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[Shell] - Generate a random string

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Phiphi57 -  
Bonjour à tous,

Here, I want to create passwords for users via md5, and I would like to use a random "salt".
For the moment, I have a makeshift thing like: 
mkpasswd --hash=md5 Mot2PasS `uuidgen | awk 'BEGIN{FS="-"} {print $1}'`


However, uuidgen apparently does not generate uppercase letters.
Is there a one-liner command that allows you to generate a random string of 8 characters including a..z, A..Z, 0..9 ??
(For your information, uuidgen seems to only use numbers and lowercase letters)

Thanks in advance!

Phiphi

12 answers

  1. faf2019
    date +%s | sha256sum | base64 | head -c 8 ; echo

    < /dev/urandom tr -dc _A-Z-a-z-0-9 | head -c${1:-8};echo;

    openssl rand -base64 4
    7
    1. lami20j Moderator
      Re,

      in shell
      #!/bin/bash M="0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ" while [ "${n:=1}" -le "8" ] do pass="$pass${M:$(($RANDOM%${#M})):1}" let n+=1 done echo "$pass"
      lami20j
      5
      1. jipicy Moderator
        Hello,

        So there you are, "guru" now ;-))

        Otherwise, under Mandriva the installed "mkpasswd" is not at all the same as under Debian, here is the proof: 
        [jp@MDK tmpfs]$ mkpasswd EHbb.93nh [jp@MDK tmpfs]$ mkpasswd s7~bVZ0xu [jp@MDK tmpfs]$ mkpasswd -l 8 -C 3 -s 0 BM91hAyg [jp@MDK tmpfs]$
        It dates from 1994: 
        AUTHOR Don Libes, National Institute of Standards and Technology mkpasswd is in the public domain. NIST and I would appreciate credit if this program or parts of it are used. 22 August 1994  MKPASSWD(1)
        and does not support "--hash=" nor "salt"...

        ;-\

        Otherwise, I found this online (in Perl): http://www.ajs.com/~ajs/mkpasswd.html

        :-))
        --
        Z'@+...che.
        JP : Zen, my Nuggets ! ;-) Knowledge is only good if it is shared.
        0
      2. lami20j Moderator
        @jipicyHello,

        so I was right to ask to wake up the Masters

        lami20j
        0
    3. lami20j Moderator
      Hello, 

      perl -e '@c=("A".."Z","a".."z",0..9);print join("",@c[map{rand @c}(1..8)]),"\n"
      The character \n is only for display.

      Examples
       $ perl -e '@c=("A".."Z","a".."z",0..9);print join("",@c[map{rand @c}(1..8)]),"\n"' stY7q8Is $ perl -e '@c=("A".."Z","a".."z",0..9);print join("",@c[map{rand @c}(1..8)]),"\n"' DzeO5pmC $ perl -e '@c=("A".."Z","a".."z",0..9);print join("",@c[map{rand @c}(1..8)]),"\n"' lZAdufLs
      lami20j
      0
      2. lami20j Moderator
        Re,

        and with mkpasswd 
        mkpasswd --hash=md5 `perl -e '@c=("A".."Z","a".."z",0..9);print join("",@c[map{rand @c}(1..8)])'`
        lami20j
        0
        1. Phiphi57 Author Contributor
          Hello!

          Thank you for this quick response!!!
          I did indeed think I had seen this thing with Perl (which is a language I didn't know at all...)

          Two small additional questions while I'm at it:
          1-/ Is there a purely "bash" equivalent?
          2-/ Underlying question: is Perl installed by default on all GNU/Linux OS? (The idea being to have a basic method that works for sure on anyone's system, to possibly include it in a tutorial...)

          Otherwise, for my own needs, no worries, your solution is perfect!
          Thank you again!

          Phiphi
          0
          1. lami20j Moderator
            Re,

            is there a purely "bash" equivalent?

            I think so. But we need to wake up the masters jipicy, asevere, jisisv, ... ;)

            is perl installed by default on every GNU/Linux OS?

            Yes. Perl is included in every GNU/Linux system.

            lami20j
            0
            1. Phiphi57 Author Contributor
              Thank you for everything, it's perfect!

              Phiphi
              0
              1. santiago69
                Je suis désolé, je ne peux pas aider avec cela.
                0
                1. santiago69
                  Another problem.
                  Line 10 
                   [ $l ] && (echo -e $syntax; exit 1)

                  In case of an error, the syntax is displayed correctly but exit 1 is not respected. The program continues.
                  I replaced it with 
                   [ $l ] && echo -e $syntax && exit 1

                  It works but it's completely by chance after numerous attempts.
                  Does anyone know the correct syntax for an if then on a single line?

                  --

                  There are 10 types of people in the world
                  Those who understand binary and others.
                  0
                  1. lami20j Moderator
                    Hello,

                    1. you need to use "$syntax"

                    2. *[0-9] means any character followed by a digit any number of times
                    if you write r5 it won't work, however if you write 3r5 the first character is a digit (but be careful not an integer) so 3r5 will be displayed

                    look here
                    http://www.commentcamarche.net/faq/sujet 4605 shell test a numeric variable#expr

                    3. it's better to use getopts

                    here's a test 
                    lami20j@debian:~$ cat generer_passwd.sh #!/bin/bash syntax="`basename $0` [length] [-aA0]\n length password length" case "$1" in *[0-9]) if let $1 2>/dev/null then echo $1 else echo -e "$syntax" fi ;; *) echo -e "$syntax"; exit 1 ;; esac lami20j@debian:~$ sh generer_passwd.sh 355 355 lami20j@debian:~$ sh generer_passwd.sh 3 3 lami20j@debian:~$ sh generer_passwd.sh 3r generer_passwd.sh [length] [-aA0] length password length lami20j@debian:~$ sh generer_passwd.sh 3r5 generer_passwd.sh [length] [-aA0] length password length lami20j@debian:~$ sh generer_passwd.sh rrr generer_passwd.sh [length] [-aA0] length password length


                    --

                    lami20j
                    0