Shell Programming and Scripting

Hi.

I need a script (either bash or perl) that can delete previous versions of files.

For instance, from our continuous build process I get directories such as

build5_dev_1.21
build5_dev_1.22
build5_dev_1.23
build5_dev_1.24

I need a script that I can run every night (using "at" command) that deletes all but the last 1 or 2, such that when the script runs tonight, if the limit is 2, I end up with only the following two directories.

build5_dev_1.23
build5_dev_1.24

if a new build is created during the day the directory ill be named

build5_dev_1.25

and after the script runs I will have directories

build5_dev_1.24
build5_dev_1.25

left.

Ideally I only need to save the "latest" build (the one with the highest "1.xx" number), but i would like the option to save "n" builds, so that I can always have the last 2 or even 3 around if I want.

I am not well versed in script building, my main job is software development, and maintaining the builds is a side task that I write scripts for when i get tired of doing the repetitive tasks by hand. As a consequence I do the scripting so seldom that I have to re-learn alot each time I start a new script. that's why I'm asking for any pointers anyone would be willing to give.

thanks !
Joe Simon

If your tail program supports the +n notation

Code:

$ ls -1t | tail +2 | xargs rm -rf

Otherwise

Code:

$ ls -1t | sed -ne '3,$p' | xargs rm -rf

Test those first using "echo" instead of "rm -rf". Wouldn't want you to delete your root fs just because of a typo.

Thanks ! Great start.

there are other files in the directory other than the ones I want to delete, so I need to do some further work, but it is a good start, and much easier than I envisioned.

Thanks Lots !

Joe

After further thought that didn't quite work.
The actual directory listing looks like this ( I "bolded" the directories of interest):

$ls -1t
suite6_dev_1.127/
latest_unit_test_log_dev_unit_tests.html@
suite6_dev_1.126/
ld@
lm@
lw@
latest_suite6_dev@
build_suite6_dev_1.126.log
suite6_dev_1.125/
latest_suite5.4@
build_suite5.4_1.53.log
suite5.4_1.53/
build_suite6_dev_1.125.log
build_suite5.4_1.52.log
suite5.4_1.52/
latest_suite5_dev@
build_suite5_dev_1.625.log
suite5_dev_1.625/

After the command
$ ls -1t | grep suite6 | sort | tail -2 | xargs echo
suite6_dev_1.126/ suite6_dev_1.127/


What i want is suite6_dev_1.125/ (and any else before, for instance suite6_dev_1.124 if it exists also), I can then use xargs rm -rf instead of xargs echo.

Any ideas of how to get the files that I want ?
The output of
$ls -1t | grep suite6 | sort

build_suite6_dev_1.125.log
build_suite6_dev_1.126.log
build_suite6_dev_1.127.log
latest_suite6_dev@
suite6_dev_1.125/
suite6_dev_1.126/
suite6_dev_1.127/

I need to modify the grep to match the file names starting with "suite6" that will get just the directories that i want. Then I need to get the names of all but the last 2.

Thanks again
Joe

OK,

$ls -1t | grep ^suite6 | sort
suite6_dev_1.125/
suite6_dev_1.126/
suite6_dev_1.127/

This gets me the list of files I am interested in.
I want to now remove all but the last 2 directories in this list.

So, I can get that list into a shell variable, and iterate over that list, removing all but the last two. Is that the best way to finish this off ?

Thanks
Joe

Why exactly did you put that grep and sort in there. What you want to do can be easily accomplished by ls alone.

Code:

$ ls -1td suite6_dev_* | sed -ne '3,$p' | xargs echo

Thanks !
that is perfect.

(sorry about the grep and sort, It didn't click off hand of how to get ls to do this.).

the line you gave gives me exactly what I want. I just have to get add it to another script that already runs every night at midnight and that will take care of that manual process.

Thanks Again !
Joe