Question

我正在尝试使用JSON创建一个注册应用程序来连接并发布到MySQL数据库。我创建了以下if / else语句，以检查空白输入框，密码匹配和正确的电子邮件字符，然后才能将其输入数据库。即使密码不匹配，输入了无效的电子邮件字符或提交了空白文本框，代码也会继续发布JSON。即使存在无效输入，为什么还要提交JSON？

    // Register Button Click event
    btnRegister.setOnClickListener(new View.OnClickListener() {         
        public void onClick(View view) {
            String name = inputFullName.getText().toString();
            String email = inputEmail.getText().toString();
            String password = inputPassword.getText().toString();
            String check = checkpass.getText().toString();
            UserFunctions userFunction = new UserFunctions();
            JSONObject json = userFunction.registerUser(name, email, password);





        try {

            if (!inputEmail.getText().toString().matches("[a-zA-Z0-9._-]+@[a-z]+.[a-z]+") && email.length() > 0)
            {
                Toast.makeText(getApplicationContext(), "Enter Valid Email Address", Toast.LENGTH_LONG).show();
                return;
            }


            else if(name.equals("") || email.equals("")|| password.equals("")||check.equals(""))
            {
                Toast.makeText(getApplicationContext(), "Field Vaccant", Toast.LENGTH_LONG).show();
                return;
            }
            // check if both password matches
            else   if(!password.equals(checkpass))
            {
                Toast.makeText(getApplicationContext(), "Password does not match", Toast.LENGTH_LONG).show();
                return;
            }



            if  (json.getString(KEY_SUCCESS) != null) {
                registerErrorMsg.setText("");
                String res = json.getString(KEY_SUCCESS); 
                if(Integer.parseInt(res) == 1){
                    // user successfully registred
                    // Store user details in SQLite Database
                    DatabaseHandler db = new DatabaseHandler(getApplicationContext());
                    JSONObject json_user = json.getJSONObject("user");

                    // Clear all previous data in database
                    userFunction.logoutUser(getApplicationContext());
                    db.addUser(json_user.getString(KEY_NAME), json_user.getString(KEY_EMAIL), json.getString(KEY_UID), json_user.getString(KEY_CREATED_AT));                        
                    // Launch Dashboard Screen
                    Intent dashboard = new Intent(getApplicationContext(), DashboardActivity.class);
                    // Close all views before launching Dashboard
                    dashboard.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
                    startActivity(dashboard);
                    // Close Registration Screen
                    finish();
                }else{
                    // Error in registration
                    registerErrorMsg.setText("User already Registered");
                }
            }
        } 
        catch (JSONException e) {

        }
    }
});

Answer 1

这样做

    if(name != null && name.trim().length() ==0 || email != null && email.trim().length() ==0|| password != null && password.trim().length() ==0||check != null && check.trim().length() ==0)
            {
                Toast.makeText(getApplicationContext(), "Field Vaccant", Toast.LENGTH_LONG).show();
                return;
            }

Answer 2

For Empty String＆amp; Email验证这可能有助于您

Boolean bstremail = (TextUtils.isEmpty(stremail));

Boolean bemail = isEmailValid(stremail);

public isEmailValid(String email) {
    boolean isValid = false;

    String expression = "^[\\w\\.-]+@([\\w\\-]+\\.)+[A-Z]{2,4}$";
    CharSequence inputStr = email;

    Pattern pattern = Pattern.compile(expression, Pattern.CASE_INSENSITIVE);
    Matcher matcher = pattern.matcher(inputStr);
    if (matcher.matches()) {
        isValid = true;
    }       

     return isValid;

    }

现在，您可以轻松Boolean轻松管理您的验证。

Android if / else无法停止发布JSON

2 个答案: