Question

我正在尝试将我的Android应用程序与后端MySQL连接，我使用php作为服务器端语言，但问题是因为我从php回显$res数组，所以在Android {{1}中无法访问它}。

我在其他应用程序中尝试了相同的代码并且它在那里工作得很好..但是这段代码没有显示正确的响应，也没有创建Json对象时的错误。

我的用户名在android代码的onPoseExecute(String res)中定义。

这是我的PHP代码：

OnCrate()

Android代码：

session_start();
$con = mysqli_connect("localhost", "root", "", "SWR");

//Function to sanitize values received from the form. Prevents SQL injection
function clean($str) {
    $str = @trim($str);
    if(get_magic_quotes_gpc()) {
        $str = stripslashes($str);
    }
    return $str;
}   

$empID = strtoupper(clean($_POST['username']));
$resu = array();
$resu["status"] = true;
$sql3="insert into t_booked_hall values ('','$empID','','','','','','1','','','0','','0','','','','','','','','0','','','0','','')";    
$result3 = mysqli_query($con,$sql3);    
echo json_encode($resu);

Answer 1

你必须删除休息;在while循环中

您可以使用getJSONfromURL（url）func从服务器获取阵列响应

 public static JSONObject getJSONfromURL(String url) {
    InputStream is = null;
    String result = "";
    JSONObject jArray = null;

    // Download JSON data from URL
    try {
        HttpParams httpParameters = new BasicHttpParams();
        HttpConnectionParams.setConnectionTimeout(httpParameters, 10000);
        HttpConnectionParams.setSoTimeout(httpParameters, 5000);

        HttpClient httpclient = new DefaultHttpClient(httpParameters);
        HttpPost httppost = new HttpPost(url);



        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();
        is = entity.getContent();

    } catch (ConnectTimeoutException e) {
        //Here Connection TimeOut excepion
    } catch (Exception e) {
    }

    // Convert response to string
    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "utf-8"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
        is.close();
        result = sb.toString();
    } catch (Exception e) {
    }

    try {

        jArray = new JSONObject(result);
    } catch (JSONException e) {
    }

    return jArray;
}

无法在android onPostExecute（String res）

1 个答案: