Android:从php网页获取空字符串


Android: Getting Empty String From Php webpage

我有两个应用程序,一个我发送长。 和纬度坐标到 PHP 文件,另一个应用程序检索长。 和纬度坐标。为了测试并查看我是否可以获得第一个工作,我创建了一个函数,我发布了纬度和长度。协调两个php服务,我将它们带回同一个应用程序中。我把它们放在吐司中,看看它是否有效。我什至实现了位置侦听器来上传坐标并在同一应用程序中检索它们以进行测试,然后再尝试在其他应用程序中接收它们。它工作正常。但是当我尝试在其他应用程序中使用相同的代码来接收坐标时,我会收到空白坐标。我调试了它,它只是空白的,就好像当我从另一个应用程序调用服务器时,它会擦除 php 服务中的当前值。

用于在应用程序 1 中放置坐标的代码:

public void postData(String longCord, String latCord) throws JSONException{  
    // Create a new HttpClient and Post Header
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost(".../android/serverFile.php");
    JSONObject json = new JSONObject();
    try {
        // JSON data:
        json.put("longitude", longCord);
        json.put("latitude", latCord);
        JSONArray postjson=new JSONArray();
        postjson.put(json);
        // Post the data:
        httppost.setHeader("json",json.toString());
        httppost.getParams().setParameter("jsonpost",postjson);
        // Execute HTTP Post Request
        //System.out.print(json);
        HttpResponse response = httpclient.execute(httppost);
        // for JSON:
        if(response != null)
        {
            InputStream is = response.getEntity().getContent();
            BufferedReader reader = new BufferedReader(new InputStreamReader(is));
            StringBuilder sb = new StringBuilder();
            String line = null;
            try {
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "'n");
                }
            } catch (IOException e) {
                e.printStackTrace();
            } finally {
                try {
                    is.close();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
                                    JSONObject jsonObj = new JSONObject(jsonStr);
                // grabbing the menu object 
                String longitudecord = jsonObj.getString("lon");
                String latitudecord = jsonObj.getString("lat");
                Toast.makeText( getApplicationContext(),longitudecord,Toast.LENGTH_SHORT).show();
                Toast.makeText( getApplicationContext(),latitudecord,Toast.LENGTH_SHORT).show();
        }
    }catch (ClientProtocolException e) {
        // TODO Auto-generated catch block
    } catch (IOException e) {
        // TODO Auto-generated catch block
    }
}

PHP文件:

<?php
$json = $_SERVER['HTTP_JSON'];
$data = json_decode($json);
$lon = $data->longitude;
$lat = $data->latitude;
$variable = array( 'lon' => "$lon", 'lat' => "$lat" );
// One JSON for both variables
echo json_encode($variable);

?>

现在,当我在其他应用程序上运行此代码时...它与上面相同,减去发布坐标...我得到 lon:" 和 lat:"。有点像通过发出请求,它以某种方式删除了其他应用程序发布的信息。是这样吗?

 public void recieveData() throws JSONException{  
        // Create a new HttpClient and Post Header
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(".../android/serverFile.php");
        try {
            HttpResponse response = httpclient.execute(httppost);
            // for JSON:
            if(response != null)
            {
                InputStream is = response.getEntity().getContent();
                BufferedReader reader = new BufferedReader(new InputStreamReader(is));
                StringBuilder sb = new StringBuilder();
                String line = null;
                try {
                    while ((line = reader.readLine()) != null) {
                        sb.append(line + "'n");
                    }
                } catch (IOException e) {
                    e.printStackTrace();
                } finally {
                    try {
                        is.close();
                    } catch (IOException e) {
                        e.printStackTrace();
                    }
                }
                String jsonStr = sb.toString();
                JSONObject jsonObj = new JSONObject(jsonStr);
                // grabbing the menu object 
                String longitudecord = jsonObj.getString("lon");
                String latitudecord = jsonObj.getString("lat");
                Toast.makeText( getApplicationContext(),longitudecord,Toast.LENGTH_SHORT).show();
                Toast.makeText( getApplicationContext(),jsonStr,Toast.LENGTH_SHORT).show();
            }

尝试在浏览器中访问您的 URL,并确保您的 PHP 脚本在以下行中没有给出任何错误(因为您在第二个应用程序中没有提供任何 JSON 输入(:

$data = json_decode($json);
$lon = $data->longitude;
$lat = $data->latitude;
另外,很高兴

听到,您如何在服务器上保存纬度/纬度,以便它们在调用之间持续存在。

例如,您可以使用以下 PHP 代码:

$db_host = "localhost";
$db_name = "********";
$db_user = "********";
$db_pass = "********";
// connect to MySQL
$db = mysql_connect($db_host, $db_user, $db_pass);
if( !$db ) {
    die( 'Unable to connect to server : ' . mysql_error() );
}
// select the DB
if( !mysql_select_db($db_name) ) {
    die( 'Unable to select DB : ' . mysql_error() );
}
// create table, if not exists
if( !mysql_query( "CREATE TABLE IF NOT EXISTS `locations` (
    `id` int(11) NOT NULL auto_increment,
    `latitude` double NOT NULL COMMENT 'latitude',
    `longitude` double NOT NULL COMMENT 'longitude',
    PRIMARY KEY  (`id`)
    ) ENGINE=MyISAM  DEFAULT CHARSET=utf8;"
) ) {
    die( 'Unable to create table : ' . mysql_error() );
}
// save values
mysql_query( "INSERT INTO locations (`latitude`, `longitude`) VALUES ( '$lat', '$lon')" );
// read them back
$track = array()
$result = mysql_query( "SELECT * FROM locations" );
while( $data = mysql_fetch_assoc($result) ) {
    $track[] = array(
        'latitude' => $data['latitude'],
        'longitude' => $data['longitude'] );
}
// here you may convert `track` into JSON / XML and send coordinates list to your application