如何验证android中的密码字段？

Question

您好，我是 android 新手，在我的应用中我有 Validations for Change password 页面。

这意味着密码必须包含至少 8 个字符，至少 1 个字母、1 个数字和 1 个特殊字符，为此我尝试了下面的代码，但它不起作用。

请帮帮我。

if(!isPasswordValidMethod(newPassword.getText().toString())){
           System.out.println("Not Valid");

                }else{

        System.out.println("Valid");
    }



     // Validate password
        private boolean isPasswordValidMethod(String password) {

            String yourString = newPassword.getText().toString();

            System.out.println("yourString is =" + yourString);

            boolean isValid = false;

            // ^[_A-Za-z0-9-\+]+(\.[_A-Za-z0-9-]+)*@[A-Za-z0-9-]+(\.[A-Za-z0-9]+)*(\.[A-Za-z]{2,})$
            // ^[\\w\\.-]+@([\\w\\-]+\\.)+[A-Z]{2,4}$

            String expression = "^(?=.*[A-Za-z])(?=.*\\\\d)(?=.*[$@$!%*#?&])[A-Za-z\\\\d$@$!%*#?&]{8,}$";
            CharSequence inputStr = password;

            Pattern pattern = Pattern.compile(expression, Pattern.CASE_INSENSITIVE);
            Matcher matcher = pattern.matcher(inputStr);
            if (matcher.matches()) {
                System.out.println("if");
                isValid = true;
            }else{
                System.out.println("else");
            }
            return isValid;
        }

Answer 1

尝试以下代码

 //*****************************************************************
public static boolean isValidPassword(final String password) {

    Pattern pattern;
    Matcher matcher;
    final String PASSWORD_PATTERN = "^(?=.*[0-9])(?=.*[A-Z])(?=.*[@#$%^&+=!])(?=\\S+$).{4,}$";
    pattern = Pattern.compile(PASSWORD_PATTERN);
    matcher = pattern.matcher(password);

    return matcher.matches();

}

把你的代码改成这个

   if(newPassword.getText().toString().length()<8 &&!isValidPassword(newPassword.getText().toString())){
        System.out.println("Not Valid");
      }else{
       System.out.println("Valid");
    }

Answer 2

public static boolean isValidPassword(String s) {
            Pattern PASSWORD_PATTERN
                    = Pattern.compile(
                    "[a-zA-Z0-9\\!\\@\\#\\$]{8,24}");

            return !TextUtils.isEmpty(s) && PASSWORD_PATTERN.matcher(s).matches();
        }

使用它，

if(isValidPassword(password)){ //password valid}

你可以在这里设置任何你允许的符号

Answer 3

public static boolean passwordCharValidation(String passwordEd) {
    String PASSWORD_PATTERN = "^(?=.*[A-Z])(?=.*[@_.]).*$";
    Pattern pattern = Pattern.compile(PASSWORD_PATTERN);
    Matcher matcher = pattern.matcher(passwordEd);
    if (!passwordEd.matches(".*\\d.*") || !matcher.matches()) {
        return true;
    }
    return false;
}

Answer 4

试试这个，效果很好

   public static boolean isPasswordValidMethod(final String password) {

    Pattern pattern;
    Matcher matcher;
    final String PASSWORD_PATTERN = "^(?=.*[A-Za-z])(?=.*\\\\d)(?=.*[$@$!%*#?&])[A-Za-z\\\\d$@$!%*#?&]{8,}$""
    pattern = Pattern.compile(PASSWORD_PATTERN);
    matcher = pattern.matcher(password);

    return matcher.matches();

}

Answer 5

  String validPassword = "12345";
                     _Password_String = Password.getText().toString();
                    Matcher matcher = Pattern.compile(validPassword).matcher(_Password_String);
                    if (matcher.matches()) {
                        Log.e("d11", _Password_String);
                        Toast.makeText(getActivity(), "Password Match", Toast.LENGTH_LONG).show();
                        getFragmentManager().popBackStack();


                    } else {
                        Password.setError("Password");
                        Toast.makeText(getActivity(), "Password not Match", Toast.LENGTH_LONG).show();
                    }

Answer 6

用于电子邮件字段验证和密码验证的简单代码，并检查密码字段中至少 8 个字符。

  if (isValidEmail(et_regemail.getText().toString())&&etpass1.getText().toString().length()>7){
      if (validatePassword(etpass1.getText().toString())) {
      Toast.makeText(getApplicationContext(),"Go Ahead".....
      }
      else{

       Toast.makeText(getApplicationContext(),"InvalidPassword".....
       }

}else{

 Toast.makeText(getApplicationContext(),"Invalid Email".....
}


public boolean validatePassword(final String password){
    Pattern pattern;
    Matcher matcher;
    final String PASSWORD_PATTERN = "^(?=.*[0-9])(?=.*[A-Z])(?=.* 
    [@#$%^&+=!])(?=\\S+$).{4,}$";
    pattern = Pattern.compile(PASSWORD_PATTERN);
    matcher = pattern.matcher(password);

    return matcher.matches();
}

public final static boolean isValidEmail(CharSequence target) {
    if (target == null)
        return false;

    return android.util.Patterns.EMAIL_ADDRESS.matcher(target).matches();
}

Answer 7

简单和排序，只需使用以下方法检查有效密码，

public static boolean isValidPassword(String password) {
    Matcher matcher = Pattern.compile("((?=.*[a-z])(?=.*\\d)(?=.*[A-Z])(?=.*[@#$%!]).{4,20})").matcher(password);
    return matcher.matches();
}

您可以使用以下方法

          if (!isValidPassword(edtPassword.getText().toString())) {
                errorDialog("Password must contain mix of upper and lower case letters as well as digits and one special charecter(4-20)");
                }

Answer 8

密码不匹配

if  (password.getText().toString() != reEnteredPassword.getText().toString()) {
//not match
}

对于空字段

if (password.getText().toString().isEmpty() || reEnteredPassword.getText().toString().isEmpty()) {
//Fieds empty error message

少于 8 个字符

if ((password.getText().toString().length() < 8) || (reEnteredPassword.getText().toString().length() < 8)) {
// less than 8 characters error message
}

没有特殊字符

if (!password.getText().toString().matches( "^(?=.*[0-9])(?=.*[a-z])(?=.*[!@#$%^&*+=?-]).{8,15}$") || !reEnteredPassword.getText().toString().matches( "^(?=.*[0-9])(?=.*[a-z])(?=.*[!@#$%^&*+=?-]).{8,15}$")){
//not having special characters error message
}