How to handle TextField validation in password in Flutter

Question

I created a login page and I need to add these things to my password. How do I do it with validation alert message?

• Minimum 1 Upper case

Answer 1

You have to use TextFormField widget with validator property.

TextFormField(
   validator: (value) {
      // add your custom validation here.
      if (value.isEmpty) {
        return 'Please enter some text';
      }
      if (value.length < 3) {
        return 'Must be more than 2 charater';
      }
   },
),

Take a look on official docs: https://flutter.dev/docs/cookbook/forms/validation

Answer 2

here is the complete answer

Write a Dart program to check whether a string is a valid password. a. A password must have at least ten characters. b. A password consists of only letters and digits. c. A password must contain at least two digits.

import 'dart:io';

main() {
  var password;
  stdout.write("Enter You'r Password: ");
  password=stdin.readLineSync();

   if(password.length>=10 && !password.contains(RegExp(r'\W')) && RegExp(r'\d+\w*\d+').hasMatch(password))
   {
     print(" \n\t$password is Valid Password");
   }
   else
   {
     print("\n\t$password is Invalid Password");

   }

Answer 3

Your regular expression should look like:

r'^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[!@#\$&*~]).{8,}$

Here is an explanation:

r'^
  (?=.*[A-Z])       // should contain at least one upper case
  (?=.*[a-z])       // should contain at least one lower case
  (?=.*?[0-9])          // should contain at least one digit
 (?=.*?[!@#\$&*~]).{8,}  // should contain at least one Special character
$

Match above expression with your password string.Using this method-

String validatePassword(String value) {
    Pattern pattern =
        r'^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[!@#\$&*~]).{8,}$';
    RegExp regex = new RegExp(pattern);
    print(value);
    if (value.isEmpty) {
      return 'Please enter password';
    } else {
      if (!regex.hasMatch(value))
        return 'Enter valid password';
      else
        return null;
    }
  }

Answer 4

You need to use Regular Expression to validate the structure.

 bool validateStructure(String value){
        String  pattern = r'^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[!@#\$&*~]).{8,}$';
        RegExp regExp = new RegExp(pattern);
        return regExp.hasMatch(value);
  }

output: 

    Vignesh123! : true
    vignesh123 : false
    VIGNESH123! : false
    vignesh@ : false
    12345678? : false

This function will validate the passed value is having the structure or not.

    var _usernameController = TextEditingController();
    String _usernameError;

    ...

    @override
    Widget build(BuildContext context) {
        return
        ...
        TextFormField(
          controller: _usernameController,
          decoration: InputDecoration(
              hintText: "Username", errorText: _usernameError),
          style: TextStyle(fontSize: 18.0),
        ),
        Container(
          width: double.infinity,
          height: 50.0,
          child: RaisedButton(
            onPressed: validate,
            child: Text(
              "Login",
              style: TextStyle(color: Colors.white),
            ),
            color: Theme.of(context).primaryColor,
            shape: RoundedRectangleBorder(
              borderRadius: BorderRadius.circular(50.0),
            ),
          ),
        ),
        ...
    }

    ...

    validate(){
        if(!validateStructure(_usernameController.text)){
            setState(() {
                _usernameError = emailError;
                _passwordError = passwordError;
            });
            // show dialog/snackbar to get user attention.
            return;
        }
        // Continue 
    }

Answer 5

You can achieve this using below flutter plugin.

wc_form_validators

You can use it something like this:

          TextFormField(
            decoration: InputDecoration(
              labelText: 'Password',
            ),
            validator: Validators.compose([
              Validators.required('Password is required'),
              Validators.patternString(r'^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[!@#\$&*~]).{8,}$', 'Invalid Password')
            ]),
          ),

Its documentation is really good. You can read it for more util functions like this.