awk not printing variables as expected - variables

Question

Trying really hard to get awk print out the following variables. But no matter how I tried,
awk -F, -v x=$CLIENT_ID -v y=$BRANCH -v z=$UUID -v b=$HERMES_GROUP_CSV_ID 'BEGIN {
OFS = ","; ORS = "\n"
} {
if (length($3) == 0) {
printf "\nCLIENT $x at $y Linux System Time: $z Pacific Time: $b #####: Column 3, Row "; printf NR; printf " data missing in the Client $x group input csv. Please check\n"
}
}' ${INPUT_FILE}
it always prints out
CLIENT $x at $y Linux System Time: $z Pacific Time: $b #####: Column 3, Row 249 data missing in the Client $x group input csv. Please check
Could any guru enlighten? Thanks.

Answer 1

You are using $x as a variable reference, but $ in awk is to reference fields in the input. Variables are used without decoration, like x. So:
awk -F, -v x=$CLIENT_ID -v y=$BRANCH -v z=$UUID -v b=$HERMES_GROUP_CSV_ID 'BEGIN {
OFS = ","; ORS = "\n"
} {
if (length($3) == 0) {
print "\nCLIENT "x" at "y" Linux System Time: "z" Pacific Time: "b" #####: Column 3, Row "; printf NR; printf " data missing in the Client "x" group input csv. Please check\n"
}
}' ${INPUT_FILE}
It looks like x is quoted here, but it is not: the point is to have x appear NOT in the quoted string, so that it can be expanded as a variable.