I have a lot of data with file name (just for example):
dt_upd_global_merged_madt_uv_20100801_20100801_20110721.bil
dt_upd_global_merged_madt_uv_20100802_20100802_20110721.bil
dt_upd_global_merged_madt_uv_20100803_20100803_20110721.bil
What should I do if I want to rename that files into the following?
20100801.bil
20100802.bil
20100803.bil
Try this:
for f in *.bil; do
n=$(echo "${f}"|sed -r 's/^.+([0-9]{8})_\1_[0-9]{8}[.]bil$/\1/'|grep -Ev '.bil$')
#or:
#n=`echo "${f}"|sed -r 's/^.+([0-9]{8})_\1_[0-9]{8}[.]bil$/\1/'|grep -Ev '.bil$'`
if [ -n "${n}" ]; then
mv "${f}" "${n}.bil"
fi
done
Easiest way with Zsh (by calling zsh first, obviously):
autoload -U zmv
zmv 'dt_upd_global_merged_madt_uv_(*)_(*)_(*).bil' '$1.bil'
Or with the Perl rename:
rename 's/dt_upd_global_merged_madt_uv_(.*)_(.*)_(.*).bil/$1.bil/' *.bil
Using awk, -F option to specify _ as the delimiter. The 7th field output plus ".bil" (for example) places 20100801.bil in the fnew variable.
Then mv the original name to the new name.
for f in *.bil
do
fnew=$(awk -F_ '{print $7 ".bil"}' <<< $f);
mv $f $fnew
done
