Stuck with shell scripting...

[mpua]

Hi, Im trying to write a shell script that performs the following actions:

- Check in a given directory how many occurrences of a file called server* (the * representes a date, obviously different for each file).

- Leave untouched the two more recent server* entries and delete the older.

How can be that done?

Regards.

[jack challen_1 · Accepted solution]

===
NUMFILES=$(ls server* | wc -l)
STARTLINE=$(expr $NUMFILES - 2)
ls server* | sed -n "$STARTLINE,\$p" | xargs echo rm
===

That'll only echo the "rm" commands; remove the word "echo" if you're happy with the results of that test.

[Dennis Handly]

By "recent" do you mean by last modification or do you mean a date encoded in the filename?

If it is by last modification, you could do:
ll -rt server* | tail -n +3
(This should give you the ones to remove.)

[jack challen_1]

Ah yes. "tail", I forgot that... whoops.

[muruganantham raju]

Hi,
Following command remove everything except recent two entries.
# ls -lt server* | rm `awk ' NR > 3 { print $NF }'`

HTH
Muru

[Dennis Handly]

>Muru: Following command remove everything except recent two entries.
> ls -lt server* | rm $(awk 'NR > 3 { print $NF }')

Your edge arithmetic is off by one, you need: NR > 2

[mpua]

Hi again,

I mean modification time.
The tail and awk code you gave to me prints (or removes, in this case) the two more recent items of server*.
Is there any way of printing (in my case, removing) every server* occurrence EXCEPT the two more recent ones?

[mpua]

I made a mistake and introduced the "-r" in the ls command. Its my fault :D. The code works perfect for what i was looking for.

Thanks you so much!