How do you parse a filename in bash?

I have a filename in a format like:

system-source-yyyymmdd.dat

I'd like to be able to parse out the different bits of the filename using the "-" as a delimiter.

From stackoverflow Nick Pierpoint

• Use the cut command.

  e.g. echo "system-source-yyyymmdd.dat" | cut -f1 -d'-'

  will extract the first bit.

  Change the value of the -f parameter to get the appropriate parts.

  Here's a guide on the Cut command.

  From David

• You can use the cut command to get at each of the 3 'fields', e.g.:

  $ echo "system-source-yyyymmdd.dat" | cut -d'-' -f2
  source
  

  "-d" specifies the delimiter, "-f" specifies the number of the field you require

  Jon Ericson : I'm curious why you added the # prompt. Normally, that prompt indicates the root or superuser. In generally, I'd think stuff like trying out the **cut** command would be better done as a regular user. I'd have used the $ prompt.

  Bobby Jack : Oh, yeah - good point. I must admit, I was logged in as root at the time and simply went for it - a bad habit, I know. Having said that, I think echo and cut are two of the least harmful commands :) But, for the sake of completeness, I'll certainly update the example right away. Cheers.
  From Bobby Jack

• Fantastic - an answer in 3 minutes - that's quicker than phoning a friend!

  From Nick Pierpoint

• Depending on your needs, awk is more flexible than cut. A first teaser:

  # echo "system-source-yyyymmdd.dat" \
      |awk -F- '{printf "System: %s\nSource: %s\nYear: %s\nMonth: %s\nDay: %s\n",
                $1,$2,substr($3,1,4),substr($3,5,2),substr($3,7,2)}'
  System: system
  Source: source
  Year: yyyy
  Month: mm
  Day: dd
  

  Problem is that describing awk as 'more flexible' is certainly like calling the iPhone an enhanced cell phone ;-)

  From flight

• Another method is to use the shell's internal parsing tools, which avoids the cost of creating child processes:

  oIFS=$IFS
  IFS=-
  file="system-source-yyyymmdd.dat"
  set $file
  IFS=$oIFS
  echo "Source is $2"
  

  From Shannon Nelson

• A nice and elegant (in my mind :-) using only built-ins is to put it into an array

  var='system-source-yyyymmdd.dat'
  parts=(${var//-/ })
  

  Then, you can find the parts in the array...

  echo ${parts[0]}  ==> system
  echo ${parts[1]}  ==> source
  echo ${parts[2]}  ==> yyyymmdd.dat
  

  Caveat: this will not work if the filename contains "strange" characters such as space, or, heaven forbids, quotes, backquotes...

  From Colas Nahaboo