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Old 09-07-2007, 03:58 PM   #1
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Best tools to use for this script? Bash? Perl?

Hi All!

I am having trouble determining which are the best tools to get this job done.

I need to find files in a directory structure older than X days and move those files to another server (leave the files younger than X days). The tricky part is that I need to preserve part of the directory structure. For example:

On server1, the file /data/a/anderson/anderson_2007-08-02.gz is found to be over 30 days old. (I need to leave files alone younger than 30 days.)

I need to move it to server2 into /data/backups/a/anderson/. The first subdirectory under backups is the first letter of the file name and the subdirectory under that is the person’s last name (letters up to the first ‘_’ in the filename). So in this example, anderson_2007-08-02.gz, that’s how we get the /a/anderson/ subdirectories. I don’t believe the directory structure is necessary the way they’ve done it but it’s one of those “this is how we do it” things they want to preserve.

My line of thinking so far:

Use the find command to get my list of files and then move them to /tmp/workingdir somehow preserving the “a/anderson/” portion of directory structure so that I have /tmp/workingdir/a/anderson/anderson_2007-08-02.gz.
Then I can rsync this working directory structure to the other server using the –remove-sent-files option to delete the files afterward.
Lastly I’ll check to be sure there are no files left hanging around and use find /tmp/workingdir –type d –empty –exec rmdir {} \; to cleanup the directory structure since the –remove-sent-files only deletes files.

My questions are, is this the smartest way to do this or is there a better way? If this is ok, is Perl or bash better? And lastly, how do I pull out the first letter and last name from the filename? I know how to do that in Perl but not bash. Of course, the other portions of this script I know how to do in bash but not Perl. =) So, I’m kinda stuck and wondering if I’m going about this the right way at all.

THANK YOU for any suggestions!

P.S. If you need to know, this is on Ubuntu server (Dapper).

Last edited by mstarap; 09-07-2007 at 03:59 PM.
Old 09-07-2007, 08:21 PM   #2
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I would have thought either would be o.k. If you have most of it in bash, use that.
`basename <something>` (in back-ticks) works in bash. As for stripping the bit in the middle, if you have the regex already (for perl), surely similar parsing would work with something like sed.
Old 09-10-2007, 02:10 PM   #3
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Thanks. For those who may be interested, here are a couple tools I ended up using:

I did my find (redirected to a temporary file) and got the list of file names that looks something like this: aarode_2007-08-14_16-56-41.041_1258.upload.lsx.gz

Then I used this code to get just the last name (the part up to the first "_"):
for file in `cat /tmp/list.$$`; do
NAME=${file/_*/} # this substitutes nothing for everything after the first _

Then I used this piece to get the first letter
FIRST=`perl -e "print substr($NAME,0,1)"`

Doesn't feel pretty but it works.


bash, linux, perl

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