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awk to grep ".do"

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# 1  
Old 02-12-2014
awk to grep ".do"
awk command to search dot
i have a file with the following contents
Code:
22.152.25.36 - K##### [03/Jul/2013:18:05:00 -0900] "GET /GMAIL/CheckMail.do HTTP/1.1" 200 44948
22.152.25.36 - K##### [03/Jul/2013:18:05:00 -0900] "GET /GMAIL/getdoMail.gif HTTP/1.1" 200 44948

Now i use the following to return only the first line which matches ".do"
Code:
awk -F\" '($2 ~ ".do\"){print $2}' filename.log | awk '{print $2}' | sort | uniq -c | sort -r

But it returns me both the line. i tried giving \.do also but didint work.
Any help is appreciated.
Last edited by Scrutinizer; 02-12-2014 at 08:16 PM.. Reason: code tags
# 2  
Old 02-12-2014
A regex can be used as a pattern by enclosing it in slashes:
Code:
awk '/\.do/' file

OR
Code:
awk -F\" '$2 ~ /\.do/' file

# 3  
Old 02-12-2014
I sometimes prefer [.] over \. Easier to read when it's surrounded by other slashes.
# 4  
Old 02-12-2014
Actually
Code:
awk -F\" '($2 ~ "\.do"){print $2}' filename.log

does work. If your log file's structure is always the same, try also
Code:
awk  '($7 ~ "\.do"){print $7}' filename.log

in lieu of your two awk commands.
# 5  
Old 02-14-2014
Code:
awk '$7 ~ /.do$/ { print }' file

Last edited by Franklin52; 02-14-2014 at 04:57 AM.. Reason: Please use code tags
# 6  
Old 02-14-2014
If I understand correctly what the OP was trying to do, the simple fix would be to change:
Code:
awk -F\" '($2 ~ ".do\"){print $2}' filename.log | awk '{print $2}' | sort | uniq -c | sort -r

to:
Code:
awk -F\" '($2 ~ /\.do/){print $2}' filename.log | awk '{print $2}' | sort | uniq -c | sort -r

A slightly more complex awk script would eliminate the need for the 2nd invocation of awk, the 1st invocation of sort, and the invocation of uniq:
Code:
awk -F\" '
$2 ~ /[.]do/ {
        split($2, f2f, /[ \t]+/)
        c[f2f[2]]++
}
END {   for(i in c) printf("%4d %s\n", c[i], i)
}' filename.log | sort -r

and still produce the same output.
# 7  
Old 03-19-2014
Quote:
Postaed by: suindar1982
awk to grep ".do"
awk command to search dot
i have a file with the following contents

Code:
22.152.25.36 - K##### [03/Jul/2013:18:05:00 -0900] "GET /GMAIL/CheckMail.do HTTP/1.1" 200 4494822.152.25.36 - K##### [03/Jul/2013:18:05:00 -0900] "GET /GMAIL/getdoMail.gif HTTP/1.1" 200 44948

Now i use the following to return only the first line which matches ".do"


Code:
awk -F\" '($2 ~ ".do\"){print $2}' filename.log | awk '{print $2}' | sort | uniq -c | sort -r

But it returns me both the line. i tried giving \.do also but didint work.
Any help is appreciated.

Following solution may help if we need to get only complete line which matches .do in it.

Code:
awk '{match($7,/\..*/); a=substr($7, RSTART, RLENGTH);} (a==".do"){print}'  file_name

Output will be as follows.


Code:
22.152.25.36 - K##### [03/Jul/2013:18:05:00 -0900] "GET /GMAIL/CheckMail.do HTTP/1.1" 200 44948

Thanks,
R. Singh
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