repetative grep from user specified start file...


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.


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...

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TrickyAuthor Commented:
Please mail
(other address is dead)...


If u have perl:

try command
perl UKLIrad055232

------- file --------------

($sec,$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]");
TrickyAuthor Commented:
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 fact I know very little...

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

Talk soon...THX..

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In what way is maxkir's solution not seamless or complete enough for you?
If you don't have perl:

end=`date +%Y%m%d`
end=`expr $end + 1`
echo $end
for i in `ls 2000[01][0-9][0-9][0-9]*`
if expr $end \<= $i~ >/dev/null ; then break; fi
zcat $i | grep $1

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end=`date +%Y%m%d`
ls -1 2000[01][0-9][0-9][0-9]* | sed "/^$end/q" | xargs zcat | grep -e "$1"
TrickyAuthor Commented:
Thanks ozo... this is exactly what was required. A UNIX script, that did exactly what I asked...perfect.

Kind Regards,

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