CODEWITHSUNDEEP
bash
DenCowboy
DenCowboy

Reputation: 15066

Remove last 2 lines of 'ls' output

I want to delete some revisions of docker images. I want to delete the last 2 lines. I'm able to print the last 2 lines with:

ls -lt | tail -n 2

gives my the 2 last lines.

drwxr-xr-x 2 root root 4096 Nov  9 10:56 541a303d3c82785293f89a401038ac33ef2b54b6aeb09efd3d3bda7xxxx
drwxr-xr-x 2 root root 4096 Oct 25 12:07 c74e1399c99de0c23517abc95bc9b16d09df5c4d518776e77d9ae67xxxx

Now is my question. How do I have to delete them?

I tried ls -lt | tail -n 2 | rm -r * but than I deleted everything (the whole output of ls)

Upvotes: 1

Views: 2910

Answers (3)

Questionmark
Questionmark

Reputation: 884

You could get that to work. I would probably use something like rm -rf $(ls -t | tail -n2), but parsing ls is really not recommended.

A cleaner way to do this would be to use find. You can use that to delete everything before a certain time. Something like this: find . -mtime -180 -exec rm -f {} \; would delete everything newer than 180 days ago.

I would highly recommend testing whatever you are planning to run before you actually do the delete!

Upvotes: 2

V. Michel
V. Michel

Reputation: 1619

Just for test :

 ls -t | tail -n 2 | xargs -i -t echo {}

-t, --verbose Print the command line on the standard error output before executing it.

After the test, you can delete them with : 

ls -t | tail -n 2 | xargs -i -t rm -fr {}

rm -fr 541a303d3c82785293f89a401038ac33ef2b54b6aeb09efd3d3bda7xxxx 
rm -fr c74e1399c99de0c23517abc95bc9b16d09df5c4d518776e77d9ae67xxxx

Upvotes: 0

ibstevieb123
ibstevieb123

Reputation: 96

You have the right idea; however, whatever comes after 'rm' gets deleted, which is why * deleted all instead of what you were trying to pipe into it.

This is a pretty clean way to do it though

    rm `ls | tail -n 2`

Upvotes: 1

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