2

我需要通过创建 JSON 对象将数据从 MYSQL 数据库传递到 android 应用程序。我只能从 MYSQL 数据库传递第一个值。如何将所有值一起传递给 android 应用程序。（我的数据库包含超过 10 个位置的纬度和经度值。）

以下是我仅将第一行值从数据库传递到 android 应用程序的代码。

        try {

        HttpClient client = new DefaultHttpClient();
        URI website = new URI("http://192.168.1.15/latlonret1.php");
        HttpGet request = new HttpGet();
        request.setURI(website);
        HttpResponse response = client.execute(request);
        in = new BufferedReader(new InputStreamReader(response
                .getEntity().getContent()));
        StringBuffer sb = new StringBuffer("");
        String l = "";
        String nl = System.getProperty("line.separator");

        while ((l = in.readLine()) != null) {
            sb.append(l + nl);
        }
        in.close();
        data = sb.toString();
        // return data;
    } catch (Exception e) {
        Log.e("log_tag", "Error converting result " + e.toString());
    }
        try {
        String returnString;
        String returnString1;
        JSONArray jArray = new JSONArray(data);
        for (int i = 0; i < jArray.length(); i++) {
            JSONObject json_data = jArray.getJSONObject(i);
            JSONObject json_data1 = jArray.getJSONObject(i);


            returnString=json_data.getString("lat") + "\n";
            returnString1=json_data1.getString("lon") + "\n";
            System.out.println(returnString);
            System.out.println(returnString1);

            Intent viewIntent =new Intent(Androidmap.this,Mapview.class);

            Bundle bundle = new Bundle();
            bundle.putString("stuff", returnString); 
            viewIntent.putExtras(bundle);

            Bundle bundle1 = new Bundle();
            bundle1.putString("stuff1", returnString1); 
            viewIntent.putExtras(bundle1); 
            startActivity(viewIntent);
      }

这是我的服务器端编码：

<?php
 ob_start();
 $host="localhost"; // Host name 
 $username=""; // Mysql username 
 $password=""; // Mysql password 
 $db_name="test"; // Database name 
 $tbl_name="manu"; // Table name 

 mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
 mysql_select_db("$db_name")or die("cannot select DB");
 $sql = "select lat , lon from manu ";
 $result=mysql_query($sql);

 if(! $result )
 {
 die('Could not get data: ' . mysql_error());
 }

 while($row=mysql_fetch_array($result , MYSQL_ASSOC)) 
{

$output[]=$row;

}

 echo json_encode($output);
 mysql_close();
?>
4

2 回答 2

2

您将需要在前一个字符串中附加下一个纬度和经度值：

ArrayList<String> arrlan=new ArrayList<String>();
ArrayList<String> arrlat=new ArrayList<String>(); 
       for (int i = 0; i < jArray.length(); i++) {
            JSONObject json_data = jArray.getJSONObject(i);
            JSONObject json_data1 = jArray.getJSONObject(i);


            returnString  =json_data.getString("lat") + "\n";
            returnString1  =json_data1.getString("lon") + "\n";

            arrlat.add(returnString);  
            arrlan.add(returnString1);   

            System.out.println(returnString);
            System.out.println(returnString1);
      }
 Intent viewIntent =new Intent(Androidmap.this,Mapview.class);


 viewIntent.putStringArrayListExtra("returnString",arrlat);
 viewIntent.putStringArrayListExtra("returnString1",arrlan);


 startActivity(viewIntent);

并在 Mapview 中将这些数组列表获取为：

ArrayList<String> arrlan=new ArrayList<String>();
ArrayList<String> arrlat=new ArrayList<String>(); 

arrlat = getIntent().getStringArrayListExtra("returnString");
arrlan = getIntent().getStringArrayListExtra("returnString1");
于 2012-12-18T05:16:13.093 回答
1

试试这个，你只需要创建一个对象。

    for (int i = 0; i < jArray.length(); i++) 
       {
        JSONObject json_data = jArray.getJSONObject(i);

        returnString=json_data.getString("lat") + "\n";
        returnString1=json_data.getString("lon") + "\n";
       }
于 2012-12-18T05:09:22.683 回答