【发布时间】:2017-06-29 19:29:48
【问题描述】:
我是在 Android Studio 中编程的新手,我正在尝试将我的应用程序连接到 Web 服务器上的 MySQL 数据库,echo "login success !!!!! Welcome $user"; 工作正常,正如我所希望的,但我想获得一个像 @987654322 这样的 PHP 变量@ 也回到我的 android studio 开发,以便我可以使用它在应用程序中进行会话,但我无法获取变量
我得到了回声,我在其他地方使用它来显示登录成功。需要帮助谢谢
<?php
require "pconn.php";
$username = $_POST["username"];
$password = $_POST["password"];
$username = stripslashes($username);
$password = stripslashes($password);
$hash = md5($password);
$qry = "SELECT * FROM UserLogin WHERE Email='".$username."' and BINARY Password ='".$hash."'";
$result = mysqli_query($conn ,$qry);
if(mysqli_num_rows($result) > 0) {
$row = mysqli_fetch_assoc($result);
$user = $row['Username'] ;
echo "login success !!!!! Welcome $user";
echo $user;
}
else {
echo "login not success";
}
?>
我的 Java 类是:
protected String doInBackground(String... params) {
String type = params[0];
String login_url = "http://xaas.com/login.php";
if(type.equals("login")) {
try {
String username = params[1];
String password = params[2];
URL url = new URL(login_url);
HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
httpURLConnection.setRequestMethod("POST");
httpURLConnection.setDoOutput(true);
httpURLConnection.setDoInput(true);
OutputStream outputStream = httpURLConnection.getOutputStream();
BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream, "UTF-8"));
String post_data = URLEncoder.encode("username","UTF-8")+"="+URLEncoder.encode(username,"UTF-8")+"&"
+URLEncoder.encode("password","UTF-8")+"="+URLEncoder.encode(password,"UTF-8");
bufferedWriter.write(post_data);
bufferedWriter.flush();
bufferedWriter.close();
outputStream.close();
InputStream inputStream = httpURLConnection.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream,"iso-8859-1"));
String result="";
String line="";
while((line = bufferedReader.readLine())!= null) {
result += line;
}
bufferedReader.close();
inputStream.close();
httpURLConnection.disconnect();
return result;
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (JSONException e) {
e.printStackTrace();
}
}
return null;
}
【问题讨论】:
-
How to get a php variable back to android studio?错误的问题。我认为您希望在 Android 设备上运行的客户端应用程序中接收该变量的值。当您的应用在您或其他人的设备上运行时,它与 Android Studio 几乎没有关系。 -
@greenapps 我认为您的意思是它与 Android Studio完全没有关系。