bash: String Operator with regular expression - regex

Question

I want to list all files in my home folder, and remove the # in the filename.
For example:
#.emacs# should be printed out as .emacs
This is my code
for dir in $(ls ~)
do
# trim trailing
filename="${dir#\#}"
echo ${filename}
done
But it's still showing files preceding with # though I managed the regular expression ${dir#\#} in terminal.
Can you tell me where's the smell in my code?

Answer 1

remove # from filename, should be :
filename="${dir//#/}"
edit: in some systems (such as Solaris), above command doesn't work, you need escape.
filename="${dir//\#/}"
The rest are fine for both cygwin and Solaris.
If you need remove all contents before #
filename="${dir##*#}"
If you need remove all contents after #
filename="${dir%%#*}"
Here are full explanation I copy and paste from bash Substring Replacement
${string/substring/replacement}
Replace first match of $substring with $replacement.
${string//substring/replacement}
Replace all matches of $substring with $replacement.
${string%substring}
Deletes shortest match of $substring from back of $string.
${string%%substring}
Deletes longest match of $substring from back of $string.
${string#substring}
Deletes shortest match of $substring from front of $string.
${string##substring}
Deletes longest match of $substring from front of $string.

Answer 2

Don't parse ls. You can just you bash wildcard expansion instead. Also, your use of parameter expansion is wrong, ${word#something} removes something from the prefix not the suffix. So try
#!/bin/bash
for dir in ~/*
do
# trim trailing
filename="${dir%#}"
echo "${filename}"
done

Answer 3

Here's a - hopefully - instructive version:
#!/usr/bin/env bash
# Make pathname expansion match files that start with '.', too.
shopt -s dotglob
# Loop over all files/dirs. in the home folder.
for f in ~/*; do
# Exit, if no files/dirs match at all (this test may
# not be necessary if `shopt -s nullglob` is in effect).
# Use -f to only match files, -d to only match dirs.
[[ -e $f ]] || break
# Remove the path component ...
filename=$(basename "$f")
# ... and then all '#' chars. from the name.
filename="${filename//#/}"
# Process result
echo "${filename}"
done
As others have noted, you should not parse ls output - direct pathname expansion of globs (wildcard patterns) is always the better choice.
shopt -s dotglob ensures that files or dirs whose name starts with . are included in pathname expansion.
Pathname expansions occurs with the path component intact, so to get the mere filename from the loop variable, basename must be applied (first), in order to strip the path component.
Probably not an issue here, but unless shopt -s nullglob is in effect (not by default), a glob that matches nothing is left untouched, so the loop is entered with an invalid filename - hence the [[ -e ... ]] test.

Answer 4

Hi you just echo the filename but not to rename it. So first you need to cd to home directory from scripts then rename files. Please below scripts to find file name which contain # char and removed # from file name.
#! /bin/bash
cd ~
for i in $(ls ~ )
do
if [[ "${i}" == *#* ]]
then
var=$(echo "$i" | sed 's/#//' )
printf "%s\n" "$var" #to print only
#mv "$i" "$var" #to renmae
fi
done

Answer 5

You didn’t state that your files have # at the start and end of filename earlier. Try something like:
for dir in ~/*; do
filename="${dir#\#}"
filename="${filename%\#}"
echo "$dir ---> ${filename}"
done
or use what BMW has shown as his first example:
for dir in ~/*; do
filename="${dir//#/}"
echo "$dir ---> ${filename}"
done
Once you are satisfied with echo’s output. You can replace that with mv.
P.S: Re-iterating what BroSlow stated. Don’t parse ls.