Question

我正在尝试从数据库中返回多行。我有两个要检索的坐标存储在我的数据库中，即经度和纬度。因为我的asynctask目前只能返回一行。我应该如何更改代码，以便能够从数据库中返回多行？

感谢您的帮助。

以下是我的代码：

php code

 $user=$_POST["username"];

  $query = "SELECT longitude,latitude FROM friends INNER JOIN coordinates ON friends.username = coordinates.username WHERE friends.friend_of='$user'";
  $sql=mysqli_query($conn, $query);

    if (!$sql) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}

    while($row = mysqli_fetch_array($sql)>0){
    $response["longitude"] = $row[0];
    $response["latitude"] = $row[1];
    die(json_encode($response));
    }

Asynctask类。

class Findfriends extends AsyncTask<String, String, JSONObject> {
    protected JSONObject doInBackground(String... args) {
        // TODO Auto-generated method stub
        // here Check for success tag
        try {
            HashMap<String, String> params = new HashMap<>();
            params.put("username", args[0]);
            JSONObject json = jsonParser.makeHttpRequest(
                    GET_FRIENDS, "POST", params);

            if (json != null) {
                Log.d("JSON result", json.toString());

                return json;
            }

        } catch (Exception e) {
            e.printStackTrace();
        }

        return null;
    }

    protected void onPostExecute(JSONObject json) {

        if (json != null) {
            Toast.makeText(Borrower_AP.this, json.toString(),
                    Toast.LENGTH_LONG).show();

            try {
                Longitude = json.getDouble("longitude");
                Latitude = json.getDouble("latitude");
            } catch (JSONException e) {
                e.printStackTrace();
            }
        }
    }
}

Answer 1

我认为你必须在这里犯错，第一个是你的PHP代码只是返回一行，而你必须使用这样的东西：

 $user=$_POST["username"];

  $query = "SELECT longitude,latitude FROM friends INNER JOIN coordinates ON friends.username = coordinates.username WHERE friends.friend_of='$user'";
  $sql=mysqli_query($conn, $query);

    if (!$sql) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}

$rows = array();
while($r = mysqli_fetch_assoc($sql)) {
    $rows[] = $r;
}
print json_encode($rows);

另一方面，在您的Android代码中，如果您使用的是Google Maps Api，则可以返回一个ArrayList，如下所示：

class Findfriends extends AsyncTask<String, String, ArrayList<LatLong> {

  protected ArrayList<LatLong> doInBackground(String... args) {
        // TODO Auto-generated method stub
        // here Check for success tag
    ArrayList<LatLong> geoPos=new ArrayList<LatLong>();
        try {
            HashMap<String, String> params = new HashMap<>();
            params.put("username", args[0]);
            JSONObject json = jsonParser.makeHttpRequest(
                    GET_FRIENDS, "POST", params);

            if (json != null) {
                Log.d("JSON result", json.toString());

                //Complete here the code in which you iterate over the json and add each point to your ArrayList<LatLong>
            }

        } catch (Exception e) {
            e.printStackTrace();
        }

        return geoPos;
    }


}

Answer 2

请改变这样的php脚本，

$query = "SELECT longitude,latitude FROM friends INNER JOIN coordinates ON friends.username = coordinates.username WHERE friends.friend_of='$user'";
$sql=mysqli_query($conn, $query);
if(mysqli_num_rows($sql)>0)
{
while($row = mysqli_fetch_array($sql)
   {
   $latlongArray =array();
   $latlongArray ["longitude"] = $row[0];
   $latlongArray ["latitude"] = $row[1];
   array_push($response["DATA"],$latlongArray) 
   }
$response["success"] = "1";
echo json_encode($response);
}
else
{ 
 $response["success"] = "0";
 echo json_encode($response);
}

在android代码中，在Post执行粘贴此代码，在logcat中，您将看到数据库中的lat / long。

 JSONArray ja = json.getJSONArray("DATA");
 for (int r = 0; r < ja.length(); r++) {
 JSONObject obj = ja.getJSONObject(r);
 Log.i("Lat / Long",obj.getString("latitude")+" , "+obj.getString("longitude"));
 }

试试这可能对你有所帮助..而且它正在运行代码。谢谢，

Answer 3

像这样创建自己的处理程序：

//import android.os.Handler; this import is required.
public class MyHandler extends Handler{

   public void handleMessage(Message msg) {
       super.handleMessage(msg);
       List<YourObject> objs=(List<YourObject>)msg.obj;     
   }

}

在您的活动中创建异步对象，如下所示：

MyDataFetcher mdf=new MyDataFetcher(new MyHandler());

像这样写你的异步：

class MyDataFetcher extends AsyncTask<String, String, JSONObject> {
    private Handler handler
    public MyDataFetcher(Handler handler){
        this.handler=handler;
    }

    //here is code you need to write in onPost
   protected void onPostExecute(JSONObject json) {
        if (json != null) {
            Toast.makeText(Borrower_AP.this, json.toString(),
                    Toast.LENGTH_LONG).show();
            try {
                List<YourObject> objs=new ArrayList<>();
                // populate your info from json into objs
                //------your code here for above logic
                //Now create new Message 
                Message msg=new Message();
                msg.obj=objs;
                handler.handleMessage(msg);
            } catch (JSONException e) {
                e.printStackTrace();
            }
        }
    }
}

使用asynctask返回多个值

3 个答案: