auto-generated method stub
            try {
                json = RedData();
                //Toast.makeText(OnlineTestActivity.this, json.getString(arg0[0]), Toast.LENGTH_LONG);
                return json.getString(arg0[0]);

            } catch (ClientProtocolException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            return null;
        }

        @Override
        protected void onPostExecute(String data) {
            // TODO Auto-generated method stub
            resultView.setText(data);
        }
    }
}

这是php文件

<?php
header('Content-type: application/json');

$server = "XXXXXXXX";
$username = "XXXXXXX";
$password = "XXXXXXXX";
$database = "XXXXXXX";

$con = mysql_connect($server, $username, $password) or die ("No se puede conectar: " . mysql_error());

$db_selected = mysql_select_db($database, $con);
if (!$db_selected)
{
  die ("Error al seleccionar la Base de datos: " . mysql_error());
}

$sql = "select XXXXXXXXXXX)";
$result = mysql_query($sql) or die ("Query error: " . mysql_error());

$records = array();

while($row = mysql_fetch_assoc($result)) {
    $records[] = $row;
}

mysql_close($con);

print(json_encode($records));
?>

谢谢大家

4

2 回答 2

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查看链接。AsyncTask 可以忽略，如示例中所示。

于 2012-08-16T10:10:19.407 回答
0

问题出在 PHP 文件中：

print(json_encode($records));

必须先去：

mysql_close($con);

JSON 无法理解，它失败了，但更改顺序可以正常工作

于 2012-08-29T10:58:46.960 回答