CODEWITHSUNDEEP
java
SteveYo
SteveYo

Reputation: 47

Java Random Password generator issue

So I need to know is there a way to show the final password of this code to jTextField control, cuz when I tried I got issue char couldnt convert to String.

import java.util.Random;

Random r = new Random();

String alphabet = "qwertyuiopasdfghjklzxcvbnm";
for (int i = 0; i < 50; i++) {
    System.out.println(alphabet.charAt(r.nextInt(alphabet.length())));
}

Upvotes: 0

Views: 310

Answers (3)

gonczor
gonczor

Reputation: 4146

If you are trying to provide secure random-generated password use SecureRandom with proper methdos. Please refer to this. I do not have any articles about it in english, but if you could get on with some translator please read this one to see how one could hack such algorithm.

EDIT:

More on difference between Random and SecureRandom

Upvotes: 0

Maurice Perry
Maurice Perry

Reputation: 9650

    Random r = new Random();

    String alphabet = "qwertyuiopasdfghjklzxcvbnm";
    StringBuilder buf = new StringBuilder();
    for (int i = 0; i < 50; i++) {
        buf.append(alphabet.charAt(r.nextInt(alphabet.length())));
    }
    jTextField.setText(buf.toString());

Upvotes: 1

Sanket Makani
Sanket Makani

Reputation: 2489

Currently in your code you are just generating the characters randomly and printing it. You need to form a String from all these generated Characters and then you can set it as a Text in a TextField.

You can have a StringBuilder which gets appended with every random character.

String alphabet = "qwertyuiopasdfghjklzxcvbnm";
StringBuilder password=new StringBuilder();

for (int i = 0; i < 50; i++) {
    password.append(alphabet.charAt(r.nextInt(alphabet.length())));
} 
String password_str=password.toString();
System.out.println(password_str);

Suppose that you have a JTextField then you can set there password_str as its value.

JTextField password_field = new JTextField();
password_field.setText(password_str);

Upvotes: 1

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