How do you parse a filename in bash?
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]} ==> systemecho ${parts[1]} ==> sourceecho ${parts[2]} ==> yyyymmdd.datCaveat: this will not work if the filename contains "strange" characters such as space, or, heaven forbids, quotes, backquotes...
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: systemSource: sourceYear: yyyyMonth: mmDay: ddProblem is that describing awk as 'more flexible' is certainly like calling the iPhone an enhanced cell phone ;-)