Generate random passwords in shell with one special character

问题

I have the following code:

</dev/urandom tr -dc 'A-Za-z0-9@#$%&_+=' | head -c 16

which is randomly generating passwords perfectly.

I want two changes:

1. It should only contain one special character listed above
2. It should choose a random length

I tried with length = $(($RANDOM%8+9))

then putting length as

</dev/urandom tr -dc 'A-Za-z0-9@#$%&_+=' | head -c$length

but got no positive result.

回答1:

#! /bin/bash
chars='@#$%&_+='
{ </dev/urandom LC_ALL=C grep -ao '[A-Za-z0-9]' \
        | head -n$((RANDOM % 8 + 9))
    echo ${chars:$((RANDOM % ${#chars})):1}   # Random special char.
} \
    | shuf \
    | tr -d '\n'

• LC_ALL=C prevents characters like ř from appearing.
• grep -o outputs just the matching substring, i.e. a single character.
• shuf shuffles the lines. I originally used sort -R, but it kept the same characters together (ff1@22MvbcAA).

来源：https://stackoverflow.com/questions/30947873/generate-random-passwords-in-shell-with-one-special-character