如何在成功注册后返回登录页面

Question

  @Override
    public void onClick(View v) {
         if(v == buttonRegister){
            registerUser();
          }
     }
private void registerUser() {
    ?*some code here*/   
}
private void register(String name, String email, String password, String phone ) {
    String urlSuffix = "?name="+name+"&email="+email+"&password="+password+"&phone="+phone;
    class RegisterUser extends AsyncTask<String, Void, String> {
        ProgressDialog loading;
        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            loading = ProgressDialog.show(SignUpActivity.this, "Please Wait",null, true, true);
        }
        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            loading.dismiss();
            Toast.makeText(getApplicationContext(), s, Toast.LENGTH_SHORT).show();


        }

        @Override
        protected String doInBackground(String... params) {
            String s = params[0];
            BufferedReader bufferedReader = null;
            try {
                URL url = new URL(REGISTER_URL+s);
                HttpURLConnection con = (HttpURLConnection) url.openConnection();
                bufferedReader = new BufferedReader(new InputStreamReader(con.getInputStream()));

                String result;

                result = bufferedReader.readLine();

                return result;
            }catch(Exception e){
                return null;
            }
        }
    }

    RegisterUser ru = new RegisterUser();
    ru.execute(urlSuffix);
}

Answer 1

If you moved from login activity to this screen then you can use:

 @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            loading.dismiss();
            Toast.makeText(getApplicationContext(), s, Toast.LENGTH_SHORT).show();
            this.finish();


        }


Otherwise you can start new activity like :

@Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            loading.dismiss();
            Toast.makeText(getApplicationContext(), s, Toast.LENGTH_SHORT).show();
            startActivity(new Intent(SignUpActivity.this,LoginActivity.class));
            this.finish();

        }

Answer 2

这完全取决于你的REST API和它返回的内容，如果你能从API中发布json结果，这将会有很大的帮助。如果它返回的消息类似于“电子邮件已经存在”，那么您可以检查消息中是否存在该字符串，并执行相应的操作

将您的onPostExecute方法更改为：

@Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            loading.dismiss();
            Toast.makeText(getApplicationContext(), s, Toast.LENGTH_SHORT).show();
            if(s.has("email already exists")){
                //Show some error
            }else if(s.has("Registration successfull"))
            startActivity(new Intent(SignUpActivity.this,LoginActivity.class));
            this.finish();

        }

Answer 3

 if($_SERVER['REQUEST_METHOD']=='GET'){
 $name = $_GET['name'];
 $password = $_GET['password'];
 $email = $_GET['email'];
 $phone = $_GET['phone'];

 if($name == '' || $password == '' || $email == '' ||  $phone == '')
   {
     echo 'please fill all values';
       }
         else{
        $check =$db->query("SELECT * FROM register WHERE email='".$_GET['email']."'"; 
   $row=$check->num_rows;
        if($row > 0)
 {
               echo 'email already exist';
             }
       else
      { 
             $insert = $db->query("INSERT INTO register(`name`,`password`,`email`,`phone`) VALUES('".$_GET['name']."','".$_GET['password']."','".$_GET['email']."','".$_GET['phone']."')";
             if($insert)
    {
  echo 'successfully registered';

} else{ echo 'oops！请重试！‘；} else { echo "Error"；}