I have a text file that has a list of paths to various files. Is there a command I can use that will iterate through each line and delete the file at the stated path?
6 Answers
Use xargs
:
xargs rm < file # or
xargs -a file rm
But that will not work if the file names/paths contain characters that should be escaped.
If your filenames don't have newlines, you can do:
tr '\n' '\0' < file | xargs -0 rm # or
xargs -a file -I{} rm {}
Alternatively, you can create the following script:
#!/bin/bash
if [ -z "$1" ]; then
echo -e "Usage: $(basename $0) FILE\n"
exit 1
fi
if [ ! -e "$1" ]; then
echo -e "$1: File doesn't exist.\n"
exit 1
fi
while read -r line; do
[ -n "$line" ] && rm -- "$line"
done < "$1"
Save it as /usr/local/bin/delete-from
, grant it execution permission:
sudo chmod +x /usr/local/bin/delete-from
Then run it with:
delete-from /path/to/file/with/list/of/files
-
2Really clean answer, but
cat
is not required, you can usestdin
redirection:< file xargs rm
– kosMar 13, 2015 at 21:12 -
Its splitting names of folders up and then looking for them. E.g. MY FOLDER is being interpreted as "MY" and "FOLDER"?– olfekMar 13, 2015 at 21:25
-
-
If the files don't have spaces then you could just "rm -- $(cat file)" in bash or "rm --
cat file
" in (ba)sh or csh.– shooperMar 14, 2015 at 2:02 -
@shooper
cat
is unuseful, as just stated, take advantage fromstdin
! Look at his updated answer– kosMar 14, 2015 at 3:55
Here's one way that can deal with file names with whitespace, backslashes and other strange characters:
while read -r file; do rm -- "$file"; done < list.txt
That will read each line of list.txt
, save it as $file
and run rm
on it. The -r
ensures that backslashes are read literally (so that \t
matches a \
and a t
and not a TAB). The --
ensures that it also deals with file names starting with -
.
You could also do this in Perl:
perl -lne '$k{$_}++; END{unlink for keys(%k)}' list.txt
This one wil read each file name into the %k
hash and then use unlink
to delete each of them.
Through python.
import sys
import os
fil = sys.argv[1]
with open(fil) as f:
for line in f:
os.remove(line.rstrip('\n'))
Save the above script in a file named like script.py
and then execute the script by firing the below command on terminal.
python3 script.py file
file
is an input file where the path of the files you actually want to remove are stored.
Another way to do this:
You can 'prepare' the file by making it a shell script:
$ sed -E "s/^(.*)$/rm '\1'/" input_file
rm 'file1'
rm 'file2'
rm 'file with some spaces.ext'
If your filenames may have a single quote ('
), you can use this slightly expanded version to escape them first:
$ sed -E "s/'/'\\\''; s/^(.*)$/rm '\1'/" input_file
rm 'file1'
rm 'file2'
rm 'file with some spaces.ext'
rm 'a file with "quotes"'
rm 'a file with '\''quotes'\'''
And you can run this by piping it to sh
:
$ sed -E "s/'/'\\\''; s/^(.*)$/rm '\1'/" input_file | sh
As I understand it, you have a text file with the files with the complete paths. There are two possibilities:
Your list has the filenames separated by newlines, i.e. each line has the complete path to a file. in this case: here is a simple way out:
for i in $(cat listOfFiles.txt); do rm -f $i done
If your list has one or more lines of filenames separated by spaces or tabs, then here is the drill:
sed -i 's/\s\+/\n/g' listOfFiles.txt
this will convert all whitespaces to newlines
for i in $(cat listOfFiles.txt); do rm -f $i done
Yes, there are many ways to get it done, but this is a very simple approach.