help with a bash script

[sco21]

I want some help to make a script
this is my log file

2003-01-29 23:20:02 24138915 4841823
2003-01-29 23:25:02 23023245 4179976
2003-01-29 23:30:01 50114161 3065804
2003-01-29 23:35:01 24771375 3501016
2003-01-29 23:40:01 27893795 4199491
2003-01-29 23:45:01 26523593 3705775
2003-01-29 23:50:01 23440011 3372306
2003-01-29 23:55:01 23125174 4037042
2003-01-30 00:00:02 24366334 3715173
2003-01-30 00:05:01 26824279 3663193
2003-01-30 00:10:01 28317816 4285886
2003-01-30 00:15:01 28170460 6281458
2003-01-30 00:20:01 25076290 3312461
2003-01-30 00:25:01 27465629 3989601
........
........
........
How can I parse this file with a bash script
and give as arguments two dates

exm. script 2003-01-29 2003-01-30

and the output to give me only those dates from the file


THANKS

 

[biker1]

sco21 -

Have you considered using the grep command?

EXAMPLE -

grep 2003-01-29 logfile

You can also redirect the results to a new file...

grep 2003-01-29 logfile > newfile

If you wanted to add the second search results to the same file, use...

grep 2003-01-30 logfile >> newfile

">>" will append where ">" will overwrite.

Hope this helps!


Biker
Systems/Network Administrator
LiveFire Labs - Online UNIX and Linux Training
with Hands-on Lab Exercises

http://www.LiveFireLabs.com

 

[marsd]

function date_grep() {
logfile="/var/log/logfile"
start=$1
end=$2

awk -v a=$start -v b=$end ' {
if ($1 == a || $1 == b) {
print
}
}' $logfile

return $?
}

 

[sco21]

THANKS FOR THE HELP FOR THIS FUNCTION
BUT IS THERE A WAY GETING THE RANGE OF TWO DATES

THANKS AGAIN.

 

[marsd]

Do you mean you would like the line numbers of the lines
constituting start and end patterns?
Please don't YELL, there is no need to.