Remove certain characters in a filename in Linux - linux

I have files in a directory such as
FILE1.docx.txt
FILE2.docx.txt
FILE3.docx.txt
FILE4.docx.txt
FILE5.docx.txt
And I would like to remove .docx from all of them to make the final output such as
FILE1.txt
FILE2.txt
FILE3.txt
FILE4.txt
FILE5.txt
How do I do this?

With Parameter Expansion and mv
for f in *.docx.txt; do
echo mv -vn "$f" "${f%%.*}.${f##*.}"
done
The one-liner
for f in *.docx.txt; do echo mv -vn "$f" "${f%%.*}.${f##*.}"; done
Remove the echo if you think the output is correct, to rename the files.
Should work in any POSIX compliant shell, without any script.
With bash, enable the nullglob shell option so the glob *.docx.txt will not expand as literal *.docx.txt if there are no files ending with .docx.txt
#!/usr/bin/env bash
shopt -s nullglob
for f in *.docx.txt; do
echo mv -vn "$f" "${f%%.*}.${f##*.}"
done
UPDATE: Thanks to #Léa Gris add nullglob change the glob to *.docx.txt and add -n to mv, Although -n and -v is not defined by POSIX as per https://pubs.opengroup.org/onlinepubs/9699919799/utilities/mv.html It should be in both GNU and BSD mv

Just run this python script in the same folder that contains files:
import os
for file in os.listdir(os.getcwd()):
aux = file.split('.')
if len(aux) == 3:
os.rename(file, aux[0] + '.' + aux[2])

you can make use of sed and bash like this:
for i in *.docx.txt
do
mv "$i" "`echo $i | sed 's/.docx//'`"
done

Related

How do I copy files at same location ending with "*100000.prm" with different name "*full.prm" in linux?

#!/bin/bash
for FILE in *1000000.wgt; do
BASE=${FILE%1000000.wgt}
[[ -e $BASE.trs && -e $BASE.1000000.wgt ]] && cp "$FILE" "$BASE.trs" "$BASE.wav" /some/dir
done
This script does what you need according to your commment.
eg: 'xyz_100000.prm' is to be copied with name 'xyz_full.prm' at the same location.
#!/bin/sh
IFS=$'\n'
for FILE in *1000000.prm; do
new_name=$(echo "$FILE" | sed "s/1000000.prm$/full.prm/")
cp "$FILE" "$new_name"
done
Demonstration:
➜ ls
a1000000.prm b1000000.prm copy.sh
➜ ./copy.sh
➜ ls
afull.prm bfull.prm copy.sh
I'd suggest this:
for i in *1000000.prm; do mv $i ${i%1000000.prm}full.prm; done
Read Parameter expansion section from bash man page.

Deleting all files except ones mentioned in config file

Situation:
I need a bash script that deletes all files in the current folder, except all the files mentioned in a file called ".rmignore". This file may contain addresses relative to the current folder, that might also contain asterisks(*). For example:
1.php
2/1.php
1/*.php
What I've tried:
I tried to use GLOBIGNORE but that didn't work well.
I also tried to use find with grep, like follows:
find . | grep -Fxv $(echo $(cat .rmignore) | tr ' ' "\n")
It is considered bad practice to pipe the exit of find to another command. You can use -exec, -execdir followed by the command and '{}' as a placeholder for the file, and ';' to indicate the end of your command. You can also use '+' to pipe commands together IIRC.
In your case, you want to list all the contend of a directory, and remove files one by one.
#!/usr/bin/env bash
set -o nounset
set -o errexit
shopt -s nullglob # allows glob to expand to nothing if no match
shopt -s globstar # process recursively current directory
my:rm_all() {
local ignore_file=".rmignore"
local ignore_array=()
while read -r glob; # Generate files list
do
ignore_array+=(${glob});
done < "${ignore_file}"
echo "${ignore_array[#]}"
for file in **; # iterate over all the content of the current directory
do
if [ -f "${file}" ]; # file exist and is file
then
local do_rmfile=true;
# Remove only if matches regex
for ignore in "${ignore_array[#]}"; # Iterate over files to keep
do
[[ "${file}" == "${ignore}" ]] && do_rmfile=false; #rm ${file};
done
${do_rmfile} && echo "Removing ${file}"
fi
done
}
my:rm_all;
If we assume that none of the files in .rmignore contain newlines in their name, the following might suffice:
# Gather our exclusions...
mapfile -t excl < .rmignore
# Reverse the array (put data in indexes)
declare -A arr=()
for file in "${excl[#]}"; do arr[$file]=1; done
# Walk through files, deleting anything that's not in the associative array.
shopt -s globstar
for file in **; do
[ -n "${arr[$file]}" ] && continue
echo rm -fv "$file"
done
Note: untested. :-) Also, associative arrays were introduced with Bash 4.
An alternate method might be to populate an array with the whole file list, then remove the exclusions. This might be impractical if you're dealing with hundreds of thousands of files.
shopt -s globstar
declare -A filelist=()
# Build a list of all files...
for file in **; do filelist[$file]=1; done
# Remove files to be ignored.
while read -r file; do unset filelist[$file]; done < .rmignore
# Annd .. delete.
echo rm -v "${!filelist[#]}"
Also untested.
Warning: rm at your own risk. May contain nuts. Keep backups.
I note that neither of these solutions will handle wildcards in your .rmignore file. For that, you might need some extra processing...
shopt -s globstar
declare -A filelist=()
# Build a list...
for file in **; do filelist[$file]=1; done
# Remove PATTERNS...
while read -r glob; do
for file in $glob; do
unset filelist[$file]
done
done < .rmignore
# And remove whatever's left.
echo rm -v "${!filelist[#]}"
And .. you guessed it. Untested. This depends on $f expanding as a glob.
Lastly, if you want a heavier-weight solution, you can use find and grep:
find . -type f -not -exec grep -q -f '{}' .rmignore \; -delete
This runs a grep for EACH file being considered. And it's not a bash solution, it only relies on find which is pretty universal.
Note that ALL of these solutions are at risk of errors if you have files that contain newlines.
This line do perfectly the job
find . -type f | grep -vFf .rmignore
If you have rsync, you might be able to copy an empty directory to the target one, with suitable rsync ignore files. Try it first with -n, to see what it will attempt, before running it for real!
This is another bash solution that seems to work ok in my tests:
while read -r line;do
exclude+=$(find . -type f -path "./$line")$'\n'
done <.rmignore
echo "ignored files:"
printf '%s\n' "$exclude"
echo "files to be deleted"
echo rm $(LC_ALL=C sort <(find . -type f) <(printf '%s\n' "$exclude") |uniq -u ) #intentionally non quoted to remove new lines
Test it online here
Alternatively, you may want to look at the simplest format:
rm $(ls -1 | grep -v .rmignore)

Renaming a set of files in linux

I have 100 files named sub1.txt, sub2.txt,..., sub100.txt. I like to rename them as all1.txt, all2.txt,..., all100.txt. How can one do this in linux (unix)?
Thanks for your help.
The script below will rename an arbitrary number of file with the pattern sub*.txt. This also does a dry-run thanks to the echo. Simply remove the echo once you are satisfied with the results.
#!/bin/bash
for file in sub*.txt; do
echo mv "$file" "all${file#sub}"
done
Using the util-linux-ng version of rename:
rename sub all sub*.txt
Using the Perl script version of rename:
rename 's/^sub/all/' sub*.txt
for i in `seq 1 100`; do mv sub$i.txt all$i.txt; done
or
for i in sub*.txt; do j=`echo $i|sed -e s/sub/all/`; mv $i $j; done
for F in sub*.txt ; do mv $F all${F#sub}; done
Ruby(1.9+)
ruby -e 'Dir["sub*.txt"].each {|x| File.rename(x, x.gsub(/^sub/,"all") ) }'
There is always a new way to do it:
$ ls sub*.txt | tr -d "sub" | xargs -I{} mv sub{} all{}
Hope it helps.

Help with Replacing Strings on Solaris

I am on a Solaris 8 box that does not support -i option for sed, so I am using the following from a google search on the topic:
# find . -name cancel_submit.cgi | while read file; do
> sed 's/ned.dindo.com\/confluence\/display\/CESDT\/CETS+DocTools>DOC Team/wwwin-dev.dindo.com\/Eng\/CntlSvcs\/InfoFrwk\/GblEngWWW\/Public\/index.html>EDCS Team/g' ${file} > ${file}.new
> mv ${file}.new ${file}
> done
This works except it messes up file permissions and group:owner.
How can I retain the original information?
You may use 'cat'.
cat ${file}.new > ${file} && rm ${file}.new
cp -p preserves the stuff you want. Personally I would do this (to imitate sed -i.bak):
...
cp -p ${file} ${file}.bak
sed 's/..../g' ${file}.bak > ${file}
...
You could add rm ${file}.bak to the end if desired, in which case you wouldn't technically need the -p in the cp line above. But with the above you can do mv ${file}.bak ${file} to recover if the replacement goes awry.

Linux Write Something on multiple files

I have a file "atest.txt" that have some text..
I want to print this text at files "asdasd.txt asgfaya.txt asdjfusfdgh.txt asyeiuyhavujh.txt"
This files is not exist on my server..
I'm running Debian.. What can i do?
Use the tee(1) command, which duplicates its standard input to standard output and any files specified on the command line. E.g.
printf "Hello\nthis is a test\nthank you\n"
| tee test1.txt test2.txt $OTHER_FILES >/dev/null
Using your example:
cat atest.txt
| tee asdasd.txt asgfaya.txt asdjfusfdgh.txt asyeiuyhavujh.txt >/dev/null
From your bash prompt:
for f in test1.txt test2.txt test3.txt; do echo -e "hello\nworld" >> $f; done
If the text lives in atest.txt then do:
for f in test1.txt test2.txt test3.txt; do cat atest.txt >> $f; done
Isn't it simply:
cp atest.txt asdasd.txt
cp atest.txt asgfaya.txt
cp atest.txt asdjfusfdgh.txt
cp atest.txt asyeiuyhavujh.txt
?
In bash you can write
#!/bin/bash
$TEXT="hello\nthis is a test\nthank you"
for i in `seq 1 $1`; do echo -e $TEXT >text$i.txt; done
EDIT (in response of question change)
If you can't determine programmatically the names of the target files then you can use this script it:
#!/bin/bash
ORIGIN=$1;
shift
for i in `seq $#`; do cp "$ORIGIN" "$1"; shift; done
you can use it this way:
script_name origin_file dest_file1 second_dest_file 'third file' ...
If you are wondering why there are the double quotes into the cp command, it is for cope with filename containing spaces
If anyone would like to write same thing to all files in dir:
printf 'your_text' | tee *

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