How to remove all not single files Linux directory?

Question

I have duplicated files in directory on Linux machine which are listed like that:

ltulikowski@lukasz-pc:~$ ls -1
abcd
abcd.1
abcd.2
abdc
abdc.1
acbd

I want to remove all files witch aren't single so as a result I should have:

ltulikowski@lukasz-pc:~$ ls -1
acbd

Answer 1

here is one way to do it

for f in *.[0-9]; do rm ${f%.*}*; done

may get exceptions since some files will be deleted more than once (abcd in your example). If versions always start with .1 you can restrict to match to that.

Answer 2

The function uses extglob , so before execution, set extglob : shopt -s extglob

rm_if_dup_exist(){ 
    arr=()
    for file in *.+([0-9]);do
        base=${file%.*};
        if [[ -e $base ]]; then
            arr+=("$base" "$file")
        fi
    done
    rm -f -- "${arr[@]}"
}

This will also support file names with several digits after the . eg abcd. 250 is also acceptable.

---

Usage example with your input:

$ touch abcd abcd.1 abcd.2 abdc abdc.1 acbd
$ rm_if_dup_exist
$ ls
  acbd

---

Please notice that if, for example, abcd.1 exist but abcd does not exist, it won't delete abcd.1 .

Answer 3

You can use:

while read -r f; do
   rm "$f"*
done < <(printf "%s\n" * | cut -d. -f1 | uniq -d)

printf , cut and uniq are used to get duplicate entries (part before dot) in current directory.

Answer 4

The command

 rm *.*

Should do the trick if I understand you correctly

Use ls to confirm first