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Old 12-08-2016, 06:27 AM   #1
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Pass generator question

Im trying to make a bash script that will ask the name of the user, favorite food and the current date and save these to the variables food, date and name a random password will then be generated. When the user forgets the password all he has to do is enter the same name (userrecovery), favorite food (foodrecovery) and the date (daterecovery) when the password was generated and the password should be shown. The code i use to generate a random password:

choose() { echo ${1:RANDOM%${#1}:1} $RANDOM; }
  pass="$({ choose '!@#$%^\&'
   choose '0123456789'
   choose 'abcdefghijklmnopqrstuvwxyz'
   for i in $( seq 1 $(( 4 + RANDOM % 8 )) )
         choose '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
  } | sort -R | awk '{printf "%s",$1}')"
I tried something like:

if [ $food=$foodrecovery -a $date=$daterecovery -a.... ]
   echo $pass
This works but when i close the script and reopen it will generate a new password which it shouldnt.
Old 12-08-2016, 07:00 AM   #2
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How are you storing/retrieving your data?

Who is going to remember when a password was generated?
Old 12-08-2016, 09:55 PM   #3
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I'm not very bash literate, but it doesn't look like you're seeding RANDOM anywhere in there. I assume that's a necessary step.
Old 12-09-2016, 08:54 AM   #4
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Yep, you're going to need to save that in a DB of some kind that BASH can read later on.
Where are you calling the function to generate a new key?
Maybe have a line in there somewhere to test if one is already created then do not create another one.

I am not an expert, but logically that sounds sound. It is hard to see why you are having this problem only seeing two snippets of your code.


bash, shell

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