Question

我目前在带有php脚本的linux机器上安装了SSL安全服务器。

我目前有一个Android应用程序，登录页面将与我的服务器建立安全连接，并通过使用php脚本引用数据库中的详细信息来验证用户。如果详细信息正确，则会向我的应用返回特定值。

将网址对象分配到“ https ：// mydomain / myscript”时，返回的值为“null”，但是当我将网址对象分配给“ http 时：// mydomain / myscript“返回正确的值，应用程序内的身份验证可以开始。

我有 -

检查我的图书馆导入，我正在使用“javax.net.ssl.HttpsURLConnection”
检查我的安全链接“https://mydomain/myscript”是否可供开放世界访问
正确安装SSL证书并在apache服务器中全面应用

这是我的Android代码：

protected String doInBackground(String... params) {
    String login_url = "https://xxx.xxxxx.com/json/json_login.php";
    String method = params[0];
    if(method.equals("login")){
        String database_name = params[1];
        String password_user = params[2];
        try {
            URL url = new URL(login_url);
            HttpsURLConnection httpsURLConnection = (HttpsURLConnection) url.openConnection();
            httpsURLConnection.setRequestMethod("POST");
            httpsURLConnection.setDoOutput(true);
            httpsURLConnection.setDoInput(true);
            OutputStream outputStream = httpsURLConnection.getOutputStream();
            BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream, "UTF-8"));
            String data = URLEncoder.encode("database_name","UTF-8")+"="+URLEncoder.encode(database_name,"UTF-8")+"&"+
                    URLEncoder.encode("password_user","UTF-8")+"="+URLEncoder.encode(password_user,"UTF-8");
            bufferedWriter.write(data);
            bufferedWriter.flush();
            bufferedWriter.close();
            outputStream.close();
            InputStream inputStream = httpsURLConnection.getInputStream();
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream,"iso-8859-1"));
            response = "";
            String line = "";
            while((line = bufferedReader.readLine())!=null){
                response += line;
            }
            bufferedReader.close();
            inputStream.close();
            httpsURLConnection.disconnect();
            return response;

        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return null;
}

@Override
    protected void onPostExecute(String result) {
        if(result.equals("Login Failed")){
            Toast.makeText(ctx, "Database password incorrect", Toast.LENGTH_SHORT).show();
        }else{
            //Create shared preferences module called "database" and write result to shared preferences variable "database_name"
            SharedPreferences preferences = ctx.getSharedPreferences("database", Context.MODE_PRIVATE);
            SharedPreferences.Editor editor = preferences.edit();
            editor.putString("database_name", result);
            editor.apply();
            Intent intent = new Intent(ctx, MainActivity.class);
            ctx.startActivity(intent);
            ((Activity)ctx).finish();
}

这是我的PHP代码：

<?php
require "../init_accounts.php";

$database_name = $_POST["database_name"];
$password_user = $_POST["password_user"];

$sql_query = "SELECT database_name FROM xxx_Table WHERE database_name LIKE '$database_name' and password_user LIKE '$password_user';";

$result = mysqli_query($con,$sql_query);


if (mysqli_num_rows($result)>0)
{
    $row = mysqli_fetch_assoc($result);
    $database_name = $row["database_name"];
    echo "$database_name";
}
else
{
    echo "Login Failed" . mysqli_error($con);
}

?>

以下是使用HTTPS URL时的调试结果（返回值不正确）：

以下是使用HTTP URL（正确的返回值）时的调试结果：

如果有人能够在正确的方向上发出一些亮点，那就太棒了！提前致谢

2017年7月19日更新：

现在已修复此问题，我使用Trusting all certificates using HttpClient over HTTPS来帮助解决问题！

HttpsURLConnection使用POST请求方法返回空值

0 个答案: