Rename Multiple Files - Shell script - shell

Question

I have 66 files in my folder and I want to rename all of them by using their current names' parts.
For example;
My file: Subscriber_ID_List_2015_11_14.csv
I want it like this -> TheFile_20151114.csv
I tried the code below but it did not work. How could I do this?
Thank you
#!/bin/bash
FILES=/my/path/
for f in $FILES
do
cd /my/path/
b=`ls "$f" |awk -F" " '{print $5}' |cut -c6-9`
c=`ls "$f" |awk -F" " '{print $5}' |cut -c11-12`
d=`ls "$f" |awk -F" " '{print $5}' |cut -c14-15`
cp "$f" TheFile_${b}${c}${d}.csv
done

Answer 1

Can u try this one.
#!/bin/bash
for files in `ls Subscriber_ID_List_*.csv`
do
cd /my/path/
datePrefix=`echo $files | awk -F '.' '{ print $1}'| awk -F '_' '{ print $4$5$6}'`
cp "$files" "TheFile_${datePrefix}.csv"
done

Answer 2

FILES=/my/path/
This puts only the path into the variable, not the filenames. Provided that you want to process all files in /my/path/, you can use FILES=/my/path/*.
b=`ls "$f" |awk -F" " '{print $5}' |cut -c6-9`
c=`ls "$f" |awk -F" " '{print $5}' |cut -c11-12`
d=`ls "$f" |awk -F" " '{print $5}' |cut -c14-15`
This doesn't take into account that $f still contains /my/path/, uses space as a field separator without discernible cause, and cuts at the wrong character positions. Here's an easier approach:
#!/bin/bash
FILES=my/path/*
for f in $FILES
do newf=`sed 's/Subscriber_ID_List_\(....\)_\(..\)_\(..\).csv/TheFile_\1\2\3.csv/' <<<$f`
cp "$f" "$newf"
done