Help to write a shell script in HPUX

[R.O.]

Hi,

I am trying to write a program in shell script that does the following:

- check if three files (with its date of creation in its name) of today, yesterday and three days ago, exists
- If they exists, remove all the other files older than three days ago. If they do not
exists, do nothing.

I don

[R.O.]

I don´t know so much about shell script, so any help will be appreciated .

Regards,

(the messages was cu

[R.O.]

I don´t know so much about shell script, so any help will be appreciated .

Regards,

(the messages was cu

[Jannik]

If you assign points to all you old questions we might just help you!

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I have assigned points to 46 of 134 responses to my questions.

[Thierry Poels_1]

hi,

can you give examples of this filenames.

regards,
Thierry.

[R.O.]

Hi,

Jannik : Sorry for the points not assigned; i use to forget to assign, but not in my lasts threads.

The names are in the form:

- filexxx--.Z

Regards,

[Peter Godron]

Hi,
if the file names are in the format *YYYYMMDD*
then somthing like this will get you started.

Some of the guys here have prewritten scripts that do this sort of thing. Use the keyword 'yesterday'in the forum search.
Regards

[Thierry Poels_1]

hi,

as you still didn't give real examples, how about something general:

#! /usr/bin/ksh
nofiles=`find /yourdir -type f -name "file*.Z" -mtime -4 | wc -l`
if [ $nofiles -ge 4 ]
then
find /yourdir -type f -name "file*.Z" -mtime +4 -exec rm {} \;
fi



notes:
- adapt "name" and "-mtime" to your needs
- test this first without the "-exec rm {} \;"
- find will also work through subdirs, so don't use this if you have subdirs.

regards,
Thierry.

[R.O.]

Hi,

The exactly names of the files are:

backup_01-02-05_00_00.dmp.Z , where "01-02-05_00_00" is "day-month-year_hour_minute"

I have already a script with "mtime" but I think it is a bit "dangerous". The script that have sent Peter, works fine to calculate the date; I only need to convert to this date and to remove he older...

Regards,

R.O.

[R.O.]

Hi,

Peter, how can I convert four digits of year in two? (tried printf("%02d... but it does not work)

In the other hand, how can I select the files older than the three firsts to delete them??

Regards,

[Thierry Poels_1]

hi,

1. something like:
two = `echo $four | cut -c3-4`

2.
find /yourdir ! -newer file3daysago ! -name file3daysago

or specify all 3 files:
find /yourdir ! -newer file1 ! -name file1 ! -newer file2 ! -name file2 ! -newer file3 ! -name file3

regards,
Thierry.

[R.O.]

Hello,

I want to avoid to select files based upon in its modification time, but in the date inserted in its name. It is possible that someone modifies a file, so the modification time is not valid to do this.

Regards,

[Peter Godron · Accepted solution]

R.O.,
to answer your question re 4/2 digits:

firstdate=`perl -e '($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(time-2*86400); $fix_year = $year - 100; $fix_m
on = $mon + 1;printf("%02d%02d%02d\n",$fix_year,$fix_mon,$mday)'`

The difference is rather than adding 1900 you subtract 100 as year is number of years since 1900. The format can then be changed from %04 to %02. Problems MAY occurs with date before 1/1/2000, but this should not be problem for what you want to do.
Regards