compare logfile content

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# 1  
Old 08-22-2009
compare logfile content

Hi folks

I have some logfiles like this:


each one contains a timestamp from "date +'%s'"

now i need function which finds the logfile with the greatest number in it and returns (echos) the filename.

how can this be done?

I know this is not it..but a start..

function find_most_recent_entire_backup(){
    # read the logfile of the most recent entire_backup and 
    for f in `find ./ -type f -regex ".*entire.*.log" `; do
        timestamp=`grep [1,2,3,4,5,6,7,8,9,0] $f`
        echo $timestamp

thank you already
# 2  
Old 08-22-2009
Originally Posted by latenite
each one contains a timestamp from "date +'%s'"
Post the file contents within CODE tags.

# 3  
Old 08-22-2009
ok i will ...

did i make my point? did i discribe right what i want to do?

any ideas?
# 4  
Old 08-22-2009
Something like this?

awk -F"./" '{d=substr($2,1,16);gsub("_","",d)}d>g{g=d;f=$0}END{print f}' file

# 5  
Old 08-22-2009
ok time for me to learn awk...
any good tutorials on hand??

embaressing ..but i do not understand this oneliner at all Smilie

just running it at the prompt does not do it:

0:521:root@x301 /lvdata/backupdir [0]# find ./ -type f -regex ".*entire.*.log"
0:522:root@x301 /lvdata/backupdir [0]# for f in `find ./ -type f -regex ".*entire.*.log" `; do timestamp=`grep [1,2,3,4,5,6,7,8,9,0] $f`; echo $timestamp; done
0:523:root@x301 /lvdata/backupdir [0]# awk -F"./" '{d=substr($2,1,16);gsub("_","",d)}d>g{g=d;f=$0}END{print f}' file
awk: Kommandozeile:1: Fatal: Kann Datei 'file' nicht zum Lesen öffnen (Datei oder Verzeichnis nicht gefunden).
0:524:root@x301 /lvdata/backupdir [2]#

thank you
# 6  
Old 08-22-2009
In that case:

find ./ -type f -regex ".*entire.*.log" |
awk -F"./" '{d=substr($2,1,16);gsub("_","",d)}d>g{g=d;f=$0}END{print f}'

# 7  
Old 08-22-2009
man this feels funny... using some code i do not have a cloue
well it does it s thing..thnak you for that
and i am learning awk now...promise!!!

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