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Using "awk" with other variables from the same script

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# 1  
Old 03-07-2012
Using "awk" with other variables from the same script
Hello all,

I'm just start using "awk" for my log processing script, actually I just use awk for grep things:

Code:
awk '/projectA/ {print;}' /usr/local/tomcat/log/server_access_log.txt > /home/lunaticdawn/log/projectA/process.log

but now I have to put it into a script so that I can add more project and add to cronjob, the original script is:
log_process.sh

Code:
# !bin/sh
# usage: log_process.sh <project name>
path=/usr/local/tomcat/log/
filename=server_access_log.txt
destpath=/home/lunaticdawn/log/

awk '/$1/ {print;}' $path$filename > $destpath$1/process.log

I searched around in internet know that need to add -v for awk, so i tried:

Code:
# !bin/sh
# usage: log_process.sh <project name>
path=/usr/local/tomcat/log/
filename=server_access_log.txt
destpath=/home/lunaticdawn/log/

awk -v VAR1=$path VAR2=$filename VAR3=$destpath '/$1/ {print;}' $VAR1$VAR2 > $VAR3$1/process.log

but error appeared: awk: VAR2=server_access_log.txt syntax error

Actually I got another errors beside this syntax error but I forgot to mark them down, I think the method I use to passing variables to awk is wrong, would you please help on advice me to use awk in a shell script, thanks a lot.

p.s. Although using grep can do the same thing, but I really want to try awk, thank you so much.
# 2  
Old 03-07-2012
Something like this?
Code:
#!/bin/sh
# usage: log_process.sh <project name>
path=/usr/local/tomcat/log/
filename=server_access_log.txt
destpath=/home/lunaticdawn/log/

awk -v Var1="$1" '$0 ~ Var1' $path$filename > $destpath${1}/process.log


Or maybe faster with grep:

Code:
grep $1 $path$filename > $destpath${1}/process.log

This User Gave Thanks to Franklin52 For This Post:
lunaticdawn
# 3  
Old 03-09-2012
Thanks Franklin,

It really solved my problem, I noticed:

Code:
awk -v Var1="$1" '$0 ~ Var1' $path$filename > $destpath${1}/process.log

I'm not so understand the
Code:
'$0 ~ Var1'

what is $0 for? Is it contain the script name?

Thanks.
# 4  
Old 03-09-2012
Quote:
Originally Posted by lunaticdawn
I'm not so understand the
Code:
'$0 ~ Var1'

what is $0 for?
$0 in awk refers to the line being read currently.

'~' is the match operator. $0 ~ Var1 is a check to see whether $0 (i.e., the current line being read) contains the value in variable Var1.
This User Gave Thanks to balajesuri For This Post:
lunaticdawn
# 5  
Old 03-10-2012
Using weak quoting we could do something like this:
Code:
awk "/$1/" "$path$filename" > "$destpath$1/process.log"

If you do not want to use grep.
Last edited by Scrutinizer; 03-10-2012 at 06:31 AM..
This User Gave Thanks to Scrutinizer For This Post:
lunaticdawn
# 6  
Old 03-20-2012
Thanks for the help guys, I think using "awk" to do the things that "grep" is a little bit stupid but I wanna try to learn more about "awk"

Now I'm gonna upgrade it to do more complicated task, my log is in /usr/local/tomcat/log/log.txt, I want to find different browsers from the log and put them in different files, e.g. find IE from the log and put IE's log in IE.txt by using awk.

I watched some tutorial, and know that I can use BEGIN & END on this task, so I tried:

Code:
awk -f browser.awk log.txt

browser.awk contains:
Code:
/Mozilla/ {print $0} > /home/lunaticdawn/log/mozilla.txt
/IE/ {{print $0} > /home/lunaticdawn/log/IE.txt
/chrome/{print $0} > /home/lunaticdawn/log/chrome.txt
END

i think it should work... but it doesn't, it gives syntax error, and I'm seeking for help again Smilie... thanks so much
# 7  
Old 03-20-2012
Hi, there are two {{ on line 2... Leave out the END statement...
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