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Insert Date timestamp from .gz files.

Hello Friends,

I have an inquiry where i have a directory of .gz files.

rw-r--r-- 1 Wootie Wootie 175 Sep 2 00:30 bk.2012.09.01.gz
-rw-r--r-- 1 Wootie Wootie 20311 Sep 3 00:30 bk.2012.09.02.gz
-rw-r--r-- 1 Wootie Wootie 534140 Sep 4 01:00 bk.2012.09.03.gz

The files do not have timestamps in them and i'm trying to figure out a way to insert the associated timestamp to the correct file into one file.

So the output would be something as such from reading bk.2012.09.01.gz

Sep 2
83.60300,0/4750000,2459405781485463,83.603@0
83.60300,0/4750000,2459405781485463,83.603@0,

Sep 3

83.60300,0/4750000,2459405781485463,83.603@1
83.60300,0/4750000,2459405781485463,83.603@1

I have tried this but to no avail.

#!/bin/bash

cd ~tech/reports/production

TIMESTAMP=$(ls -l | grep -i bk.2012.09 |awk '{print $6,$7}')
FILE="/home/resource/public_html/reports/production/bk.2012.09*.gz"

while IFS= read -r line

do gunzip -c $line |grep -i alphamale > /tmp/test.xls

done <"$FILE"

Please assist,

Regards,

Liketheshell

Something like this?

Code:

SRCPATH="/home/resource/public_html/reports/production"

find $SRCPATH -type f -iname bk.\*.gz -printf "%f\n"| while read ITEM; do

 F=(${ITEM//./ }); M=$(/bin/date +"%b" --date="${F[2]}"); echo -en "${M} ${F[3]}\n\n"

 gunzip "${SRCPATH}/${ITEM}"; echo; done > "${SRCPATH}/output.txt"

Quote:

Originally Posted by unSpawn (Post 4789274)

Something like this?

Code:

SRCPATH="/home/resource/public_html/reports/production"

find $SRCPATH -type f -iname bk.\*.gz -printf "%f\n"| while read ITEM; do

 F=(${ITEM//./ }); M=$(/bin/date +"%b" --date="${F[2]}"); echo -en "${M} ${F[3]}\n\n"

 gunzip "${SRCPATH}/${ITEM}"; echo; done > "${SRCPATH}/output.txt"

====

Unspawn,

i will try your suggestion and many thanks for the feedback. The only inquiry i do have is the timestamp needs to come from the time the file(s) where created and not the current system time. I will give it a go though.

Many Thanks,

Alfred (Liketheshell)... :)