Here is the situation.

I have a directory which contains many files with different extensions. I want to delete all files except one with a specific name.

This could be easily done using the GUI by selecting all and pressing ctrl and deselecting the file in question.

That is exactly what I want to, but how can I do it from the command line?

For example: dirA contains the following files:

a.txt b.txt c.php d.html a.db b.db e.html

I want to delete all files keeping only the file named a.txt.

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5 Answers

I've come with this easy simple great command:

rm !(a.txt)

you can use ! as a negation

Test the glob with echo first i.e. 

echo !(a.txt)

If it doesn't work, for bash you may need to enable this with 

shopt -s extglob

If you wanted to keep both a.txt and b.txt, you can use !(a.txt|b.txt) or !([ab].txt).

Edit:

to make rm working recursively just add -r like

rm -r !(a.txt)

and also, it is working with folder. just need to change the name to the dir name, such as for a_dir

rm -r !(a_dir)
10

You can try this command:

find . \! -name 'a.txt' -delete

But you need be careful because find command is recursive.

1

You can do this in terminal:

cd dirA export GLOBIGNORE=a.txt rm * export GLOBIGNORE=
1

Use find and xargs

find folder -type f -not -name 'a.txt' -print0 | xargs -0 rm

To exclude multiple things:

find folder -type f -not -name 'a.txt' -not -name 'b.txt' -print0 | xargs -0 rm

This also works with wildcards:

find folder -type f -not -name '*.png' -not -name 'b.txt' -print0 | xargs -0 rm

To search in the current folder, use . in place of 'folder'.

Base source

2

You can use the command :

find ! -name 'a.txt' -type f -exec rm -f {} +

This will look for files (-type f) in the current directory except for file a.txt (! -name 'a.txt) and then will remove them (-exec rm -f {} +)

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