How to generate a random string without repeated characters? - linux

Question

Im trying to generate a password in Bash that matches MacOS password requirements and one of them is that it can't have repeated characters (aa, bb, 44, 00, etc).
I know i can use openssl rand -base64 or /dev/urandom and use tr -d to manipulate the output string. I use grep -E '(.)\1{1,}' to search for repeated characters but if i use this regex to delete (tr -d (.)\1{1,}'), it deletes the entire string. I even tried tr -s '(.)\1{1,}' to squeeze the characters to just one occurrence but it keep generating repeated characters in some attempts. Is it possible to achieve what i'm trying to?
P.S.: that's a situation where i cant download any "password generator tool" like pwgen and more. It must be "native"

Answer 1

Sorry I have no bash at hands to, but trying to help you.
What about iteratively grabbing the unique chars, eg. by
chars=$(openssl rand -base64)
pwd=
for (( i=0; i<${#chars}; i++ )); do
if [[ "$(echo $pwd | grep "${chars:$i:1}")" == "" ]]; then
pwd=$pwd${chars:$i:1}
fi
done

Answer 2

The issue might be that you have non-printable characters, so it's not actually repeated. If you first get the first e.g. 30 characters, then delete any non-alphanumeric, non punctuation characters, then squeeze any of those characters, then from whatever is left get the first 20 characters, it seems to work:
cat /dev/urandom | tr -dc '[:alnum:][:punct:]' | fold -w ${1:-30} | head -n 1 | tr -s '[:alnum:][:punct:]' | cut -c-20
Output e.g.:
]'Zc,fs6m;wUo%wLIG%K
2O3Ff4dzi30~L.RH8jR0
sU?,WkK]&I;z'|eTSLjY
5gK]\H51i#Rtux.{bdC=
:g"\?5JsjBd1r])2^WR+
;{cR:jY\rIc&Q(2yo:|-
fFykmxvZ|ATX_l6L(8h:
^Sd*,V%9}bWnTYNv"w?'
6foMgbU6:n<*cWj2W=3&
*v39FWmB#LwE5O`a3C36

Answer 3

Is there a specific size requirement? Other required characters?
How about -
openssl rand -base64 20 | sed -E 's/(.)\1+/\1/g'

Answer 4

You're getting close. tr doesn't use regex (but does use POSIX character classes).
Either of these will squeeze repeats:
tr -cs '\0'
tr -s '[:graph:][:space:]'
They differ only in how we refer to "all characters". First is "complement of null" second is all printable and all white space characters. There may be a neater way to specify "all characters".
Or using sed:
sed -E 's/(.)\1+/\1/g'
This both squeezes printable characters, and removes white space:
tr -ds '[:space:]' '[:graph:]'
Example for 32 non whitespace characters, with no repeats:
tr -ds '[:space:]' '[:graph:]' < /dev/urandom |
dd bs=32 count=1
Also, this example specifies a list of allowed characters (letters, digits, and _.), then squeezes any repeats:
tr -dc '[:alnum:]_.' < /dev/urandom |
tr -sc '\0' |
dd bs=32 count=1
Example output:
9mCEqrhHPwmq7.1qEky6qn4jqzDpRK7b
Putting dd at the end means we get 32 characters after removing repeats. You may also want to add status=none to hide dd logging on stderr.

Answer 5

It's not clear if you don't want consecutive chars repeated or no repeated chars at all (which in either case I don't think it's a good idea as it would make your passwords weaker and easier to guess), but having said that
#! /bin/bash
awk -vN=20 '
{
n=split($0,ch,"");
for (i=1; i<n; i++) {
a[ch[i]]++
}
n=0;
for (c in a) {
if (++n > N) {
break;
}
printf("%c",c)
}
printf("\n")
}
' < <(openssl rand -base64 32)
this generated N length passwords without repeated chars from 32 random bytes (that is N should be much smaller than 32)