How can I delete numbered files in a given range?

Question

I have folderA that has some files with a number sequence starting with a_000000. What I want to do is to delete files starting from a specific number: let's say a_000750 till the end of files in this folderA. Could anyone please advise how to do this using shell script?

Answer 1

Assuming you know or can guess the end of the range, you could use brace expansions:

rm a_{000750..000850}

The above will delete the 101 files between a_000750 and a_000850 inclusive (and complain about filenames that refer to non-existing files). If you have too many files for that, use find:

find . -name 'a_*' | while read -r file; do 
  [ "${file#./a_}" -gt 000749 ] && rm -v "$file" 
done

Here, the find simply lists all files matching a_*. The list is passed to a while loop where each file name is read into the variable $file. Then, using bash's string manipulation features, if the numerical part (find prints files as ./file, so ${file#./a_} prints the number only) is 000750 or greater, the file is deleted. The -v is just there so you can see what files were removed.

Note that the above assumes sane file names. If your names can have spaces, newlines or other weird characters, use this instead:

find . -name 'a_*' -print0 | while IFS= read -rd '' file; do 
  [ "${file#./a_}" -gt 000749 ] && rm -v "$file" 
done

Answer 2

You could do something like this:

find . -regextype posix-extended -iregex './a_[0-9]{6}' -execdir bash -c '[[ ${1##./a_} > 000750 ]] && echo $1' "removing: " {} \;

Or:

find . -regextype posix-extended -iregex './a_[0-9]{6}' | sort | sed '0,/000750/d' | xargs echo

The first method assumes a fixed prefix, strips it off and checks the value.

The second method assumes a fixed-length suffix (and a common fixed prefix) and relies on that fact; and that, while 201 comes before 31 in lexicographically, it doesn't before 031.

Test this out with the echo command, and once you're sure it lists the correct files, use rm instead.

Answer 3

In Bash you can run:

for i in {1001..3000} ; do rm file_"$i".dat ; done

Answer 4

POSIX shell solution

terdon's first solution relies on brace expansion, which is a property of bash and ksh, however it cannot be used in the standard /bin/sh shell, which on Ubuntu is symlinked to /bin/dash.

In cases where you have to rely /bin/sh for portability of your scripts, there's generally two ways to approach this. One would be via globbing. Just cd folderA and from there run rm a_*. The other way, would be to implement a C-style for loop alternative using while <CONDITION>;do ...done in shell language and format the numbers with printf:

$ sh -c 'i=0;while [ $i -le 750 ]; do filename=$(printf "a_%06d" $i);echo "$filename";i=$((i+1)) ;done'

Notice that here I use echo. Replace echo "$filename" with rm ./"$filename" or rm -- "$filename" when you're ready to delete the files. Also note, that this should be performed when you already cded into the desired directory.

(ab)using awk

Awk being a nice C-like language can help us in two ways: (1) we can generate filenames with for-loop and format them via sprintf function, and (2) delete said files via system() command, which will pass our generated filename and rm command to /bin/sh :

awk 'BEGIN{for(i=0;i<=750;i++){filename=sprintf("a_%06d",i);system("rm "filename);} }'

Perl

Continuing with the idea of portable approach where we "generate" filenames, we can do same in Perl:

perl -le 'for(0..750){$fd=sprintf("a_%06d",$_);unlink($fd)}'

Answer 5

As simple as

rm partialfilename* -f

In your example it is

rm a_00075* -f