delete files with partial match to another list of names

Question

I have a list of names in file1 that is a substring of the true filenames of the files I want to delete. I want to delete the files that are partially matched by the names in file1. Any idea how to specify the files to delete?

file1:

  file123
  file313
  file355

True files:

  file123.sam.out
  file313.sam.out
  file355.sam.out
  file342455.sam.out
  file34455.sam.out

Files to keep:

  file342455.sam.out
  file34455.sam.out

Answer 1

Assuming you don't have any filenames containing newline literals...

printf '%s\n' * | grep -Fvf file1 | xargs -d $'\n' rm -f --

Let's walk through this piece-by-piece:

• printf '%s\n' * generates a list of files in your current directory.
• grep -Fvf file1 removes from that list any string that contains as a substring a line from file1
• xargs -d $'\n' rm -f -- splits its stdin on newlines, and passes anything it's left as an argument to rm -f --

---

If you have GNU tools (and a shell, like bash, with process substitution support), you can use NUL delimiters and thus work with all possible filenames:

printf '%s\0' * |
  grep -zFvf <(tr '\n' '\0' <file1) |
  xargs -0 rm -f --

• printf '%s\0' * puts a NUL, instead of a newline, after each filename.
• tr '\n' '\0' <file1 emits the contents of file1 with newlines replaced with NULs
• grep -zFvf reads both its inputs as NUL-delimited, and writes NUL-delimited output, but otherwise behaves as above.
• xargs -0 rm -f -- reads content, splitting on NULs, and passes input as arguments to rm -f --.

Answer 2

#!/bin/bash

PATTERN_FILE=file1
FILE_TO_REMOVE_FOLDER=files

cat $PATTERN_FILE | while read x
do
    if [ "" != "$x" ]
    then
        echo "rm $FILE_TO_REMOVE_FOLDER/$x*"
        rm $FILE_TO_REMOVE_FOLDER/$x*
    fi
done