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Question

remove all but last 5 directories with shell script and find command

Answer 1

+2 A:

To get everything but firt 5 use tail -n +5.(I think you don't need to reverse the sort then) and to rm them you can use xargs. So you can try this

find /var/www/html/ -type d -name "build*" | sort | tail -n +5 | xargs -I % echo -rf %

(replace echo with rm to do it)

mb14 2010-07-01 11:28:49

thanks, tail +5 gives me: cannot open `+5' for reading: No such file or directory

Jorre 2010-07-01 11:35:07

just found out why, it needs tail -n +5

Jorre 2010-07-01 11:38:34

the full command gives me "xargs: %": no such file or directory

Jorre 2010-07-01 11:40:39

tail +n has been deprecated apparently. Still works on my mac and on old unix

mb14 2010-07-01 11:41:07

my mistake, thanks for the quick and perfect response guys!

Jorre 2010-07-01 11:49:56

sorry mate, I would love to mark your answer as the right one, but Philipp has a point and a solution that is slightly more efficient in the long run

Jorre 2010-07-01 14:37:15

My answer was to the question "how to take the last 5", assuming than you have already sorted everything, as you seemed to know how to do it ;)

mb14 2010-07-01 14:57:55

Answer 2

+1 A:

You can't use sort that way because it doesn't recognize that build10 > build9, for example. You have to cut off the prefix manually:

for dir in build*
do
    echo "${dir#build}"
done | sort -n | head -n -5 | xargs -I '%' echo -r 'build%'

Here is an example that shows where your original code fails:

/tmp $ mkdir build8 build9 build10
/tmp $ find . -type d -name "build*" | sort -r
./build9
./build8
./build10

Philipp 2010-07-01 11:48:17

thanks for the tip, is there a way to do this in the command given above?

Jorre 2010-07-01 11:52:28

My (corrected) code is very close to mb14's answer, the only difference being the `for` loop and the `-n` option for `sort`. I didn't use `find` because your `build` directories seem to be direct subdirectories of one common parent.

Philipp 2010-07-01 11:57:04

the code gives me an incorrent order of build directories to be deleted. it gives (with normal sort) build10001, build50, build60, ... this should be build10001, build60, build50, ...

Jorre 2010-07-01 12:15:28

The sorting works correct here; please show an example for the error. However, you have to replace `tail -n +5` by `head -n -5` to delete the oldest instead of the newest directories.

Philipp 2010-07-01 12:41:05

for dir in build*; do echo "${dir#build}"; done | sort | xargs -I '%' echo -r 'build%' GIVES -r build10001-r build1558-r build1560-r build1563-r build1564-r build1565

Jorre 2010-07-01 13:19:31

the script should list all directories, starting with the latest build nummer, but with an offset of 5 (5 most recent should not be displayed/deleted). All other directories build* must be removed

Jorre 2010-07-01 13:28:10

Yes, my code is different, look more carefully at the code and my first comment.

Philipp 2010-07-01 13:28:39

aah I see, sort -n should be sort -r, Let me try with that a bit further. Can you update your script above?

Jorre 2010-07-01 14:07:25

almost got it, you need to put sort -n -r

Jorre 2010-07-01 14:10:14

I think the `-r` is wrong if you use `head`. `head -n -5` gives every line except the last five. Isn't that what you want?

Philipp 2010-07-02 08:02:27

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remove all but last 5 directories with shell script and find command

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