Generate Random Password without these characters "l1o0" - php

Question

I need to generate random password with condition:
All characters are allowable except "l1o0"
Length of 8 to 12 in characters
Codes I've tried:
function generateRandomPassword() {
//Initialize the random password
$password = '';
//Initialize a random desired length
$desired_length = rand(8, 12);
for($length = 0; $length < $desired_length; $length++) {
//Append a random ASCII character (including symbols)
$password .= chr(rand(32, 126));
}
return $password;
}
How to avoid these 4 characters => "l1o0" ?
Reason:
These 4 characters are sometimes confused the user.
Thanks!

Answer 1

Please don't use any of the other answers currently provided for generating passwords. They're not secure by any measure.
rand() -> No
mt_rand() -> Definitely not
I'm going to pull this solution from a blog post aptly titled How to Securely Generate Random Strings and Integers in PHP.
/**
* Note: See https://paragonie.com/b/JvICXzh_jhLyt4y3 for an alternative implementation
*/
function random_string($length = 26, $alphabet = 'abcdefghijklmnopqrstuvwxyz234567')
{
if ($length < 1) {
throw new InvalidArgumentException('Length must be a positive integer');
}
$str = '';
$alphamax = strlen($alphabet) - 1;
if ($alphamax < 1) {
throw new InvalidArgumentException('Invalid alphabet');
}
for ($i = 0; $i < $length; ++$i) {
$str .= $alphabet[random_int(0, $alphamax)];
}
return $str;
}
Usage:
// Every ASCII alphanumeric except "loIO01":
$alphabet = 'abcdefghijkmnpqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ23456789';
$string = random_string(12, $alphabet);
You probably don't have random_int(), unless you're reading this in the future when PHP 7 is released. For those of us living in the present, use random_compat.

Answer 2

Try this:
function generateRandomPassword($length = 8) {
$characters = '23456789abcdefghjklmnpqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ';
$charactersLength = strlen($characters);
$randomPassword = '';
for ($i = 0; $i < $length; $i++) {
$randomPassword .= $characters[rand(0, $charactersLength - 1)];
}
return $randomPassword;
}

Answer 3

You don't need to change your code . just use str_replace to replace those word's You can try this solution :) . Just edited your code
function generateRandomPassword($length = 8) {
$characters = '23456789abcdefghjklmnpqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ';
$charactersLength = strlen($characters);
$randomPassword = '';
for ($i = 0; $i < $length; $i++) {
$randomPassword .= $characters[rand(0, $charactersLength - 1)];
}
return str_replace(['l','1','o','0'], ['A','B','C','D'], $randomPassword);
}

Answer 4

$string = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
$req_pword_len = 20;
$char_count = 0;
$password='';
$chars=str_split($string);
while ( $char_count < $req_pword_len ) {
$char = mt_rand(0,61);
$password .= (string) $chars[$char];
$char_count++;
}
Change the value for
$string to be just the characters you want to allow
$req_pword_len to the required length of the password

Answer 5

Try this:
function generateRandomPassword($length = 8) {
$randomPassword = '';
for ($i = 0; $i < $length; $i++) {
while (true) {
//remove 0,1,I,O,l,o
while(in_array(($number = rand(65, 122)), array(48, 49, 73, 79, 108, 111)));
if ($number <= 90 or $number >= 97) {
$randomPassword .= chr($number);
break ;
}
}
}
return $randomPassword;
}
echo generateRandomPassword();