Question

我的应用程序是一个简单的定位系统，每个用户位置都更新到服务器上，当需要帮助时，它会发回给定范围内所有人的移动号码（大约）。该应用程序可以定期向服务器定期更新位置，并使用近似方法查找某个范围内的人，我现在面临的问题是将查询的数字列表发送回设备。我已经尝试了一些教程，但没有运气，应用程序一直在崩溃。这是我的代码，我发送数据和PHP代码。如果有人帮我弄清楚如何从服务器接收数字。

private void serverConnection() {
    // TODO Auto-generated method stub


    GPSTracker gps1 = new GPSTracker(MainActivity.this);
    double latitude = gps1.getLatitude();
    double longitude = gps1.getLongtitude();
    Toast.makeText(getApplicationContext(),"Your Location is -\nLat:"+latitude+"\nLon:"+longitude,Toast.LENGTH_LONG).show();
    String lat = String.valueOf(latitude);
    String log = String.valueOf(longitude);

    String svdNum;

    RegistrationActivity gsm = new RegistrationActivity();
    SharedPreferences sharedData = getSharedPreferences(gsm.filename,0);
    svdNum = sharedData.getString("Mobile Number", "No Number Registered");


    List<NameValuePair>  nameValuePairs = new ArrayList<NameValuePair>(1);

    nameValuePairs.add(new BasicNameValuePair("Lat",lat));
    nameValuePairs.add(new BasicNameValuePair("Long",log));
    nameValuePairs.add(new BasicNameValuePair("Number",svdNum));

    try{
        if(distress == 0){
            HttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost("http://10.0.2.2/tech/serverConnection.php");
            httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpClient.execute(httpPost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();
        }else{
            HttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost("http://10.0.2.2/tech/serverConnection2.php");
            httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpClient.execute(httpPost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();

        }
    }
    catch(ClientProtocolException e){
        Log.e("ClientProtocol", "Log_tag");
        e.printStackTrace();
    }
    catch(IOException e){
        Log.e("Log_tag", "IOException");
        e.printStackTrace();
    }

php代码

<?php
$conn = mysql_connect(localhost, 'root', '');
mysql_select_db('finalyearproject');
if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}

$Number=$_POST['Number'];
$LatitudeS=$_POST['Lat'];
$LongitudeS=$_POST['Long'];

$Latitude = floatval($LatitudeS);
$Longitude = floatval($LongitudeS);

$sql = "INSERT INTO app_backup(Number,Latitude, Longitude) VALUES('$Number', '$Latitude', '$Longitude') ON DUPLICATE KEY UPDATE Latitude=VALUES(Latitude), Longitude=VALUES(Longitude)";


$sql = "UPDATE app_backup SET Lat_diff=Latitude-$Latitude ,Long_diff=Longitude-$Longitude";

$query= mysql_query("SELECT Number FROM app_backup WHERE Lat_diff <= 30 AND Long_diff <=30 AND Long_diff <> 0 AND Lat_diff <> 0 AND Long_diff >= -30 AND Lat_diff >= -30"); 


$column = array();

while ( $row = mysql_fetch_array($query, MYSQL_ASSOC) ) {
  $column[] = $row['Number'];
}

echo json_encode(column);


$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($conn);

?>

先谢谢你。

编辑：应用程序一直崩溃的代码块

HttpEntity entity2 = response.getEntity();
                InputStream nums = entity2.getContent();
                try{
                BufferedReader reader = new BufferedReader(new InputStreamReader(nums,"iso-8859-1") );
                StringBuilder sb = new StringBuilder();
                String line = null;
                while ((line = reader.readLine()) != null){
                    sb.append(line + "\n");
                     }
            nums.close();
            res = sb.toString(); 
            }
            catch(IOException e){
                Log.e("Log_tag", "IOException");
                e.printStackTrace();
            }

Json循环

try{
    JSONArray jarray = new JSONArray(res); 
    JSONObject json_data = null;
    for(int i=0; i < jarray.length(); i++)
    {
        json_data = jarray.getJSONObject(i);
    }}
catch(JSONException e){
    Log.e("error parsing data", "Log_tag");
    e.printStackTrace();}

添加这两个块之后，logcat说'关闭VM'并且应用程序崩溃了。我不太确定我的方法在检索方面是否正确。

Answer 1

要使用JSON将数据发送到您的应用，您必须发送标头，最后使用json_encode发送JSON。另一个缺陷是PHP文件的编码（必须是没有BOM的UTF-8），否则会破坏json。

header('Cache-Control: no-cache, must-revalidate');
header('Expires: Mon, 26 Jul 1997 05:00:00 GMT');
header('Content-type: application/json');//*/
echo json_encode(column);

但是为了完成这项工作，有一个轻型图书馆Simple JSON for PHP它使你能够'＃34;锻造＆＃34;您可以随心所欲地使用JSON并将其发送（包括标题）。

在你的情况下，它会像：

// At the beginning
$Json = new json();

// Add the data
while ( $row = mysql_fetch_array($query, MYSQL_ASSOC) ) {
  $Json->addContent(new arrayJson("data",$row));
}
$Json->addContent(new propertyJson('message', 'Success'));

// At the end, send the json
json_send($Json)

在JAVA部分：

要获取JSON，您应该制作类似here

的方法

public String getJSON(String url, int timeout) {
    HttpURLConnection c = null;
    try {
        URL u = new URL(url);
        c = (HttpURLConnection) u.openConnection();
        c.setRequestMethod("GET");
        c.setRequestProperty("Content-length", "0");
        c.setUseCaches(false);
        c.setAllowUserInteraction(false);
        c.setConnectTimeout(timeout);
        c.setReadTimeout(timeout);
        c.connect();
        int status = c.getResponseCode();

        switch (status) {
            case 200:
            case 201:
                BufferedReader br = new BufferedReader(new InputStreamReader(c.getInputStream()));
                StringBuilder sb = new StringBuilder();
                String line;
                while ((line = br.readLine()) != null) {
                    sb.append(line+"\n");
                }
                br.close();
                return sb.toString();
        }

    } catch (MalformedURLException ex) {
        Logger.getLogger(getClass().getName()).log(Level.SEVERE, null, ex);
    } catch (IOException ex) {
        Logger.getLogger(getClass().getName()).log(Level.SEVERE, null, ex);
    } finally {
       if (c != null) {
          try {
              c.disconnect();
          } catch (Exception ex) {
             Logger.getLogger(getClass().getName()).log(Level.SEVERE, null, ex);
          }
       }
    }
    return null;
}

您应该使用Google GSON将JSON转换为数据Java对象，如StackOverflow所示。

import java.util.List;
import com.google.gson.Gson;

public class Test {

    public static void main(String... args) throws Exception {
        String json = 
            "{"
                + "'title': 'Computing and Information systems',"
                + "'id' : 1,"
                + "'children' : 'true',"
                + "'groups' : [{"
                    + "'title' : 'Level one CIS',"
                    + "'id' : 2,"
                    + "'children' : 'true',"
                    + "'groups' : [{"
                        + "'title' : 'Intro To Computing and Internet',"
                        + "'id' : 3,"
                        + "'children': 'false',"
                        + "'groups':[]"
                    + "}]" 
                + "}]"
            + "}";

        // Now do the magic.
        Data data = new Gson().fromJson(json, Data.class);

        // Show it.
        System.out.println(data);
    }

}

class Data {
    private String title;
    private Long id;
    private Boolean children;
    private List<Data> groups;

    public String getTitle() { return title; }
    public Long getId() { return id; }
    public Boolean getChildren() { return children; }
    public List<Data> getGroups() { return groups; }

    public void setTitle(String title) { this.title = title; }
    public void setId(Long id) { this.id = id; }
    public void setChildren(Boolean children) { this.children = children; }
    public void setGroups(List<Data> groups) { this.groups = groups; }

    public String toString() {
        return String.format("title:%s,id:%d,children:%s,groups:%s", title, id, children, groups);
    }
}

希望有所帮助

从我的数据库android中检索数据？

1 个答案: