asked on

repetative grep from user specified start file...

Hi,

Thanks for reading...

The filenames I need to grep start as zcat 20000101*, this obviously being the 1st of Jan this year, and need to grep to present day, for a user defined string, much like UKLIrad055232. This is the exact format, four letters caps, three small, six numbers.

~SUMMARY~

I need the grep to be automated, from user input of string to search for (which I assume is simply $1), from the 1st of Jan (ie zcat 20000101*) to present day (today being the 14th, would be zcat 20000114*), without entering finish date.

I'm using Sun OS 5.5.1, and use VI as my editor program, and will give 300 points for a working solution!!!

Thanks for your time...appreciate it very much...

Tricky.

Tricky

ASKER

Please mail lafarge4@hotmail.com
(other address is dead)...

Thx!

Tricky.

maxkir

If u have perl:

try command
perl grep.pl UKLIrad055232

------- file grep.pl --------------
#!/usr/local/bin/perl

($sec,$min,$hour,$mday,$mon,$year) = localtime(time);

$mon ++;
$year += 1900;
$last = sprintf("$year%02d%02d", $mon, $mday);

for ($i = 20000101; $i< $last; $i++)
{
system("zcat $i* |grep $ARGV[1]");
}
---------------------------------------

Tricky

ASKER

Hi Maxkir :)

I need a seamless, complete, solution.

I need the grep string to be defined upon user input.

I understand unix, but am not fluent...in fact I know very little...

All I need to use is UNIX itself, nothing else...

Talk soon...THX..

Tricky

ozo

In what way is maxkir's solution not seamless or complete enough for you?

ASKER CERTIFIED SOLUTION

ozo

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ecw

#!/bin/sh
end=`date +%Y%m%d`
ls -1 2000[01][0-9][0-9][0-9]* | sed "/^$end/q" | xargs zcat | grep -e "$1"

Tricky

ASKER

Thanks ozo... this is exactly what was required. A UNIX script, that did exactly what I asked...perfect.

Kind Regards,

Tricky.