如何将用户信息从android studio注册到000webhost.com数据库（php）？

我正在尝试从android studio向000webhost数据库phpMyAdmin插入用户信息。我的数据库已连接，但我无法在数据库中发布任何内容。有人能帮我吗

    registerbutton.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            register();
});
}

public void register() {

    FirstName = firstname.getText().toString();
    LastName = lastname.getText().toString();
    Email = email.getText().toString();
    Password = pass.getText().toString();
    Password2 = pass2.getText().toString();
    BackGround bg = new BackGround();
    bg.execute(FirstName, LastName, Email, Password, Password2);


}

class BackGround extends AsyncTask<String, String, String> {

    @Override
    protected String doInBackground(String... params) {

        String firstname = params[0];
        String lastname = params[1];
        String email = params[2];
        String pass = params[3];
        String pass2 = params[5];
        String data = "";
        int temp;

        try {
            URL url = new URL("https://lovethesnipers.000webhostapp.com/register.php");
            String urlParams = "firstname=" + firstname + "&lastname=" + lastname +
                    "&email=" + email + "&pass=" + pass + "&pass2=" + pass2;

            HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
            httpURLConnection.setDoOutput(true);

            OutputStream os = httpURLConnection.getOutputStream();
            os.write(urlParams.getBytes());
            os.flush();
            os.close();

            InputStream is = httpURLConnection.getInputStream();
            while ((temp = is.read()) != -1) {
                data += (char) temp;

            }
            is.close();
            httpURLConnection.disconnect();

            return data;

        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
            return "Exception: " + e.getMessage();

        }
        return null;
    }

    @Override
    protected void onPostExecute(String s) {
        if (s.equals("")) {
            s = "Register Successful.";
        }
        Toast.makeText(AppRegisterActivity.this, s, Toast.LENGTH_LONG).show();
    }
}

}

这是服务器端的register.php

<?php
error_reporting(0);
require "init.php";

$firstname = $_POST["firstname"];
$lastname = $_POST["lastname"];
$email = $_POST["email"];
$pass = $_POST["pass"];
$pass2 = $_POST["pass2"];
//$statement = mysqli_prepare($con, "INSERT INTO users_info (id,firstname,  
  lastname, email, pass, pass2) VALUES (null,?, ?, ?, ?,?)");

$sql = "INSERT INTO users_info(id, firstname, lastname, email, pass, pass2)  VALUES (?, ?, ?,?,?)";

mysqli_stmt_bind_param($sql, "siss", $firstname, $lastname, $email, $pass, $pass2);
mysqli_stmt_execute($sql);

if(!mysqli_query($conn, $sql)){
echo '{"Unable to save the data to the database"}';

}


?>

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我的数据库名是table name是user_info，字段是id、firstname、lastname、email、pass和pass2。它是连接的，但每次我点击register按钮，它都不会将值传递给数据库。提前谢谢你的帮助

如果要在url中发送数据，请使用$\u GET方法而不是$\u POST。

---

所以只要把$\u POST改成$\u GET。

000webhost你说的是免费的000webhost主机吗？由phpMyAdmin？？你是说mysql吗？嗨，Matas Lesinskas，是的，我说的是免费网络主机000webhost。我想使用phpMyAdmin数据库。有什么想法吗？试着像这样引用pass，如果您使用绑定等，也可以使用with prepare等。就这样做吧，我不这样使用sintax，因为它很难阅读，所以我不能直接指出错误。