[SOLVED] How to convert decimal number to Base58 in a shell script

Issue

myNumber=$(date +%s) # big number in decimal
myNumberInB58=$(toBase58 $myNumber)

toBase58() {
  # <your answer here>
}

What is the most elegant and/or concise way to encode an integer in Base58?

Solution

Here's another version:

# Order-Preserving Base58 (OPB58).
# Also supports negative numbers.
int2b58() {
  # Omit IOlo
  local n="$1" i BASE58=$(echo {0..9} {A..H} {J..N} {P..Z} {a..k} {m..n} {p..z} | tr -d ' ')
  ((n < 0 )) && printf -- '-' && n=$((-n))
  for i in $(echo "obase=58; $n" | bc); do
    printf ${BASE58:$(( 10#$i )):1}
  done; echo
}

Answered By - Elifarley

Answer Checked By - David Goodson (WPSolving Volunteer)