shell script でのファイル名の部分取得

#!/bin/sh f=/home/xxx/yyy-012.eps echo $f f0="yyy-" f1=${f##*/} # remove path ("basename $f" と同じ) echo ${f1} f2=${f1#${f0}} # remove ${f0} from ${f1} echo ${f2} NUM=${f2%.eps} # remove ".eps" echo ${NUM} FNUM=`printf "%06d" ${NUM}` echo ${FNUM}