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Script to find duplicate pattern in a file irrespective of case

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# 8  
Old 10-17-2012
Is it GNU sed? I tried /.../Ip on cygwin and it worked fine.
# 9  
Old 10-17-2012
Quote:
Originally Posted by johnjs
Thanks Don. it works. I would like to ignore spaces or # as the start of a configuration. It should take as the start of the record, only if the record starts with a character/number and line ends with :\

Code:
ABCD:\

I'm sorry it has taken me so long to get back to you. This should do what you want and be portable to almost any version of awk (on Solaris, however, use nawk or /usr/xpg4/bin/awk rather than just awk):
Code:
awk -F":" '
$0 ~ /^[ #]/ {next}
$1 ~ /^[[:alnum:]]*$/ && $NF ~ /^\\$/ {
        if(list[toupper($1)]++)
                printf("%s on line %d has been seen %d times\n",
                        $1, NR, list[toupper($1)])
}' in

Last edited by Don Cragun; 10-17-2012 at 11:38 PM.. Reason: Added missing ^ in /^\\$/
# 10  
Old 10-18-2012
Maybe too old a GNU sed ? From "sed --version" I get 4.2.1; what do you get?
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