CODEWITHSUNDEEP
bash
Jamie Chou
Jamie Chou

Reputation: 71

Bash script: variable output to rm with single quotes

I'm trying to pass parameter to rm in bash script to clean my system automatically. For example, I want to remove everything except the *.doc files. So I wrote the following codes. 

#!/bin/bash
remove_Target="!*.txt"
rm $remove_Target

However, the output always say

rm: cannot remove ‘!*.txt’: No such file or directory

It is obviously that bash script add single quotes for me when passing the variable to rm. How can I remove the single quotes?

Upvotes: 2

Views: 826

Answers (1)

John1024
John1024

Reputation: 113814

Using Bash

Suppose that we have a directory with three files

$ ls
a.py  b.py  c.doc

To delete all except *.doc:

$ shopt -s extglob
$ rm !(*.doc)
$ ls
c.doc

!(*.doc) is an extended shell glob, or extglob, that matches all files except those ending in .doc.

The extglob feature requires a modern bash.

Using find

Alternatively:

find . -maxdepth 1 -type f ! -name '*.doc' -delete

Upvotes: 3

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