I want to be able to call a shell script contining the following command.
awk -F $'\xE7' '{OFS ="¬"} { $3 = sprintf("%010s". $3) ; print $0 }' > outputfile
But I want to be able to pass a shell script variable into this command which will be used to replace the $3 within the AWK command.
How do I pass a shell script variable into this AWK command?
Single quotes inhibit expansion. Switch them up.
awk ... '...'"$foo"'...' ...
foo, surrounded by double quotes, as a piece of awk script. If $foo contains double quotes or backslashes, all hell may break loose. See Dennis's answer for a solution that works.
Oct 27, 2010 at 20:23
foo becomes a fragment of awk script — but then I don't see how it answers the question. @mattm123: I don't understand exactly what you're trying to do, but if you were thinking of using the contents of a shell variable as a string in awk, see Dennis's answer.
Oct 27, 2010 at 20:51
$foo can contain unsanitized values, it may seriously mess up your script.
Apr 19, 2012 at 10:28
awk '$1=='"$foo"'{print $3}' when $foo evaluates to z{next} {print "garbage"} 0. And even in the ordinary case where $foo contains an ordinary string, you'll have to make sure your script places "s around it. Dennis' answer shows the right way to do this.
Apr 19, 2012 at 10:37
The correct way to do this is using AWK's variable passing feature. Also, you probably only need to set OFS once, so you should do it in the BEGIN clause.
awk -F $'\xE7' -v awkvar="$shellvar" 'BEGIN {OFS ="¬"} { $3 = sprintf("%010s". $3) ; print $0, awkvar }' input file > outputfile
You could also use -v to set OFS:
awk -F $'\xE7' -v awkvar="$shellvar" -v OFS ="¬" '{ $3 = sprintf("%010s". $3) ; print $0, awkvar }' input file > outputfile
I just added the variable to the print statement since I didn't know exactly how you wanted to use it.
awkvar="$shellvar" is the fifth word in the simple command. Without quotes, awkvar=$shellvar would undergo field splitting and pathname expansion as usual.
Oct 27, 2010 at 23:42
awk has ENVIRON array for accessing environment:
$ FOO=bar awk 'BEGIN{print ENVIRON["FOO"]}'
bar
-v technique is the preferred general solution.
Apr 19, 2012 at 10:39