How can I match a string with a regex in Bash?

Question

I am trying to write a bash script that contains a function so when given a .tar, .tar.bz2, .tar.gz etc. file it uses tar with the rel

Answer 1

A Function To Do This

extract () {
  if [ -f $1 ] ; then
      case $1 in
          *.tar.bz2)   tar xvjf $1    ;;
          *.tar.gz)    tar xvzf $1    ;;
          *.bz2)       bunzip2 $1     ;;
          *.rar)       rar x $1       ;;
          *.gz)        gunzip $1      ;;
          *.tar)       tar xvf $1     ;;
          *.tbz2)      tar xvjf $1    ;;
          *.tgz)       tar xvzf $1    ;;
          *.zip)       unzip $1       ;;
          *.Z)         uncompress $1  ;;
          *.7z)        7z x $1        ;;
          *)           echo "don't know '$1'..." ;;
      esac
  else
      echo "'$1' is not a valid file!"
  fi
}

Other Note

In response to Aquarius Power in the comment above, We need to store the regex on a var

The variable BASH_REMATCH is set after you match the expression, and ${BASH_REMATCH[n]} will match the nth group wrapped in parentheses ie in the following ${BASH_REMATCH[1]} = "compressed" and ${BASH_REMATCH[2]} = ".gz"

if [[ "compressed.gz" =~ ^(.*)(\.[a-z]{1,5})$ ]]; 
then 
  echo ${BASH_REMATCH[2]} ; 
else 
  echo "Not proper format"; 
fi

(The regex above isn't meant to be a valid one for file naming and extensions, but it works for the example)