How to ensure secure randomization for python password generator

How to ensure secure randomization for python password generator

By : Akash Malik
Date : October 25 2020, 07:10 AM
around this issue Use os.urandom(), which reads from the corresponding device.
code :

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Need a secure password generator recommendation

Need a secure password generator recommendation

By : Loan Motyl
Date : March 29 2020, 07:55 AM
I wish did fix the issue. I would not worry that much about generating incredible strong one time passwords. Make the password long and it should not be a problem with brute force granted you limit how long the password is valid. If the password is only valid for say 1 hour then it will not be a problem if the password remains unused. And in that time span it is not likely that someone will get to crack it using brute force.
It is also important that you only let the one time password work just one time. This way, if the password is intercepted the user will notice when the one time password has expired and can take appropriate actions.
Creating a secure random password generator in VB.Net

Creating a secure random password generator in VB.Net

By : Frank Louis Rosario
Date : March 29 2020, 07:55 AM
hope this fix your issue It looks like you are missing a few things. I created 2 classes and wrote this up. It give me a random password every time. Enjoy.
code :
Imports System
Imports System.Collections.Generic
Imports System.Text
Imports System.Security.Cryptography
Imports System.Windows.Forms

Public Class Form1
Private randomBytes() As Byte
Private randomInt32Value As Integer
Private possibleChars As String
Private len As Int32
Private GetRandomInt32Value As New RandomInt32Value
Private GetPasswordGenProfiler As New PasswordGenProfiler

Public Sub New()
    ' This call is required by the designer.

    ' Add any initialization after the InitializeComponent() call.
    possibleChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789!@#$%^&*()"
    len = 8
End Sub

Private Sub btnGenerate_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnGenerate.Click
        Dim cpossibleChars() As Char
        cpossibleChars = possibleChars.ToCharArray()
        If cpossibleChars.Length < 1 Then
            MessageBox.Show("You must enter one or more possible characters.")
        End If
        If len < 4 Then
            MessageBox.Show(String.Format("Please choose a password length. That length must be a value between {0} and {1}. Note: values above 1,000 might take a LONG TIME to process on some computers.", 4, Int32.MaxValue))
        End If

        Dim builder As New StringBuilder()

        For i As Integer = 0 To len - 1
            Dim randInt32 As Integer = GetRandomInt32Value.GetRandomInt()
            Dim r As New Random(randInt32)

            Dim nextInt As Integer = r.[Next](cpossibleChars.Length)
            Dim c As Char = cpossibleChars(nextInt)
        Me.Label1.Text = builder.ToString()
    Catch ex As Exception
        MessageBox.Show(String.Format("An error has occurred while trying to generate random password! Technical description: {0}", ex.Message.ToString()))
    End Try

End Sub
End Class

Public Class PasswordGenProfiler
Public Shared Function GetFrequencyDistributionOfChars(allowableChars As String, generatedPass As String) As Dictionary(Of Char, Integer)
    Dim distrib As New Dictionary(Of Char, Integer)()
    ' initialize all values to 0
    For Each c As Char In allowableChars
        ' If character is listed more than once, don't re-add it to our list.
        If Not distrib.ContainsKey(c) Then
            distrib.Add(c, 0)
        End If
    Dim val As Integer = 0
    For Each passChar As Char In generatedPass
        If distrib.TryGetValue(passChar, val) Then
            distrib(passChar) = System.Threading.Interlocked.Increment(val)
        End If

    Return distrib
End Function
End Class

Imports System.Security.Cryptography

Public Class RandomInt32Value
Public Function GetRandomInt() As Integer
    Dim randomBytes As Byte() = New Byte(3) {}
    Dim rng As New RNGCryptoServiceProvider()
    Dim randomInt As Integer = BitConverter.ToInt32(randomBytes, 0)
    Return randomInt
End Function
End Class
Improving Secure Password Generator

Improving Secure Password Generator

By : NRaju
Date : March 29 2020, 07:55 AM
like below fixes the issue The 3rd functionality is missing and will probably be way more sophisticated. But this is a simple solution to the 1st and 2nd ones.
code :
var output = document.getElementsByTagName('output')[0];

var Chars = {};
    Chars.length = 16;
    Chars.abc   = "abcdefghijklmnopqrstuvwxyz";
    Chars.Num   = "1234567890";
    Chars.NumRequired = true;
    Chars.Sym   = "@\#\-!$%^&*()_+|~=`{}\[\]:\";'<>?,.\/";

var generator = new randomPasswordGenerator(Chars);

var simpleGenerator = new randomPasswordGenerator({
    length: 6,
    abc: 'abc',
    Num: '0'

var button = document.getElementsByTagName('button')[0];
    button.addEventListener('click', clickFunction);

var checkbox = document.getElementsByTagName('input')[0];

function clickFunction () {
  if (checkbox.checked) output.textContent = simpleGenerator.randomPassword();
  else output.textContent = generator.randomPassword();

function randomPasswordGenerator(opts) {
    for(var p in opts) this[p] = opts[p];
    this.randomPassword = randomPassword;

function randomPassword() {
    var chars = (this.abc || "") +
                (this.ABE || "") + 
                (this.Num || "") +
                (this.Sym || ""),
    	pass  = [],
        PL    = this.length;
    if (this.NumRequired) {
        var r = Math.floor(Math.random() * this.Num.length);
        var i = Math.floor(Math.random() * PL);
        pass[i] = this.Num[r];
    for (var x = 0; x < PL; x++) {
        if(!pass[x]) {
          var i = Math.floor(Math.random() * chars.length);
          pass[x] = chars.charAt(i);
    return pass.join('');
output {
  margin: 12px;
  display: block;
  border-bottom: 1px solid
  <input type="checkbox">Simple
Python password generator that creates a new password on user input

Python password generator that creates a new password on user input

By : Sarang Mahajan
Date : March 29 2020, 07:55 AM
Hope this helps You have to repeat the password generation, by putting everything into the while-loop, or better, write a function to generate one password and call this function inside the while-loop:
code :
import random

alphabet = "abcdefghijklmnopqrstuvwxyz"
pw_length = 6

def generate_password(pw_length):
    mypw = ""
    for i in range(pw_length):
        next_index = random.randrange(len(alphabet))
        mypw = mypw + alphabet[next_index]

    # replace 1 or 2 characters with a number
    for i in range(random.randrange(1,3)):
        replace_index = random.randrange(len(mypw)//2)
        mypw = mypw[0:replace_index] + str(random.randrange(10)) + 

    # replace 1 or 2 letters with an uppercase letter
    for i in range(random.randrange(1,3)):
        replace_index = random.randrange(len(mypw)//2,len(mypw))
        mypw = mypw[0:replace_index] + mypw[replace_index].upper() + 
    return mypw

while True:
    print("Your password is", generate_password(pw_length))
    inp = input()
    if not inp:
import string
import random

def generate_password(pw_length):
    charsets = []
    # take 1 or 2 numbers
    for i in range(random.randrange(1,3)):
    # take 1 or 2 uppercase
    for i in range(random.randrange(1,3)):
    # fill up with lowercase up to pw_length
    while len(charsets) < pw_length:
    # generate password of charsets
    return ''.join(random.choice(cs) for cs in charsets)

while True:
    print("Your password is", generate_password(pw_length))
    inp = input()
    if not inp:
Quiz generator and randomization in Javascript

Quiz generator and randomization in Javascript

By : Tyler Alkatraz Randa
Date : March 29 2020, 07:55 AM
Any of those help You are setting the score value to 0 in your getFinalResultsString() function. Remove the line score = 0; and it should work as expected. I've also updated your JSFiddle.
The answer for the first question is 'her' as you have set the answer index to 1; don't forget that JavaScript uses 0 based indexing.
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