12-03-2009
awk search column, print line
Hello.
I've been banging my head against walls trying to search a comma delimited file, using awk. I'm trying to search a "column" for a specific parameter, if it matches, then I'd like to print the whole line.
I've read in multiple texts:
awk -F, '{ if ($4 == "string") print $0 }' filename
OR
awk -F, '{ $4 == "string" }' filename
And nothing seems to work? What am I doing wrong?
Info:
OS - Ubuntu 9.10, using bash
BTW, this is my first post on these forums, please forgive any misused etc. etc.. I've been using these forums heavily to learn how to shell script etc.. This is an amazing community, us n00bs are lucky to have something like this. Thank you all so much.
If my question has been answered in another post (which I believe it has, but it didn't work see
https://www.unix.com/unix-dummies-que...int-lines.html )... please forgive me.
Thanks again
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IGAWK(1) Utility Commands IGAWK(1)
NAME
igawk - gawk with include files
SYNOPSIS
igawk [ all gawk options ] -f program-file [ -- ] file ...
igawk [ all gawk options ] [ -- ] program-text file ...
DESCRIPTION
Igawk is a simple shell script that adds the ability to have ``include files'' to gawk(1).
AWK programs for igawk are the same as for gawk, except that, in addition, you may have lines like
@include getopt.awk
in your program to include the file getopt.awk from either the current directory or one of the other directories in the search path.
OPTIONS
See gawk(1) for a full description of the AWK language and the options that gawk supports.
EXAMPLES
cat << EOF > test.awk
@include getopt.awk
BEGIN {
while (getopt(ARGC, ARGV, "am:q") != -1)
...
}
EOF
igawk -f test.awk
SEE ALSO
gawk(1)
Effective AWK Programming, Edition 1.0, published by the Free Software Foundation, 1995.
AUTHOR
Arnold Robbins (arnold@skeeve.com).
Free Software Foundation Nov 3 1999 IGAWK(1)