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nawk regular expression issue

Posted on 2003-11-05
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Last Modified: 2013-12-26
I have a bit of code in nawk that I would like some help with. It is searching a file for User Agent strings and giving me a count of each login by browser version. The problem is when I search on Netscape/7.1 it bombs out since the string has a / character. I have tried a leading \ with no help. Here is the code;

nawk '
   found=0
     /4.79/ { count6++;found=1 }
    /Netscape/7.1/ {count7++;found=1 }
   { if(found == 0) countu++ }
END{
     print "Netscape 4.79",count6
   Print "Netscape 7.1", count7
   print "Unknown",countu
}' > $1.uaout

What character should I use to make nawk ignore the / character in Netscape/7.1? Searching on 7.1 is not an option due to the number of records with this string as part of a the time field.

Thanks!
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Question by:p1800volvo
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3 Comments
 
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Accepted Solution

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glassd earned 200 total points
ID: 9690197
Strange. I am running Solaris9 and nawk is able to use \/ as a litral /.

However:

Normally a period (.) represents any single character, so /Netscape.7.1/ should match.
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Expert Comment

by:rishisk
ID: 9690198
Your approach is correct. The one statement that will cause a problem is Print (Capitalized 'P') in

Print "Netscape 7.1", count7

Change this to

print "Netscape 7.1", count7

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LVL 1

Author Comment

by:p1800volvo
ID: 9695073
glassd has it! That did it. Thanks again!

Credit should also be given to rishisk for his answer since the "P" was a mistake that prevented it from completing in later steps.
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