I need to rename the following folders:
d_001_d d_001_h d_005_d d_005_h d_007_d d_007_h In:
d_a_d d_a_h d_b_d d_b_h d_c_d d_c_h So basically each code number correspond to a name(letter). I tried with the sed command and a for loop with an array but I'm not able to get exactly what I want.
Thanks for help
bos:$ ls d_001_d d_001_h d_005_d d_005_h d_007_d d_007_h bos:$ for x in * ; do mv $x $(echo $x | sed -e 's/001/a/;s/005/b/;s/007/c/') ; done bos:$ ls d_a_d d_a_h d_b_d d_b_h d_c_d d_c_h 
$ ls d_001_d d_001_h d_005_d d_005_h d_007_d d_007_h $ ls | awk -F_ -v OFS=_ 'BEGIN{d["001"]="a"; d["005"]="b"; d["007"]="c"}; {old=$0; $2=d[$2]; printf("mv %s %s\n", old, $0)}' | bash $ ls d_a_d d_a_h d_b_d d_b_h d_c_d d_c_h If the directory names are all the same length, you can use this:
#!/bin/bash for DIR in `ls */ -d` do INDEX=${DIR:2:3} case $INDEX in 001) LETTER="a" ;; 005) LETTER="b" ;; 007) LETTER="c" ;; esac mv "$DIR" ${DIR/$INDEX/$LETTER} done Without hardcoding the number->char mappings:
chr() { [[ ${1} -lt 256 ]] || return 1; printf \\$(printf '%03o' $1); } printf "%s\n" d_001_d d_001_h d_005_d d_005_h d_007_d d_007_h | while read name; do prefix=${name%%_*} suffix=${name##*_} [[ $name =~ ([0-9]+) ]] && num=$((96+10#${BASH_REMATCH[1]})) printf -v new_name "%s_%s_%s" $prefix $(chr $num) $suffix echo $new_name done outputs
d_a_d d_a_h d_e_d d_e_h d_g_d d_g_h chr function courtesy this answer
With the perl rename tool you can rename with regex:
rename -n 'y/[157]/[abc]/;s/00//' d_00?_? d_001_d renamed as d_a_d d_001_h renamed as d_a_h d_005_d renamed as d_b_d d_005_h renamed as d_b_h d_007_d renamed as d_c_d d_007_h renamed as d_c_h Omit the -n if the test with -n(ot really) yields a fine result.
